Samacheer Kalvi Class 11 Maths Solutions Chapter 3 Trigonometry Exercise 3.6

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.6

Question 1. Express each of the following as a sum or difference.
(i) sin 35°. cos 28°
(ii) sin 4x cos 2x
(iii) 2 sin 10\( \theta \). cos 2\( \theta \)
(iv) cos 5\( \theta \). cos 2\( \theta \)
(v) sin 5\( \theta \). sin 4\( \theta \)
Answer:
(i) We use the product-to-sum formula for `sin A cos B`. This formula helps change a multiplication of sine and cosine into an addition of sines.
We know that \( \sin A \cos B = \frac { 1 }{ 2 } [\sin (A + B) + \sin (A - B)] \)
Let \( A = 35° \) and \( B = 28° \)
\( \sin 35° \cos 28° = \frac { 1 }{ 2 }[\sin(35° + 28°) + \sin(35° - 28°)] \)
\( \sin 35° \cos 28° = \frac { 1 }{ 2 }[\sin 63° + \sin 7°] \)

(ii) We use the same product-to-sum formula. Here, the angles are given in terms of 'x'.
We know that \( \sin A \cos B = \frac { 1 }{ 2 } [\sin (A + B) + \sin (A - B)] \)
Let \( A = 4x \), \( B = 2x \)
\( \sin 4x \cos 2x = \frac { 1 }{ 2 }[\sin(4x + 2x) + \sin(4x - 2x)] \)
\( \sin 4x \cos 2x = \frac { 1 }{ 2 }[\sin 6x + \sin 2x] \)

(iii) Here we need to convert \( 2 \sin A \cos B \) into a sum. Notice the '2' is already present, simplifying the formula.
We know that \( 2 \sin A \cos B = \sin (A + B) + \sin (A - B) \)
Let \( A = 10\theta \), \( B = 2\theta \)
\( 2 \sin 10\theta \cos 2\theta = \sin (10\theta + 2\theta) + \sin (10\theta - 2\theta) \)
\( 2 \sin 10\theta \cos 2\theta = \sin 12\theta + \sin 8\theta \)

(iv) For this part, we need to convert \( \cos A \cos B \) into a sum or difference. The formula is slightly different for two cosines.
We know that \( \cos A \cos B = \frac { 1 }{ 2 } [\cos (A + B) + \cos (A - B)] \)
Let \( A = 5\theta \), \( B = 2\theta \)
\( \cos 5\theta \cos 2\theta = \frac { 1 }{ 2 } [\cos (5\theta + 2\theta) + \cos (5\theta - 2\theta)] \)
\( \cos 5\theta \cos 2\theta = \frac { 1 }{ 2 } [\cos 7\theta + \cos 3\theta] \)

(v) For this last part, we convert \( \sin A \sin B \) into a sum or difference. This formula also involves cosines but with a subtraction.
We know that \( \sin A \sin B = \frac { 1 }{ 2 } [\cos (A - B) - \cos (A + B)] \)
Let \( A = 5\theta \), \( B = 4\theta \)
\( \sin 5\theta \sin 4\theta = \frac { 1 }{ 2 } [\cos (5\theta - 4\theta) - \cos (5\theta + 4\theta)] \)
\( \sin 5\theta \sin 4\theta = \frac { 1 }{ 2 } [\cos \theta - \cos 9\theta] \)
In simple words: We used different formulas to change multiplication of sine and cosine functions into either sums or differences of sine or cosine functions. Each pair of trigonometric functions (like sin A cos B or cos A cos B) has its own specific conversion formula.

🎯 Exam Tip: Remember the four main product-to-sum formulas: for \( \sin A \cos B \), \( \cos A \sin B \), \( \cos A \cos B \), and \( \sin A \sin B \). Pay close attention to the signs and whether it's a sum or difference of sines or cosines.

Question 2. Express each of the following as a product.
(i) sin 75° - sin 35°
(ii) cos 65° + cos 15°
(iii) sin 50° + sin 40°
(iv) cos 35° - cos 75°
Answer:
(i) We use the sum-to-product formula for \( \sin C - \sin D \). This formula transforms a difference of sines into a product of sine and cosine.
We know that \( \sin C - \sin D = 2 \cos \frac { C + D }{ 2 } \sin \frac { C - D }{ 2 } \)
Let \( C = 75° \), \( D = 35° \)
\( \sin 75° - \sin 35° = 2 \cos \frac { 75° + 35° }{ 2 } \sin \frac { 75° - 35° }{ 2 } \)
\( \sin 75° - \sin 35° = 2 \cos \frac { 110° }{ 2 } \sin \frac { 40° }{ 2 } \)
\( \sin 75° - \sin 35° = 2 \cos 55° \sin 20° \)

(ii) We use the sum-to-product formula for \( \cos C + \cos D \). This formula changes an addition of cosines into a product of two cosines.
We know that \( \cos C + \cos D = 2 \cos \frac { C + D }{ 2 } \cos \frac { C - D }{ 2 } \)
Let \( C = 65° \), \( D = 15° \)
\( \cos 65° + \cos 15° = 2 \cos \frac { 65° + 15° }{ 2 } \cos \frac { 65° - 15° }{ 2 } \)
\( \cos 65° + \cos 15° = 2 \cos \frac { 80° }{ 2 } \cos \frac { 50° }{ 2 } \)
\( \cos 65° + \cos 15° = 2 \cos 40° \cos 25° \)

(iii) We use the sum-to-product formula for \( \sin C + \sin D \). This converts an addition of sines into a product of sine and cosine.
We know that \( \sin C + \sin D = 2 \sin \frac { C + D }{ 2 } \cos \frac { C - D }{ 2 } \)
Let \( C = 50° \), \( D = 40° \)
\( \sin 50° + \sin 40° = 2 \sin \frac { 50° + 40° }{ 2 } \cos \frac { 50° - 40° }{ 2 } \)
\( \sin 50° + \sin 40° = 2 \sin \frac { 90° }{ 2 } \cos \frac { 10° }{ 2 } \)
\( \sin 50° + \sin 40° = 2 \sin 45° \cos 5° \)

(iv) We use the sum-to-product formula for \( \cos C - \cos D \). This one is slightly different, resulting in a product of two sines with a negative sign sometimes or D-C order.
We know that \( \cos C - \cos D = -2 \sin \frac { C + D }{ 2 } \sin \frac { C - D }{ 2 } \) or \( 2 \sin \frac { C + D }{ 2 } \sin \frac { D - C }{ 2 } \)
Let \( C = 35° \), \( D = 75° \)
\( \cos 35° - \cos 75° = 2 \sin \frac { 35° + 75° }{ 2 } \sin \frac { 75° - 35° }{ 2 } \)
\( \cos 35° - \cos 75° = 2 \sin \frac { 110° }{ 2 } \sin \frac { 40° }{ 2 } \)
\( \cos 35° - \cos 75° = 2 \sin 55° \sin 20° \)
In simple words: We changed sums or differences of sine and cosine functions into products using specific sum-to-product formulas. Each type of sum/difference (like sin C - sin D or cos C + cos D) has a unique formula to convert it into a multiplication.

