Samacheer Kalvi Class 11 Maths Solutions Chapter 3 Trigonometry Exercise 3.12

Choose The Correct Or The Most Suitable Answer:

Question 1. \( \frac { 1 }{ \cos 80^\circ } - \frac { \sqrt { 3 } }{ \sin 80^\circ } = \)
(1) \( \sqrt { 2 } \)
(2) \( \sqrt { 3 } \)
(3) 2
(4) 4
Answer: (4) 4
In simple words: The expression simplifies to 4. We combine the fractions, substitute common trigonometric values, and apply angle subtraction formulas to reach the final answer.

🎯 Exam Tip: When dealing with trigonometric expressions involving different angles, try to combine terms or use angle sum/difference formulas to simplify the expression into a known value.

Question 2. If \( \cos 28^\circ + \sin 28^\circ = k^3 \), then \( \cos 17^\circ \) is equal to
(1) \( \frac { k^3 }{ \sqrt { 2 } } \)
(2) \( -\frac { k^3 }{ \sqrt { 2 } } \)
(3) \( \pm \frac { k^3 }{ \sqrt { 2 } } \)
(4) \( -\frac { k^3 }{ \sqrt { 3 } } \)
Answer: (1) \( \frac { k^3 }{ \sqrt { 2 } } \)
In simple words: If you know the sum of cosine and sine of 28 degrees equals \( k^3 \), then the cosine of 17 degrees can be found as \( k^3 \) divided by \( \sqrt{2} \). We use trigonometric identities to link the given expression to the desired value.

🎯 Exam Tip: Remember the identity \( \sin x = \cos(90^\circ - x) \) to convert sine terms to cosine, which helps in applying sum-to-product formulas for simplification.

Question 3. The maximum value of \( 4 \sin^2x + 3 \cos^2x + \sin \frac { x }{ 2 } + \cos \frac { x }{ 2 } \) is
(1) \( 4 + \sqrt { 2 } \)
(2) \( 3 + \sqrt { 2 } \)
(4) 4
Answer: (1) \( 4 + \sqrt { 2 } \)
In simple words: To find the biggest possible value for this expression, we change parts of it using trigonometric rules. We know the maximum value of \( \sin x \) is 1. We also use the value of sine and cosine at \( 45^\circ \) to find the maximum.

🎯 Exam Tip: To maximize expressions with \( \sin^2 x \) and \( \cos^2 x \), rewrite them using the identity \( \sin^2 x + \cos^2 x = 1 \). Also, remember that \( \sin \theta + \cos \theta \) has a maximum value of \( \sqrt{2} \).

Question 4. \( \left( 1 + \cos \frac { \pi }{ 8 } \right) \left( 1 + \cos \frac { 3\pi }{ 8 } \right) \left( 1 + \cos \frac { 5\pi }{ 8 } \right) \left( 1 + \cos \frac { 7\pi }{ 8 } \right) \)
(1) \( \frac { 1 }{ 8 } \)
(2) \( \frac { 1 }{ 2 } \)
(3) \( \frac { 1 }{ \sqrt { 3 } } \)
(4) \( \frac { 1 }{ \sqrt { 2 } } \)
Answer: (1) \( \frac { 1 }{ 8 } \)
In simple words: This problem involves multiplying four terms together. We use the double angle identity for cosine, \( 1 + \cos 2\theta = 2 \cos^2 \theta \), to simplify each term. Then we pair them up and simplify further.

🎯 Exam Tip: When terms involve \( \frac{\pi}{8} \) and its multiples like \( \frac{3\pi}{8}, \frac{5\pi}{8}, \frac{7\pi}{8} \), look for complementary or supplementary angle relations to simplify the expression.

Question 5. If \( \pi < 2\theta < \frac { 3\pi }{ 2 } \), \( \sqrt { 2+\sqrt { 2+2 \cos 4 \theta } } \) equals to
(1) \( -2 \cos \theta \)
(2) \( -2 \sin \theta \)
(3) \( 2 \cos \theta \)
(4) \( 2 \sin \theta \)
Answer: (1) \( -2 \cos \theta \)
In simple words: We need to simplify the expression under the square roots. We repeatedly use the identity \( 1 + \cos 2x = 2 \cos^2 x \) to remove the square roots. The given range for \( \theta \) helps us decide if the simplified cosine term is positive or negative.

