CBSE Class 11 Mathematics Revision Worksheet Set 01

Q1 Prove by the principle of mathematical induction, that for all n in N
a) \(1 + 4 + 7 + \dots + (3n - 2) = \frac{n(3n - 1)}{2}\)
Answer: Let \(P(n)\) be the given statement: \(1 + 4 + 7 + \dots + (3n - 1) = \frac{n(3n - 1)}{2}\).
Step 1: For \(n = 1\):
LHS = \(1\)
RHS = \(\frac{1(3(1) - 1)}{2} = \frac{2}{2} = 1\)
Since LHS = RHS, \(P(1)\) is true.
Step 2: Assume \(P(k)\) is true for some positive integer \(k\):
\(1 + 4 + 7 + \dots + (3k - 2) = \frac{k(3k - 1)}{2}\)
Step 3: We must prove that \(P(k + 1)\) is true:
\(1 + 4 + 7 + \dots + (3k - 2) + (3(k + 1) - 2) = \frac{(k + 1)(3(k + 1) - 1)}{2}\)
LHS = \(\frac{k(3k - 1)}{2} + (3k + 1)\)
\(= \frac{3k^2 - k + 6k + 2}{2}\)
\(= \frac{3k^2 + 5k + 2}{2}\)
\(= \frac{(k + 1)(3k + 2)}{2} = \text{RHS}\)
Thus, \(P(k + 1)\) is true whenever \(P(k)\) is true. Hence, by mathematical induction, the statement is true for all \(n \in N\).
In simple words: We proved the formula works for the first term and then showed that if it works for any term \(k\), it automatically works for the next term \(k + 1\).
Exam Tip: Always clearly write the three steps: Base Step (\(n = 1\)), Inductive Hypothesis (\(n = k\)), and Inductive Step (\(n = k + 1\)).

Q1 b) \(1^3 + 2^3 + 3^3 + \dots + n^3 = \left(\frac{n(n + 1)}{2}\right)^2\)
Answer: Let \(P(n)\) be \(1^3 + 2^3 + 3^3 + \dots + n^3 = \left(\frac{n(n + 1)}{2}\right)^2\).
Step 1: For \(n = 1\):
LHS = \(1^3 = 1\)
RHS = \(\left(\frac{1(1 + 1)}{2}\right)^2 = 1^2 = 1\)
So, \(P(1)\) is true.
Step 2: Assume \(P(k)\) is true:
\(1^3 + 2^3 + \dots + k^3 = \left(\frac{k(k + 1)}{2}\right)^2\)
Step 3: Prove for \(n = k + 1\):
\(1^3 + 2^3 + \dots + k^3 + (k + 1)^3 = \left(\frac{(k + 1)(k + 2)}{2}\right)^2\)
LHS = \(\left(\frac{k(k + 1)}{2}\right)^2 + (k + 1)^3\)
\(= (k + 1)^2 \left[ \frac{k^2}{4} + (k + 1) \right]\)
\(= (k + 1)^2 \left[ \frac{k^2 + 4k + 4}{4} \right]\)
\(= \frac{(k + 1)^2 (k + 2)^2}{4} = \left(\frac{(k + 1)(k + 2)}{2}\right)^2 = \text{RHS}\)
Hence, by mathematical induction, the result holds for all \(n \in N\).
In simple words: Factor out the common term \((k + 1)^2\) to simplify the algebraic expression easily.

Exam Tip: Memorize the standard sum of cubes formula as it frequently appears in induction problems.