🎯 Exam Tip: There are four key sum-to-product formulas to memorize. Be careful with the signs and the order of subtraction (C-D or D-C) for cosine differences.

Question 3. Show that sin 12°. sin 48° . sin 54° = \( \frac { 1 }{ 8 } \)
Answer: This problem requires simplifying a product of three sine functions using product-to-sum formulas and specific trigonometric values.
We start with the left-hand side (LHS):
\( \text{LHS} = \sin 12° \sin 48° \sin 54° \)
Rearrange the terms: \( = \sin 48° \sin 12° \sin 54° \)
Use the formula \( \sin A \sin B = \frac { 1 }{ 2 } [\cos (A - B) - \cos (A + B)] \) for \( \sin 48° \sin 12° \):
\( = \frac { 1 }{ 2 } [\cos (48° - 12°) - \cos (48° + 12°)] \sin 54° \)
\( = \frac { 1 }{ 2 } [\cos 36° - \cos 60°] \sin 54° \)
We know that \( \cos 60° = \frac { 1 }{ 2 } \). Substitute this value:
\( = \frac { 1 }{ 2 } [\cos 36° - \frac { 1 }{ 2 }] \sin 54° \)
Now, distribute \( \sin 54° \):
\( = \frac { 1 }{ 2 } [\cos 36° \sin 54° - \frac { 1 }{ 2 } \sin 54°] \)
We also know that \( \sin 54° = \cos (90° - 54°) = \cos 36° \). So, \( \cos 36° \sin 54° \) becomes \( \cos 36° \cos 36° = \cos^2 36° \).
Substitute \( \sin 54° = \cos 36° \):
\( = \frac { 1 }{ 2 } [\cos^2 36° - \frac { 1 }{ 2 } \cos 36°] \)
We know the value of \( \cos 36° = \frac { \sqrt{5} + 1 }{ 4 } \). Substitute this:
\( = \frac { 1 }{ 2 } [(\frac { \sqrt{5} + 1 }{ 4 })^2 - \frac { 1 }{ 2 } (\frac { \sqrt{5} + 1 }{ 4 })] \)
\( = \frac { 1 }{ 2 } [\frac { (\sqrt{5})^2 + 2\sqrt{5} + 1^2 }{ 16 } - \frac { \sqrt{5} + 1 }{ 8 }] \)
\( = \frac { 1 }{ 2 } [\frac { 5 + 2\sqrt{5} + 1 }{ 16 } - \frac { \sqrt{5} + 1 }{ 8 }] \)
\( = \frac { 1 }{ 2 } [\frac { 6 + 2\sqrt{5} }{ 16 } - \frac { 2(\sqrt{5} + 1) }{ 16 }] \)
\( = \frac { 1 }{ 2 } [\frac { 6 + 2\sqrt{5} - 2\sqrt{5} - 2 }{ 16 }] \)
\( = \frac { 1 }{ 2 } [\frac { 4 }{ 16 }] \)
\( = \frac { 1 }{ 2 } \times \frac { 1 }{ 4 } \)
\( = \frac { 1 }{ 8 } \)
\( \text{LHS} = \text{RHS} \). Hence, it is proved.
In simple words: We used a formula to combine two sine terms into cosine terms. Then, we used known values for specific angles and algebra to simplify the expression. The final result matched the target value, proving the statement.

🎯 Exam Tip: When proving trigonometric identities, always start from one side (usually the more complex one) and use known formulas and values to transform it into the other side. Remember common values like \( \cos 36° \) and \( \sin 54° \).