🎯 Exam Tip: Always pay attention to the given range for \( \theta \) in trigonometry problems. This range is crucial for determining the sign when taking square roots of trigonometric functions (e.g., \( \sqrt{\cos^2 x} = |\cos x| \)).

Question 6. If \( \tan 40^\circ = \lambda \), then \( \frac { \tan 140^\circ - \tan 130^\circ }{ 1 + \tan 140^\circ \tan 130^\circ } \)
(1) \( \frac { 1-\lambda^2 }{ \lambda } \)
(2) \( \frac { 1+\lambda^2 }{ \lambda } \)
(3) \( \frac { 1+\lambda^2 }{ 2 \lambda } \)
(4) \( \frac { 1-\lambda^2 }{ 2 \lambda } \)
Answer: (4) \( \frac { 1-\lambda^2 }{ 2 \lambda } \)
In simple words: We are given the value of \( \tan 40^\circ \) as \( \lambda \). The expression we need to simplify is a form of the tangent subtraction formula. By simplifying the expression and using angle relationships, we can express the final answer in terms of \( \lambda \).

🎯 Exam Tip: Recognize the tangent subtraction formula \( \tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \). Also, use angle relationships like \( \tan(90^\circ + x) = -\cot x \) and \( \tan(180^\circ - x) = -\tan x \) to relate given angles to the known value.

Question 7. \( \cos 1^\circ + \cos 2^\circ + \cos 3^\circ + ..... + \cos 179^\circ = \)
(1) 0
(2) 1
(3) - 1
(4) 89
Answer: (1) 0
In simple words: We need to find the sum of cosines of angles from 1 degree to 179 degrees. We can pair up terms using the identity \( \cos \theta + \cos (180^\circ - \theta) = 0 \). Most terms will cancel each other out.

🎯 Exam Tip: For sums of trigonometric functions over a range, look for symmetries or identities like \( \cos (180^\circ - \theta) = -\cos \theta \) or \( \sin (180^\circ - \theta) = \sin \theta \) that cause terms to cancel or simplify.

Question 8. Let \( f_k(x) = \frac { 1 }{ k } [\sin^k x + \cos^k x] \) where \( x \in R \) and \( k \geq 1 \). Then \( f_4(x) - f_6(x) = \)
(1) \( \frac { 1 }{ 4 } \)
(2) \( \frac { 1 }{ 12 } \)
(3) \( \frac { 1 }{ 6 } \)
(4) \( \frac { 1 }{ 3 } \)
Answer: (2) \( \frac { 1 }{ 12 } \)
In simple words: We are given a function \( f_k(x) \). We need to calculate \( f_4(x) \) and \( f_6(x) \) and then find their difference. We use the identity \( \sin^2 x + \cos^2 x = 1 \) to simplify the expressions.

🎯 Exam Tip: When dealing with powers of sine and cosine, remember to use the identity \( \sin^2 x + \cos^2 x = 1 \) to simplify expressions, often by rewriting higher powers in terms of \( (\sin^2 x + \cos^2 x) \) squared or cubed.

Question 9. Which of the following is not true?
(1) \( \sin \theta = -\frac { 3 }{ 4 } \)
(2) \( \cos \theta = -1 \)
(3) \( \tan \theta = 25 \)
(4) \( \sec \theta = \frac { 1 }{ 4 } \)
Answer: (4) \( \sec \theta = \frac { 1 }{ 4 } \)
In simple words: We are looking for the statement that is mathematically impossible. We know that the values of sine and cosine are always between -1 and 1. We check each option to see if it follows this rule. The secant function is the inverse of the cosine function.