Question 4. Show that \( \cos \frac { \pi }{ 15 } \cos \frac { 2\pi }{ 15 } \cos \frac { 3\pi }{ 15 } \cos \frac { 4\pi }{ 15 } \cos \frac { 5\pi }{ 15 } \cos \frac { 6\pi }{ 15 } \cos \frac { 7\pi }{ 15 } = \frac { 1 }{ 128 } \)
Answer: This problem involves proving a product of multiple cosine terms is equal to a specific fraction. We will use the formula \( \cos A \cos 2A \cos 2^2 A \dots \cos 2^{n-1} A = \frac { \sin 2^n A }{ 2^n \sin A } \) and simplify.
Let's write the expression as:
\( \text{LHS} = \cos \frac { \pi }{ 15 } \cos \frac { 2\pi }{ 15 } \cos \frac { 3\pi }{ 15 } \cos \frac { 4\pi }{ 15 } \cos \frac { 5\pi }{ 15 } \cos \frac { 6\pi }{ 15 } \cos \frac { 7\pi }{ 15 } \)
Group terms and notice that \( \cos \frac { 3\pi }{ 15 } = \cos \frac { \pi }{ 5 } \), \( \cos \frac { 5\pi }{ 15 } = \cos \frac { \pi }{ 3 } \), and \( \cos \frac { 6\pi }{ 15 } = \cos \frac { 2\pi }{ 5 } \).
We know \( \cos \frac { \pi }{ 3 } = \frac { 1 }{ 2 } \).
So, rearrange and substitute known values:
\( \text{LHS} = (\cos \frac { \pi }{ 15 } \cos \frac { 2\pi }{ 15 } \cos \frac { 4\pi }{ 15 } \cos \frac { 7\pi }{ 15 }) (\cos \frac { 3\pi }{ 15 } \cos \frac { 6\pi }{ 15 } \cos \frac { 5\pi }{ 15 }) \)
\( = (\cos \frac { \pi }{ 15 } \cos \frac { 2\pi }{ 15 } \cos \frac { 4\pi }{ 15 } \cos \frac { 8\pi }{ 15 } \times \frac { \cos \frac { 7\pi }{ 15 } }{ \cos \frac { 8\pi }{ 15 } }) (\cos \frac { \pi }{ 5 } \cos \frac { 2\pi }{ 5 } \cos \frac { \pi }{ 3 }) \)
Use \( \cos \frac { 8\pi }{ 15 } = \cos (\pi - \frac { 7\pi }{ 15 }) = -\cos \frac { 7\pi }{ 15 } \). Thus \( \frac { \cos \frac { 7\pi }{ 15 } }{ \cos \frac { 8\pi }{ 15 } } = -1 \).
The expression becomes:
\( = -1 \times (\cos \frac { \pi }{ 15 } \cos \frac { 2\pi }{ 15 } \cos \frac { 4\pi }{ 15 } \cos \frac { 8\pi }{ 15 }) (\cos \frac { \pi }{ 5 } \cos \frac { 2\pi }{ 5 } \cos \frac { \pi }{ 3 }) \)
Now apply the formula \( \cos A \cos 2A \cos 2^2 A \dots \cos 2^{n-1} A = \frac { \sin 2^n A }{ 2^n \sin A } \).
For the first group, \( A = \frac { \pi }{ 15 } \), \( n = 4 \):
\( \cos \frac { \pi }{ 15 } \cos \frac { 2\pi }{ 15 } \cos \frac { 4\pi }{ 15 } \cos \frac { 8\pi }{ 15 } = \frac { \sin (2^4 \times \frac { \pi }{ 15 }) }{ 2^4 \sin \frac { \pi }{ 15 } } = \frac { \sin \frac { 16\pi }{ 15 } }{ 16 \sin \frac { \pi }{ 15 } } \)
\( = \frac { \sin (\pi + \frac { \pi }{ 15 }) }{ 16 \sin \frac { \pi }{ 15 } } = \frac { -\sin \frac { \pi }{ 15 } }{ 16 \sin \frac { \pi }{ 15 } } = -\frac { 1 }{ 16 } \)
For the second group, \( A = \frac { \pi }{ 5 } \), \( n = 2 \):
\( \cos \frac { \pi }{ 5 } \cos \frac { 2\pi }{ 5 } = \frac { \sin (2^2 \times \frac { \pi }{ 5 }) }{ 2^2 \sin \frac { \pi }{ 5 } } = \frac { \sin \frac { 4\pi }{ 5 } }{ 4 \sin \frac { \pi }{ 5 } } \)
\( = \frac { \sin (\pi - \frac { \pi }{ 5 }) }{ 4 \sin \frac { \pi }{ 5 } } = \frac { \sin \frac { \pi }{ 5 } }{ 4 \sin \frac { \pi }{ 5 } } = \frac { 1 }{ 4 } \)
Combining everything:
\( \text{LHS} = (-1) \times (-\frac { 1 }{ 16 }) \times (\frac { 1 }{ 4 }) \times (\cos \frac { \pi }{ 3 }) \)
\( = \frac { 1 }{ 16 } \times \frac { 1 }{ 4 } \times \frac { 1 }{ 2 } \)
\( = \frac { 1 }{ 128 } \)
Thus, \( \text{LHS} = \text{RHS} \). The statement is proven.
In simple words: We used a special formula to simplify a long chain of cosine multiplications. By grouping the terms and using known trigonometric values like \( \cos \frac { \pi }{ 3 } = \frac { 1 }{ 2 } \), we reduced the complex expression to a simple fraction.

🎯 Exam Tip: Recognize patterns like \( \cos A \cos 2A \cos 4A \dots \) to apply the general formula. Also, simplify terms like \( \cos \frac { 3\pi }{ 15 } \) to \( \cos \frac { \pi }{ 5 } \) early on to use known values and make the problem easier.

Question 5. Show that \( \frac { \sin 8x \cos x - \sin 6x \cos 3x }{ \cos 2x \cos x - \sin 3x \sin 4x } = \tan 2x \)
Answer: To prove this identity, we need to simplify the numerator and denominator separately using product-to-sum formulas and then simplify the resulting fraction.
Consider the Left-Hand Side (LHS):
\( \text{LHS} = \frac { \sin 8x \cos x - \sin 6x \cos 3x }{ \cos 2x \cos x - \sin 3x \sin 4x } \)
Numerator:
We know \( \sin A \cos B = \frac { 1 }{ 2 } [\sin (A+B) + \sin (A-B)] \).
So, \( \sin 8x \cos x = \frac { 1 }{ 2 } [\sin (8x+x) + \sin (8x-x)] = \frac { 1 }{ 2 } [\sin 9x + \sin 7x] \)
And, \( \sin 6x \cos 3x = \frac { 1 }{ 2 } [\sin (6x+3x) + \sin (6x-3x)] = \frac { 1 }{ 2 } [\sin 9x + \sin 3x] \)
Numerator \( = \frac { 1 }{ 2 } [\sin 9x + \sin 7x] - \frac { 1 }{ 2 } [\sin 9x + \sin 3x] \)
\( = \frac { 1 }{ 2 } [\sin 9x + \sin 7x - \sin 9x - \sin 3x] \)
\( = \frac { 1 }{ 2 } [\sin 7x - \sin 3x] \)
Denominator:
We know \( \cos A \cos B = \frac { 1 }{ 2 } [\cos (A+B) + \cos (A-B)] \).
So, \( \cos 2x \cos x = \frac { 1 }{ 2 } [\cos (2x+x) + \cos (2x-x)] = \frac { 1 }{ 2 } [\cos 3x + \cos x] \)
We know \( \sin A \sin B = \frac { 1 }{ 2 } [\cos (A-B) - \cos (A+B)] \).
So, \( \sin 3x \sin 4x = \frac { 1 }{ 2 } [\cos (3x-4x) - \cos (3x+4x)] = \frac { 1 }{ 2 } [\cos (-x) - \cos 7x] \)
Since \( \cos(-x) = \cos x \):
\( \sin 3x \sin 4x = \frac { 1 }{ 2 } [\cos x - \cos 7x] \)
Denominator \( = \frac { 1 }{ 2 } [\cos 3x + \cos x] - \frac { 1 }{ 2 } [\cos x - \cos 7x] \)
\( = \frac { 1 }{ 2 } [\cos 3x + \cos x - \cos x + \cos 7x] \)
\( = \frac { 1 }{ 2 } [\cos 3x + \cos 7x] \)
Now, substitute back into the LHS:
\( \text{LHS} = \frac { \frac { 1 }{ 2 } [\sin 7x - \sin 3x] }{ \frac { 1 }{ 2 } [\cos 3x + \cos 7x] } = \frac { \sin 7x - \sin 3x }{ \cos 7x + \cos 3x } \)
Apply sum-to-product formulas:
\( \sin C - \sin D = 2 \cos \frac { C+D }{ 2 } \sin \frac { C-D }{ 2 } \)
\( \cos C + \cos D = 2 \cos \frac { C+D }{ 2 } \cos \frac { C-D }{ 2 } \)
Numerator: \( \sin 7x - \sin 3x = 2 \cos \frac { 7x+3x }{ 2 } \sin \frac { 7x-3x }{ 2 } = 2 \cos \frac { 10x }{ 2 } \sin \frac { 4x }{ 2 } = 2 \cos 5x \sin 2x \)
Denominator: \( \cos 7x + \cos 3x = 2 \cos \frac { 7x+3x }{ 2 } \cos \frac { 7x-3x }{ 2 } = 2 \cos \frac { 10x }{ 2 } \cos \frac { 4x }{ 2 } = 2 \cos 5x \cos 2x \)
\( \text{LHS} = \frac { 2 \cos 5x \sin 2x }{ 2 \cos 5x \cos 2x } \)
Cancel out \( 2 \cos 5x \):
\( \text{LHS} = \frac { \sin 2x }{ \cos 2x } = \tan 2x \)
Thus, \( \text{LHS} = \text{RHS} \). The statement is proven.
In simple words: We used special formulas to change all the multiplications in the top and bottom parts of the fraction into additions or subtractions. Then we used other formulas to turn those additions/subtractions back into multiplications. This helped us cancel out common terms and simplify the whole fraction to get \( \tan 2x \).

🎯 Exam Tip: When dealing with complex trigonometric fractions, simplify the numerator and denominator separately first. Look for opportunities to apply product-to-sum and sum-to-product formulas, and remember \( \tan x = \frac { \sin x }{ \cos x } \).

Question 6. Show that \( \frac { (\cos \theta - \cos 3\theta) (\sin 8\theta + \sin 2\theta) }{ (\sin 5\theta - \sin \theta) (\cos 4\theta - \cos 6\theta) } = 1 \)
Answer: This problem requires simplifying a complex trigonometric fraction to prove it equals 1. We'll use sum-to-product formulas for each part of the expression.
Consider the Left-Hand Side (LHS):
\( \text{LHS} = \frac { (\cos \theta - \cos 3\theta) (\sin 8\theta + \sin 2\theta) }{ (\sin 5\theta - \sin \theta) (\cos 4\theta - \cos 6\theta) } \)
Apply sum-to-product formulas to each bracket:
1. \( \cos C - \cos D = 2 \sin \frac { C+D }{ 2 } \sin \frac { D-C }{ 2 } \)
\( \cos \theta - \cos 3\theta = 2 \sin \frac { \theta+3\theta }{ 2 } \sin \frac { 3\theta-\theta }{ 2 } = 2 \sin 2\theta \sin \theta \)
2. \( \sin C + \sin D = 2 \sin \frac { C+D }{ 2 } \cos \frac { C-D }{ 2 } \)
\( \sin 8\theta + \sin 2\theta = 2 \sin \frac { 8\theta+2\theta }{ 2 } \cos \frac { 8\theta-2\theta }{ 2 } = 2 \sin 5\theta \cos 3\theta \)
3. \( \sin C - \sin D = 2 \cos \frac { C+D }{ 2 } \sin \frac { C-D }{ 2 } \)
\( \sin 5\theta - \sin \theta = 2 \cos \frac { 5\theta+\theta }{ 2 } \sin \frac { 5\theta-\theta }{ 2 } = 2 \cos 3\theta \sin 2\theta \)
4. \( \cos C - \cos D = 2 \sin \frac { C+D }{ 2 } \sin \frac { D-C }{ 2 } \)
\( \cos 4\theta - \cos 6\theta = 2 \sin \frac { 4\theta+6\theta }{ 2 } \sin \frac { 6\theta-4\theta }{ 2 } = 2 \sin 5\theta \sin \theta \)
Substitute these simplified expressions back into the LHS:
\( \text{LHS} = \frac { (2 \sin 2\theta \sin \theta) (2 \sin 5\theta \cos 3\theta) }{ (2 \cos 3\theta \sin 2\theta) (2 \sin 5\theta \sin \theta) } \)
Now, cancel out common terms from the numerator and denominator:
\( \text{LHS} = \frac { (2 \times 2) \sin 2\theta \sin \theta \sin 5\theta \cos 3\theta }{ (2 \times 2) \cos 3\theta \sin 2\theta \sin 5\theta \sin \theta } \)
All terms \( \sin 2\theta, \sin \theta, \sin 5\theta, \cos 3\theta \) appear in both the numerator and denominator, so they cancel out, along with the numerical factor \( 4 \).
\( \text{LHS} = 1 \)
Thus, \( \text{LHS} = \text{RHS} \). The statement is proven.
In simple words: We changed each part of the top and bottom of the fraction from additions or subtractions into multiplications using special trigonometric rules. After that, we saw that all the parts on top were the same as the parts on the bottom, so they all canceled each other out, leaving us with just 1.

🎯 Exam Tip: Systematically apply the correct sum-to-product formula to each term. Write out each transformation clearly. Look for opportunities to cancel terms after all transformations are done.