🎯 Exam Tip: Always remember the range of trigonometric functions: \( -1 \leq \sin \theta \leq 1 \), \( -1 \leq \cos \theta \leq 1 \). For \( \sec \theta \) and \( \csc \theta \), their absolute values must be greater than or equal to 1, i.e., \( |\sec \theta| \geq 1 \) and \( |\csc \theta| \geq 1 \).

Question 10. \( \cos 2\theta \cos 2\Phi + \sin^2(\theta - \Phi) - \sin^2(\theta + \Phi) \) is equal to
(1) \( \sin 2(\theta + \Phi) \)
(2) \( \cos 2(\theta + \Phi) \)
(3) \( \sin 2(\theta - \Phi) \)
(4) \( \cos 2(\theta - \Phi) \)
Answer: (2) \( \cos 2(\theta + \Phi) \)
In simple words: We need to simplify the given trigonometric expression. We use the identity \( \sin^2 A - \sin^2 B = \sin(A-B)\sin(A+B) \) and then apply product-to-sum and angle addition formulas to get the final simplified form.

🎯 Exam Tip: Use the identity \( \sin^2 A - \sin^2 B = \sin(A-B)\sin(A+B) \) for terms with squared sines. Also, remember the cosine addition formula \( \cos(A+B) = \cos A \cos B - \sin A \sin B \).

Question 11. \( \frac { \sin (A - B) }{ \cos A \cos B } + \frac { \sin (B - C) }{ \cos B \cos C } + \frac { \sin (C - A) }{ \cos C \cos A } \) is
(1) \( \sin A + \sin B + \sin C \)
(2) 1
(3) 0
(4) \( \cos A + \cos B + \cos C \)
Answer: (3) 0
In simple words: We need to simplify the sum of three fractions. Each fraction can be separated into two tangent terms using the expansion of \( \sin(X-Y) \). Then, notice how the terms cancel each other out in the sum.

🎯 Exam Tip: For fractions like \( \frac{\sin(A-B)}{\cos A \cos B} \), expand \( \sin(A-B) = \sin A \cos B - \cos A \sin B \). Dividing by \( \cos A \cos B \) will give \( \tan A - \tan B \), which then simplifies nicely in a cyclic sum.

Question 12. If \( \cos p\theta + \cos q\theta = 0 \) and if \( p \neq q \) then \( \theta \) is equal to (n is any integer)
(1) \( \frac { \pi (3n + 1) }{ p - q } \)
(2) \( \frac { \pi (2n + 1) }{ p \pm q } \)
(3) \( \pi (2n + 1) \)
(4) \( \frac { \pi (n + 2) }{ p+q } \)
Answer: (2) \( \frac { \pi (2n + 1) }{ p \pm q } \)
In simple words: When the sum of two cosine terms is zero, it means \( \cos A = -\cos B \). This can be rewritten as \( \cos A = \cos(\pi - B) \). We use the sum-to-product formula for cosines and set the expression to zero to find the general solution for \( \theta \).

🎯 Exam Tip: When \( \cos A + \cos B = 0 \), use the sum-to-product formula \( \cos A + \cos B = 2 \cos \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right) = 0 \). This means either \( \cos \left( \frac{A+B}{2} \right) = 0 \) or \( \cos \left( \frac{A-B}{2} \right) = 0 \), leading to two sets of solutions for \( \theta \).

Question 13. If \( \tan \alpha \) and \( \tan \beta \) are the roots of \( x^2 + ax + b = 0 \) then \( \frac { \sin (\alpha+\beta) }{ \sin \alpha \sin \beta } \) is equal to
(1) \( \frac { b }{ a } \)
(2) \( \frac { a }{ b } \)
(3) \( -\frac { a }{ b } \)
(4) \( -\frac { b }{ a } \)
Answer: (3) \( -\frac { a }{ b } \)
In simple words: For a quadratic equation, if \( \tan \alpha \) and \( \tan \beta \) are the roots, we know their sum and product. We use these to simplify the given trigonometric expression. The expression can be rewritten in terms of \( \cot \alpha \) and \( \cot \beta \).