Question 7. Prove that \( \sin x + \sin 2x + \sin 3x = \sin 2x (1 + 2 \cos x) \)
Answer: To prove this identity, we'll start from the Left-Hand Side (LHS) and use sum-to-product formulas and algebraic manipulation to transform it into the Right-Hand Side (RHS).
Consider the Left-Hand Side (LHS):
\( \text{LHS} = \sin x + \sin 2x + \sin 3x \)
Rearrange the terms to group \( \sin x \) and \( \sin 3x \):
\( \text{LHS} = (\sin 3x + \sin x) + \sin 2x \)
Apply the sum-to-product formula \( \sin C + \sin D = 2 \sin \frac { C+D }{ 2 } \cos \frac { C-D }{ 2 } \) to \( \sin 3x + \sin x \):
Let \( C = 3x \), \( D = x \)
\( \sin 3x + \sin x = 2 \sin \frac { 3x+x }{ 2 } \cos \frac { 3x-x }{ 2 } \)
\( = 2 \sin \frac { 4x }{ 2 } \cos \frac { 2x }{ 2 } \)
\( = 2 \sin 2x \cos x \)
Now substitute this back into the LHS expression:
\( \text{LHS} = 2 \sin 2x \cos x + \sin 2x \)
Notice that \( \sin 2x \) is a common factor in both terms. Factor it out:
\( \text{LHS} = \sin 2x (2 \cos x + 1) \)
Rearrange the terms inside the bracket:
\( \text{LHS} = \sin 2x (1 + 2 \cos x) \)
This is equal to the Right-Hand Side (RHS).
Thus, \( \text{LHS} = \text{RHS} \). The statement is proven.
In simple words: We grouped the first and third sine terms and used a formula to change their addition into a multiplication. After that, we noticed that \( \sin 2x \) was common in both parts of our expression, so we pulled it out. This made the expression look exactly like what we needed to prove.

🎯 Exam Tip: When proving identities with three terms, try grouping two terms and applying a sum-to-product formula. Look for common factors that can be extracted to match the target expression. Often, the middle term will be the common factor.

Question 8. Prove that \( \frac { \sin 4x + \sin 2x }{ \cos 4x + \cos 2x } = \tan 3x \)
Answer: To prove this identity, we will simplify the Left-Hand Side (LHS) using sum-to-product formulas for both the numerator and the denominator.
Consider the Left-Hand Side (LHS):
\( \text{LHS} = \frac { \sin 4x + \sin 2x }{ \cos 4x + \cos 2x } \)
Apply the sum-to-product formula to the numerator:
\( \sin C + \sin D = 2 \sin \frac { C+D }{ 2 } \cos \frac { C-D }{ 2 } \)
Let \( C = 4x \), \( D = 2x \)
Numerator \( = 2 \sin \frac { 4x+2x }{ 2 } \cos \frac { 4x-2x }{ 2 } \)
\( = 2 \sin \frac { 6x }{ 2 } \cos \frac { 2x }{ 2 } \)
\( = 2 \sin 3x \cos x \)
Apply the sum-to-product formula to the denominator:
\( \cos C + \cos D = 2 \cos \frac { C+D }{ 2 } \cos \frac { C-D }{ 2 } \)
Let \( C = 4x \), \( D = 2x \)
Denominator \( = 2 \cos \frac { 4x+2x }{ 2 } \cos \frac { 4x-2x }{ 2 } \)
\( = 2 \cos \frac { 6x }{ 2 } \cos \frac { 2x }{ 2 } \)
\( = 2 \cos 3x \cos x \)
Now substitute the simplified numerator and denominator back into the LHS:
\( \text{LHS} = \frac { 2 \sin 3x \cos x }{ 2 \cos 3x \cos x } \)
Cancel out the common terms \( 2 \) and \( \cos x \):
\( \text{LHS} = \frac { \sin 3x }{ \cos 3x } \)
We know that \( \frac { \sin A }{ \cos A } = \tan A \).
So, \( \text{LHS} = \tan 3x \)
This is equal to the Right-Hand Side (RHS).
Thus, \( \text{LHS} = \text{RHS} \). The statement is proven.
In simple words: We used special formulas to change both the top and bottom parts of the fraction from additions into multiplications. After doing that, we saw that some parts were the same on both the top and the bottom, so we canceled them out. What was left was \( \frac { \sin 3x }{ \cos 3x } \), which is the same as \( \tan 3x \).

🎯 Exam Tip: This is a common pattern for identities leading to tan. Always remember the two sum-to-product formulas for \( \sin C + \sin D \) and \( \cos C + \cos D \) as they often appear together in such problems. The arguments for sine and cosine will usually be the same after applying the formulas.

Question 9. Prove that \( 1 + \cos 2x + \cos 4x + \cos 6x = 4 \cos x \cos 2x \cos 3x \)
Answer: To prove this identity, we will start from the Right-Hand Side (RHS) and use product-to-sum formulas and algebraic manipulation to transform it into the Left-Hand Side (LHS). This approach is often easier when the RHS is a product.
Consider the Right-Hand Side (RHS):
\( \text{RHS} = 4 \cos x \cos 2x \cos 3x \)
Rearrange the terms to group \( \cos 3x \) and \( \cos 2x \):
\( \text{RHS} = 4 \cos x (\cos 3x \cos 2x) \)
Apply the product-to-sum formula \( \cos A \cos B = \frac { 1 }{ 2 } [\cos (A+B) + \cos (A-B)] \) to \( \cos 3x \cos 2x \):
Let \( A = 3x \), \( B = 2x \)
\( \cos 3x \cos 2x = \frac { 1 }{ 2 } [\cos (3x+2x) + \cos (3x-2x)] \)
\( = \frac { 1 }{ 2 } [\cos 5x + \cos x] \)
Substitute this back into the RHS expression:
\( \text{RHS} = 4 \cos x \times \frac { 1 }{ 2 } [\cos 5x + \cos x] \)
\( = 2 \cos x [\cos 5x + \cos x] \)
Distribute \( 2 \cos x \):
\( \text{RHS} = 2 \cos x \cos 5x + 2 \cos^2 x \)
Apply the product-to-sum formula again to \( 2 \cos x \cos 5x \). This time, it's \( 2 \cos A \cos B = \cos(A+B) + \cos(A-B) \).
Let \( A = 5x \), \( B = x \)
\( 2 \cos 5x \cos x = \cos (5x+x) + \cos (5x-x) \)
\( = \cos 6x + \cos 4x \)
Substitute this back into the RHS:
\( \text{RHS} = (\cos 6x + \cos 4x) + 2 \cos^2 x \)
We know the double angle identity \( \cos 2A = 2 \cos^2 A - 1 \), which means \( 2 \cos^2 A = 1 + \cos 2A \).
So, \( 2 \cos^2 x = 1 + \cos 2x \).
Substitute this identity into the RHS:
\( \text{RHS} = \cos 6x + \cos 4x + (1 + \cos 2x) \)
Rearrange the terms to match the LHS:
\( \text{RHS} = 1 + \cos 2x + \cos 4x + \cos 6x \)
This is equal to the Left-Hand Side (LHS).
Thus, \( \text{LHS} = \text{RHS} \). The statement is proven.
In simple words: We started with the right side, which had three cosine terms multiplied together. We combined them two at a time using a multiplication-to-addition formula. We kept doing this until we had only additions. Finally, we used another known rule about \( 2 \cos^2 x \) to change it into \( 1 + \cos 2x \), which gave us the left side of the equation.