🎯 Exam Tip: Remember Vieta's formulas for roots of a quadratic equation: if \( r_1, r_2 \) are roots of \( Ax^2 + Bx + C = 0 \), then \( r_1 + r_2 = -B/A \) and \( r_1 r_2 = C/A \). Apply this to \( \tan \alpha \) and \( \tan \beta \) to find their sum and product.

Question 14. In a triangle ABC, \( \sin^2 A + \sin^2 B + \sin^2 C = 2 \) then the triangle is .
(1) equilateral triangle
(2) isosceles triangle
(3) right triangle
(4) scalene triangle
Answer: (3) right triangle
In simple words: For any triangle, the sum of sine squares of its angles is equal to 2 plus twice the product of cosines of its angles. If this sum is 2, it means the product of the cosines must be zero. This happens when one of the angles is 90 degrees, making it a right-angled triangle.

🎯 Exam Tip: A useful identity for triangles is \( \sin^2 A + \sin^2 B + \sin^2 C = 2 + 2 \cos A \cos B \cos C \). If this sum equals 2, it implies \( 2 \cos A \cos B \cos C = 0 \), which means at least one angle is \( 90^\circ \), making the triangle a right-angled triangle.

Question 15. If \( f(\theta) = |\sin \theta| + |\cos \theta| \), \( \theta \in R \) then \( f(\theta) \) is in the interval
(1) [0, 2]
(2) [1, \( \sqrt { 2 } \)]
(3) [1, 2]
(4) [0, 1]
Answer: (2) [1, \( \sqrt { 2 } \)]
In simple words: We need to find the smallest and largest values that \( |\sin \theta| + |\cos \theta| \) can take. The smallest value is 1 (when one is 0 and the other is 1). The largest value is \( \sqrt{2} \) (when both \( |\sin \theta| \) and \( |\cos \theta| \) are \( \frac{1}{\sqrt{2}} \)). The graph shows how this value changes.

🎯 Exam Tip: To find the range of \( |\sin \theta| + |\cos \theta| \), you can square the expression or use calculus. Squaring it gives \( 1 + |2 \sin \theta \cos \theta| = 1 + |\sin 2\theta| \). Since \( 0 \leq |\sin 2\theta| \leq 1 \), the range of \( 1 + |\sin 2\theta| \) is [1, 2]. Taking the square root gives the range [1, \( \sqrt{2} \)].

Question 15. If \( f(\theta) = |\sin \theta| + |\cos \theta| \), \( \theta \in R \) then \( f(\theta) \) is in the interval
(1) [0, 2]
(2) [1, \( \sqrt{2} \)]
(3) [1, 2]
(4) [0, 1]
Answer: (2) [1, \( \sqrt{2} \)]
We need to find the range of the function \( f(\theta) = |\sin \theta| + |\cos \theta| \). First, we consider key points where sine and cosine values are known. At \( \theta = 0 \) radians, \( f(0) = |\sin 0| + |\cos 0| = 0 + 1 = 1 \). At \( \theta = \frac{\pi}{4} \) radians, \( f(\frac{\pi}{4}) = |\sin \frac{\pi}{4}| + |\cos \frac{\pi}{4}| = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \). At \( \theta = \frac{\pi}{2} \) radians, \( f(\frac{\pi}{2}) = |\sin \frac{\pi}{2}| + |\cos \frac{\pi}{2}| = 1 + 0 = 1 \). For any real number \( \theta \), the value of \( |\sin \theta| + |\cos \theta| \) will always be between 1 and \( \sqrt{2} \). This calculation helps us understand the behavior of sums of absolute trigonometric functions.
In simple words: The smallest value of this function is 1, and the largest is \( \sqrt{2} \). So, the answer will always be found in the range from 1 up to \( \sqrt{2} \).

🎯 Exam Tip: Remember that the absolute values \( |\sin \theta| \) and \( |\cos \theta| \) ensure the function is always positive or zero, impacting its minimum value.

Question 16. \( \frac{\cos 6x + 6\cos 4x + 15\cos 2x + 10}{\cos 5x + 5\cos 3x + 10\cos x} \) is equal to
(1) cos 2x
(2) cos x
(3) cos 3x
(4) 2 cos x
Answer: (4) 2 cos x
We need to simplify the given trigonometric expression. The expression is \( \frac{\cos 6x + 6\cos 4x + 15\cos 2x + 10}{\cos 5x + 5\cos 3x + 10\cos x} \).
We can rewrite the numerator:
\( \cos 6x + 6\cos 4x + 15\cos 2x + 10 \)
\( = (\cos 6x + \cos 4x) + 5(\cos 4x + \cos 2x) + 10(\cos 2x + 1) \)
Using the sum-to-product formula \( \cos A + \cos B = 2\cos(\frac{A+B}{2})\cos(\frac{A-B}{2}) \) and \( 1 + \cos 2x = 2\cos^2 x \):
\( = 2\cos(\frac{6x+4x}{2})\cos(\frac{6x-4x}{2}) + 5 \left[ 2\cos(\frac{4x+2x}{2})\cos(\frac{4x-2x}{2}) \right] + 10(2\cos^2 x) \)
\( = 2\cos 5x \cos x + 10\cos 3x \cos x + 20\cos^2 x \)
Now, we can factor out \( 2\cos x \) from the numerator:
\( = 2\cos x (\cos 5x + 5\cos 3x + 10\cos x) \)
So, the original expression becomes:
\( \frac{2\cos x (\cos 5x + 5\cos 3x + 10\cos x)}{\cos 5x + 5\cos 3x + 10\cos x} \)
We can cancel out the common factor \( (\cos 5x + 5\cos 3x + 10\cos x) \) from both numerator and denominator, assuming it's not zero.
\( = 2\cos x \)
This shows how trigonometric identities help simplify complex expressions into a simpler form.
In simple words: We broke down the top part of the fraction using special math rules. Then, we found a common part in both the top and bottom of the fraction and cancelled it out. What was left was \( 2\cos x \).

🎯 Exam Tip: Look for opportunities to use sum-to-product or double-angle formulas to simplify complex trigonometric expressions, and always factor out common terms.

Question 17. The triangle of the maximum area with a constant perimeter of 12m
(1) is an equilateral triangle with a side of 4m
(2) is an isosceles triangle with sides 2m, 5m, 5m
(3) is a triangle with sides 3m, 4m, 5m
(4) does not exist.
Answer: (1) is an equilateral triangle with a side of 4m
For a fixed perimeter, a triangle will have the largest possible area when all its sides are equal, meaning it is an equilateral triangle. If the perimeter is 12m, an equilateral triangle would have sides of \( \frac{12}{3} = 4 \)m each. This is a fundamental property in geometry.
In simple words: To get the biggest area from a triangle with a set boundary length, it must be a triangle where all three sides are the same length.

🎯 Exam Tip: Remember that for a fixed perimeter, an equilateral triangle maximizes the area among all triangles.

Question 18. A wheel is spinning at 2 radians/second. How many seconds will it take to make 10 complete rotations?
(1) 10 \( \pi \) seconds
(3) 5 \( \pi \) seconds
(4) 15 \( \pi \) seconds
Answer: (1) 10 \( \pi \) seconds
One complete rotation of a wheel is equal to \( 2\pi \) radians. For 10 complete rotations, the total angle covered will be \( 10 \times 2\pi = 20\pi \) radians. The wheel is spinning at a rate of 2 radians per second. To find the time taken, we divide the total radians by the speed: \( \text{Time} = \frac{\text{Total radians}}{\text{Speed}} = \frac{20\pi \text{ radians}}{2 \text{ radians/second}} = 10\pi \) seconds. Understanding rotational speed is key in physics problems.
In simple words: One full turn is \( 2\pi \) radians. Ten turns mean \( 20\pi \) radians. Since it spins 2 radians every second, it will take \( 10\pi \) seconds to do all ten turns.

🎯 Exam Tip: Pay attention to units and ensure you convert rotations to radians correctly (1 rotation = \( 2\pi \) radians).