🎯 Exam Tip: When faced with a product on one side and a sum on the other, it's generally easier to start with the product side and expand using product-to-sum formulas. Remember the identity \( 2 \cos^2 x = 1 + \cos 2x \) as it's crucial for simplifying expressions involving squared cosine terms.

Question 10. Prove that \( \sin \frac { \theta }{ 2 } \sin \frac { 7\theta }{ 2 } + \sin \frac { 3\theta }{ 2 } \sin \frac { 11\theta }{ 2 } = \sin 2\theta \sin 5\theta \)
Answer: To prove this identity, we'll start with the Left-Hand Side (LHS) and use the product-to-sum formula for sine-sine terms. This will convert the products into differences of cosines, which can then be simplified.
Consider the Left-Hand Side (LHS):
\( \text{LHS} = \sin \frac { \theta }{ 2 } \sin \frac { 7\theta }{ 2 } + \sin \frac { 3\theta }{ 2 } \sin \frac { 11\theta }{ 2 } \)
We use the product-to-sum formula: \( \sin A \sin B = \frac { 1 }{ 2 } [\cos (A - B) - \cos (A + B)] \).
For the first term, \( \sin \frac { \theta }{ 2 } \sin \frac { 7\theta }{ 2 } \):
\( = \frac { 1 }{ 2 } [\cos (\frac { \theta }{ 2 } - \frac { 7\theta }{ 2 }) - \cos (\frac { \theta }{ 2 } + \frac { 7\theta }{ 2 })] \)
\( = \frac { 1 }{ 2 } [\cos (-\frac { 6\theta }{ 2 }) - \cos (\frac { 8\theta }{ 2 })] \)
\( = \frac { 1 }{ 2 } [\cos (-3\theta) - \cos 4\theta] \)
Since \( \cos(-x) = \cos x \):
\( = \frac { 1 }{ 2 } [\cos 3\theta - \cos 4\theta] \)
For the second term, \( \sin \frac { 3\theta }{ 2 } \sin \frac { 11\theta }{ 2 } \):
\( = \frac { 1 }{ 2 } [\cos (\frac { 3\theta }{ 2 } - \frac { 11\theta }{ 2 }) - \cos (\frac { 3\theta }{ 2 } + \frac { 11\theta }{ 2 })] \)
\( = \frac { 1 }{ 2 } [\cos (-\frac { 8\theta }{ 2 }) - \cos (\frac { 14\theta }{ 2 })] \)
\( = \frac { 1 }{ 2 } [\cos (-4\theta) - \cos 7\theta] \)
Since \( \cos(-x) = \cos x \):
\( = \frac { 1 }{ 2 } [\cos 4\theta - \cos 7\theta] \)
Now, add these two results back into the LHS:
\( \text{LHS} = \frac { 1 }{ 2 } [\cos 3\theta - \cos 4\theta] + \frac { 1 }{ 2 } [\cos 4\theta - \cos 7\theta] \)
\( = \frac { 1 }{ 2 } [\cos 3\theta - \cos 4\theta + \cos 4\theta - \cos 7\theta] \)
The \( -\cos 4\theta \) and \( +\cos 4\theta \) terms cancel each other out:
\( = \frac { 1 }{ 2 } [\cos 3\theta - \cos 7\theta] \)
Now, apply the sum-to-product formula for \( \cos C - \cos D = -2 \sin \frac { C+D }{ 2 } \sin \frac { C-D }{ 2 } \) or \( 2 \sin \frac { C+D }{ 2 } \sin \frac { D-C }{ 2 } \).
Using the second version (to avoid negative sign until the end):
Let \( C = 3\theta \), \( D = 7\theta \)
\( \cos 3\theta - \cos 7\theta = 2 \sin \frac { 3\theta+7\theta }{ 2 } \sin \frac { 7\theta-3\theta }{ 2 } \)
\( = 2 \sin \frac { 10\theta }{ 2 } \sin \frac { 4\theta }{ 2 } \)
\( = 2 \sin 5\theta \sin 2\theta \)
Substitute this back into the LHS:
\( \text{LHS} = \frac { 1 }{ 2 } [2 \sin 5\theta \sin 2\theta] \)
\( = \sin 5\theta \sin 2\theta \)
Rearranging the terms: \( = \sin 2\theta \sin 5\theta \)
This is equal to the Right-Hand Side (RHS).
Thus, \( \text{LHS} = \text{RHS} \). The statement is proven.
In simple words: We changed each multiplication of two sine terms into a subtraction of two cosine terms using a formula. When we put all these back together, some cosine terms cancelled out. Then, we used another formula to turn the remaining subtraction of cosines back into a multiplication of sines, which matched the right side of the equation.

🎯 Exam Tip: Pay close attention to the angles when applying product-to-sum formulas, especially with fractions and signs. It's often helpful to keep \( \frac { 1 }{ 2 } \) factored out initially to avoid errors in distribution.