Question 19. If \( \sin \alpha + \cos \alpha = b \), then \( \sin 2\alpha \) is equal to
(1) \( b^2 - 1 \), if \( b \le \sqrt{2} \)
(2) \( b^2 - 1 \), if \( b > \sqrt{2} \)
(3) \( b^2 - 1 \), if \( b \ge \sqrt{2} \)
(4) \( b^2 - 1 \), if \( b < \sqrt{2} \)
Answer: (1) \( b^2 - 1 \), if \( b \le \sqrt{2} \)
We are given the equation \( \sin \alpha + \cos \alpha = b \). To find \( \sin 2\alpha \), we can square both sides of the given equation.
\( (\sin \alpha + \cos \alpha)^2 = b^2 \)
Expanding the left side, we get \( \sin^2 \alpha + \cos^2 \alpha + 2\sin \alpha \cos \alpha = b^2 \).
We know that \( \sin^2 \alpha + \cos^2 \alpha = 1 \) and \( 2\sin \alpha \cos \alpha = \sin 2\alpha \).
So, \( 1 + \sin 2\alpha = b^2 \).
This means \( \sin 2\alpha = b^2 - 1 \).
Also, we know that the value of \( \sin 2\alpha \) must be between -1 and 1, inclusive.
So, \( -1 \le b^2 - 1 \le 1 \).
Adding 1 to all parts of the inequality: \( -1+1 \le b^2 - 1+1 \le 1+1 \), which gives \( 0 \le b^2 \le 2 \).
Since \( b^2 \ge 0 \) is always true, we only need \( b^2 \le 2 \). This implies \( -\sqrt{2} \le b \le \sqrt{2} \). Therefore, \( b \le \sqrt{2} \).
This identity is frequently used to relate sums of sines and cosines to double angles.
In simple words: We started with \( \sin \alpha + \cos \alpha = b \). We squared both sides to get \( \sin 2\alpha \). We also know that \( \sin 2\alpha \) can only be between -1 and 1. This helps us find the possible range for 'b'.

🎯 Exam Tip: Squaring trigonometric sums is a common method to introduce double angle identities. Always remember the range of sine and cosine functions (between -1 and 1).

Question 20. In an \( \triangle ABC \)
(i) \( \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} > 0 \)
(ii) \( \sin A \sin B \sin C > 0 \),then
(1) Both (i) and (ii) are true
(2) only (1) is true
(3) only (ii) Is true
(4) neither (i) nor (ii) is true
Answer: (1) Both (i) and (ii) are true
For any triangle \( \triangle ABC \), the sum of its internal angles is \( A + B + C = 180^\circ \).
This means each angle (A, B, C) must be greater than \( 0^\circ \) and less than \( 180^\circ \).
Also, each half-angle (\( A/2, B/2, C/2 \)) must be greater than \( 0^\circ \) and less than \( 90^\circ \) (an acute angle).
Since all angles \( A, B, C \) are between \( 0^\circ \) and \( 180^\circ \), their sine values \( \sin A, \sin B, \sin C \) will all be positive. Therefore, their product \( \sin A \sin B \sin C \) will also be positive, so \( \sin A \sin B \sin C > 0 \). This confirms statement (ii).
Similarly, since all half-angles \( A/2, B/2, C/2 \) are between \( 0^\circ \) and \( 90^\circ \), their sine values \( \sin(A/2), \sin(B/2), \sin(C/2) \) will all be positive. Therefore, their product \( \sin(A/2) \sin(B/2) \sin(C/2) \) will also be positive, so \( \sin(A/2) \sin(B/2) \sin(C/2) > 0 \). This confirms statement (i).
Both statements are true for any triangle. This relationship shows the basic properties of angles within a triangle.
In simple words: In any triangle, all angles are between 0 and 180 degrees. This means the sine of each angle and the sine of half of each angle will always be positive. If all parts are positive, then multiplying them together will also give a positive result. So, both parts of the question are correct.

🎯 Exam Tip: Remember that sine is positive for angles between \( 0^\circ \) and \( 180^\circ \). For a triangle, all angles fall within this range.