Question 11. Prove that \( \cos (30°- A) \cos (30° + A) + \cos (45° – A) \cos(45° + A) = \frac { 1 }{ 4 } + \cos 2A \)
Answer: To prove this identity, we will simplify the Left-Hand Side (LHS) by applying the product-to-sum formula for \( \cos A \cos B \) to both terms.
Consider the Left-Hand Side (LHS):
\( \text{LHS} = \cos (30°- A) \cos (30° + A) + \cos (45° – A) \cos(45° + A) \)
We use the product-to-sum formula: \( \cos A \cos B = \frac { 1 }{ 2 } [\cos (A + B) + \cos (A - B)] \).
For the first term, \( \cos (30°- A) \cos (30° + A) \):
Let \( X = 30° - A \) and \( Y = 30° + A \).
Sum of angles: \( X+Y = (30°- A) + (30° + A) = 60° \)
Difference of angles: \( X-Y = (30°- A) - (30° + A) = -2A \)
So, \( \cos (30°- A) \cos (30° + A) = \frac { 1 }{ 2 } [\cos 60° + \cos (-2A)] \)
Since \( \cos(-x) = \cos x \):
\( = \frac { 1 }{ 2 } [\cos 60° + \cos 2A] \)
We know \( \cos 60° = \frac { 1 }{ 2 } \).
\( = \frac { 1 }{ 2 } [\frac { 1 }{ 2 } + \cos 2A] \)
For the second term, \( \cos (45° – A) \cos(45° + A) \):
Let \( P = 45° - A \) and \( Q = 45° + A \).
Sum of angles: \( P+Q = (45° - A) + (45° + A) = 90° \)
Difference of angles: \( P-Q = (45° - A) - (45° + A) = -2A \)
So, \( \cos (45° – A) \cos(45° + A) = \frac { 1 }{ 2 } [\cos 90° + \cos (-2A)] \)
Since \( \cos(-x) = \cos x \):
\( = \frac { 1 }{ 2 } [\cos 90° + \cos 2A] \)
We know \( \cos 90° = 0 \).
\( = \frac { 1 }{ 2 } [0 + \cos 2A] \)
\( = \frac { 1 }{ 2 } \cos 2A \)
Now, add these two results to find the LHS:
\( \text{LHS} = \frac { 1 }{ 2 } [\frac { 1 }{ 2 } + \cos 2A] + \frac { 1 }{ 2 } \cos 2A \)
\( = \frac { 1 }{ 4 } + \frac { 1 }{ 2 } \cos 2A + \frac { 1 }{ 2 } \cos 2A \)
\( = \frac { 1 }{ 4 } + \cos 2A \)
This is equal to the Right-Hand Side (RHS).
Thus, \( \text{LHS} = \text{RHS} \). The statement is proven.
In simple words: We used a formula to change each multiplication of two cosine terms into an addition of cosines. For each part, we calculated the sum and difference of the angles. We then put in the known values for \( \cos 60° \) and \( \cos 90° \) and added everything together. This simplified the expression to match the right side of the equation.

🎯 Exam Tip: When angles are of the form \( (X-A) \) and \( (X+A) \), their sum is \( 2X \) and their difference is \( -2A \). This pattern simplifies calculations greatly. Remember common trigonometric values like \( \cos 60° \) and \( \cos 90° \).

Question 12. Show that \( \frac { \sin x + \sin 3x + \sin 5x + \sin 7x }{ \cos x + \cos 3x + \cos 5x + \cos 7x } = \tan 4x \)
Answer: To prove this identity, we will simplify the Left-Hand Side (LHS) by grouping terms and applying sum-to-product formulas to both the numerator and the denominator.
Consider the Left-Hand Side (LHS):
\( \text{LHS} = \frac { (\sin x + \sin 3x) + (\sin 5x + \sin 7x) }{ (\cos x + \cos 3x) + (\cos 5x + \cos 7x) } \)
Apply sum-to-product formula \( \sin C + \sin D = 2 \sin \frac { C+D }{ 2 } \cos \frac { C-D }{ 2 } \):
\( \sin x + \sin 3x = 2 \sin \frac { x+3x }{ 2 } \cos \frac { x-3x }{ 2 } = 2 \sin 2x \cos (-x) = 2 \sin 2x \cos x \)
\( \sin 5x + \sin 7x = 2 \sin \frac { 5x+7x }{ 2 } \cos \frac { 5x-7x }{ 2 } = 2 \sin 6x \cos (-x) = 2 \sin 6x \cos x \)
Numerator \( = 2 \sin 2x \cos x + 2 \sin 6x \cos x \)
Factor out \( 2 \cos x \):
Numerator \( = 2 \cos x (\sin 2x + \sin 6x) \)
Apply sum-to-product formula \( \cos C + \cos D = 2 \cos \frac { C+D }{ 2 } \cos \frac { C-D }{ 2 } \):
\( \cos x + \cos 3x = 2 \cos \frac { x+3x }{ 2 } \cos \frac { x-3x }{ 2 } = 2 \cos 2x \cos (-x) = 2 \cos 2x \cos x \)
\( \cos 5x + \cos 7x = 2 \cos \frac { 5x+7x }{ 2 } \cos \frac { 5x-7x }{ 2 } = 2 \cos 6x \cos (-x) = 2 \cos 6x \cos x \)
Denominator \( = 2 \cos 2x \cos x + 2 \cos 6x \cos x \)
Factor out \( 2 \cos x \):
Denominator \( = 2 \cos x (\cos 2x + \cos 6x) \)
Substitute back into the LHS:
\( \text{LHS} = \frac { 2 \cos x (\sin 2x + \sin 6x) }{ 2 \cos x (\cos 2x + \cos 6x) } \)
Cancel out \( 2 \cos x \):
\( \text{LHS} = \frac { \sin 2x + \sin 6x }{ \cos 2x + \cos 6x } \)
Now, apply sum-to-product formulas again to the remaining terms:
Numerator: \( \sin 2x + \sin 6x = 2 \sin \frac { 2x+6x }{ 2 } \cos \frac { 2x-6x }{ 2 } = 2 \sin 4x \cos (-2x) = 2 \sin 4x \cos 2x \)
Denominator: \( \cos 2x + \cos 6x = 2 \cos \frac { 2x+6x }{ 2 } \cos \frac { 2x-6x }{ 2 } = 2 \cos 4x \cos (-2x) = 2 \cos 4x \cos 2x \)
Substitute these back:
\( \text{LHS} = \frac { 2 \sin 4x \cos 2x }{ 2 \cos 4x \ cos 2x } \)
Cancel out \( 2 \) and \( \cos 2x \):
\( \text{LHS} = \frac { \sin 4x }{ \cos 4x } = \tan 4x \)
This is equal to the Right-Hand Side (RHS).
Thus, \( \text{LHS} = \text{RHS} \). The statement is proven.
In simple words: We grouped the sine and cosine terms in pairs and changed their additions into multiplications using specific rules. We found a common part, \( 2 \cos x \), in both the top and bottom of the fraction, which we canceled out. Then, we did the same process again for the remaining terms. This simplified the whole fraction to \( \frac { \sin 4x }{ \cos 4x } \), which is the same as \( \tan 4x \).

🎯 Exam Tip: When dealing with four terms in a sum for both numerator and denominator, group them strategically (e.g., first with second, third with fourth, or first with fourth, second with third). The goal is to find common factors that can be canceled. Remember \( \cos(-A) = \cos A \).

Question 13. Prove that \( \frac { \sin (4A - 2B) + \sin (4B - 2A) }{ \cos (4A - 2B) + \cos (4B - 2A) } = \tan (A + B) \)
Answer: To prove this identity, we will apply the sum-to-product formulas for sine and cosine to the numerator and denominator of the Left-Hand Side (LHS).
Consider the Left-Hand Side (LHS):
\( \text{LHS} = \frac { \sin (4A - 2B) + \sin (4B - 2A) }{ \cos (4A - 2B) + \cos (4B - 2A) } \)
Let \( C = 4A - 2B \) and \( D = 4B - 2A \).
First, calculate \( C+D \) and \( C-D \):
\( C+D = (4A - 2B) + (4B - 2A) = 2A + 2B = 2(A + B) \)
\( C-D = (4A - 2B) - (4B - 2A) = 4A - 2B - 4B + 2A = 6A - 6B = 6(A - B) \)
Now apply the sum-to-product formula to the numerator:
\( \sin C + \sin D = 2 \sin \frac { C+D }{ 2 } \cos \frac { C-D }{ 2 } \)
Numerator \( = 2 \sin \frac { 2(A+B) }{ 2 } \cos \frac { 6(A-B) }{ 2 } \)
\( = 2 \sin (A+B) \cos 3(A-B) \)
Apply the sum-to-product formula to the denominator:
\( \cos C + \cos D = 2 \cos \frac { C+D }{ 2 } \cos \frac { C-D }{ 2 } \)
Denominator \( = 2 \cos \frac { 2(A+B) }{ 2 } \cos \frac { 6(A-B) }{ 2 } \)
\( = 2 \cos (A+B) \cos 3(A-B) \)
Substitute these simplified expressions back into the LHS:
\( \text{LHS} = \frac { 2 \sin (A+B) \cos 3(A-B) }{ 2 \cos (A+B) \cos 3(A-B) } \)
Cancel out the common terms \( 2 \) and \( \cos 3(A-B) \):
\( \text{LHS} = \frac { \sin (A+B) }{ \cos (A+B) } \)
We know that \( \frac { \sin X }{ \cos X } = \tan X \).
So, \( \text{LHS} = \tan (A+B) \)
This is equal to the Right-Hand Side (RHS).
Thus, \( \text{LHS} = \text{RHS} \). The statement is proven.
In simple words: We used special formulas to change the additions of sine and cosine terms in the fraction's top and bottom into multiplications. After figuring out what the sums and differences of the angles were, we put them into the formulas. This made it clear that both the top and bottom had a common multiplication part, which we then removed. What was left was \( \frac { \sin (A+B) }{ \cos (A+B) } \), which simplifies to \( \tan (A+B) \).

🎯 Exam Tip: When dealing with complex angles like \( (4A - 2B) \), simplify the sum (C+D) and difference (C-D) terms separately first. This reduces errors and makes the substitution into sum-to-product formulas much clearer.

Question 14. Prove that \( \cot (A + 15°) - \tan (A - 15°) = \frac { 4 \cos 2A }{ 1+2 \sin 2A } \)
Answer: To prove this identity, we'll start from the Left-Hand Side (LHS) and express \( \cot \) and \( \tan \) in terms of \( \sin \) and \( \cos \), then simplify.
Consider the Left-Hand Side (LHS):
\( \text{LHS} = \cot (A + 15°) - \tan (A - 15°) \)
Write in terms of sine and cosine:
\( = \frac { \cos (A + 15°) }{ \sin (A + 15°) } - \frac { \sin (A - 15°) }{ \cos (A - 15°) } \)
Combine the fractions by finding a common denominator:
\( = \frac { \cos (A + 15°) \cos (A - 15°) - \sin (A + 15°) \sin (A - 15°) }{ \sin (A + 15°) \cos (A - 15°) } \)
The numerator is in the form \( \cos X \cos Y - \sin X \sin Y = \cos (X+Y) \).
Let \( X = A + 15° \) and \( Y = A - 15° \).
\( X+Y = (A + 15°) + (A - 15°) = 2A \)
So, Numerator \( = \cos (2A) \)
Now, consider the denominator: \( \sin (A + 15°) \cos (A - 15°) \).
This is in the form \( \sin X \cos Y \). We know \( 2 \sin X \cos Y = \sin (X+Y) + \sin (X-Y) \).
So, \( \sin (A + 15°) \cos (A - 15°) = \frac { 1 }{ 2 } [\sin ((A + 15°) + (A - 15°)) + \sin ((A + 15°) - (A - 15°))] \)
\( = \frac { 1 }{ 2 } [\sin (2A) + \sin (30°)] \)
We know \( \sin 30° = \frac { 1 }{ 2 } \).
Denominator \( = \frac { 1 }{ 2 } [\sin 2A + \frac { 1 }{ 2 }] \)
Substitute the simplified numerator and denominator back into the LHS:
\( \text{LHS} = \frac { \cos 2A }{ \frac { 1 }{ 2 } [\sin 2A + \frac { 1 }{ 2 }] } \)
Multiply the numerator and denominator by 2:
\( \text{LHS} = \frac { 2 \cos 2A }{ \sin 2A + \frac { 1 }{ 2 } } \)
To get rid of the fraction in the denominator, multiply the denominator by 2:
\( = \frac { 2 \cos 2A }{ \frac { 2 \sin 2A + 1 }{ 2 } } \)
Now, multiply by the reciprocal of the denominator:
\( = 2 \cos 2A \times \frac { 2 }{ 2 \sin 2A + 1 } \)
\( = \frac { 4 \cos 2A }{ 1 + 2 \sin 2A } \)
This is equal to the Right-Hand Side (RHS).
Thus, \( \text{LHS} = \text{RHS} \). The statement is proven.
In simple words: We changed cot and tan into fractions of sin and cos. Then, we combined these fractions. The top part became \( \cos(2A) \) using a sum formula for cosines. The bottom part became \( \frac { 1 }{ 2 } [\sin(2A) + \sin(30°)] \) using a product-to-sum formula for sine and cosine. After putting in the value of \( \sin 30° \) and doing some algebra, we arrived at the target expression.

🎯 Exam Tip: When proving identities involving \( \cot \) and \( \tan \), converting them to \( \sin \) and \( \cos \) is often the first and most effective step. Look for sum/difference of angles in the numerator and denominator that match standard trigonometric identities.