Q1 c) \(1 + \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + \dots + \frac{1}{1 + 2 + 3 + \dots + n} = \frac{2n}{n + 1}\), for all \(n \in N\)
Answer: Let the \(n\)-th term of the series be \(T_n = \frac{1}{1 + 2 + 3 + \dots + n}\).
Since the denominator is the sum of the first \(n\) natural numbers, \(1 + 2 + \dots + n = \frac{n(n + 1)}{2}\).
Thus, \(T_n = \frac{2}{n(n + 1)} = 2 \left( \frac{1}{n} - \frac{1}{n + 1} \right)\).
Now, using mathematical induction or telescoping sum:
Step 1: For \(n = 1\):
LHS = \(1\)
RHS = \(\frac{2(1)}{1 + 1} = 1\)
Step 2: Assume for \(n = k\):
\(S_k = 1 + \frac{1}{1 + 2} + \dots + \frac{1}{1 + 2 + \dots + k} = \frac{2k}{k + 1}\)
Step 3: For \(n = k + 1\):
\(S_{k + 1} = S_k + T_{k + 1} = \frac{2k}{k + 1} + \frac{2}{(k + 1)(k + 2)}\)
\(= \frac{2}{k + 1} \left[ k + \frac{1}{k + 2} \right]\)
\(= \frac{2}{k + 1} \left[ \frac{k^2 + 2k + 1}{k + 2} \right]\)
\(= \frac{2(k + 1)^2}{(k + 1)(k + 2)} = \frac{2(k + 1)}{k + 2} = \text{RHS}\)
Hence proved by induction.
In simple words: Convert the denominator into a product using standard summation formulas, then split using partial fractions.

Exam Tip: Simplify the \(n\)-th term into a difference of two fractions before applying induction or summation.

Q1 d) \(\frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \dots + \frac{1}{(2n - 1)(2n + 1)} = \frac{n}{2n + 1}\)
Answer: Let \(P(n)\) be the given statement.
Step 1: For \(n = 1\):
LHS = \(\frac{1}{1 \cdot 3} = \frac{1}{3}\)
RHS = \(\frac{1}{2(1) + 1} = \frac{1}{3}\)
So, \(P(1)\) is true.
Step 2: Assume \(P(k)\) is true:
\(\frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \dots + \frac{1}{(2k - 1)(2k + 1)} = \frac{k}{2k + 1}\)
Step 3: Prove for \(n = k + 1\):
LHS = \(\frac{k}{2k + 1} + \frac{1}{(2(k + 1) - 1)(2(k + 1) + 1)}\)
\(= \frac{k}{2k + 1} + \frac{1}{(2k + 1)(2k + 3)}\)
\(= \frac{1}{2k + 1} \left[ k + \frac{1}{2k + 3} \right]\)
\(= \frac{1}{2k + 1} \left[ \frac{2k^2 + 3k + 1}{2k + 3} \right]\)
\(= \frac{1}{2k + 1} \left[ \frac{(2k + 1)(k + 1)}{2k + 3} \right]\)
\(= \frac{k + 1}{2(k + 1) + 1} = \text{RHS}\)
Hence proved by mathematical induction.
In simple words: Add the \((k + 1)\)-th term to the assumed sum for \(k\) terms and simplify algebraically.

Exam Tip: Factor quadratic expressions completely in the numerator to cancel common factors with the denominator.

Q1 e) \(7^n - 3^n\) is divisible by 4.
Answer: Let \(P(n)\) be the statement that \(7^n - 3^n\) is divisible by 4, meaning \(7^n - 3^n = 4m\) for some integer \(m\).
Step 1: For \(n = 1\):
\(7^1 - 3^1 = 4\), which is divisible by 4. So, \(P(1)\) is true.
Step 2: Assume \(P(k)\) is true:
\(7^k - 3^k = 4m\) (where \(m\) is an integer), which gives \(7^k = 4m + 3^k\).
Step 3: Prove for \(n = k + 1\):
\(7^{k + 1} - 3^{k + 1} = 7^k \cdot 7 - 3^k \cdot 3\)
Substitute \(7^k = 4m + 3^k\):
\(= (4m + 3^k)7 - 3 \cdot 3^k\)
\(= 28m + 7 \cdot 3^k - 3 \cdot 3^k\)
\(= 28m + 4 \cdot 3^k\)
\(= 4(7m + 3^k)\)
Since \(4(7m + 3^k)\) is a multiple of 4, it is divisible by 4. Hence, \(P(k + 1)\) is true.
In simple words: Express \(7^k\) in terms of \(3^k\) from the inductive hypothesis and substitute it into the expression for \(k + 1\).

Exam Tip: For divisibility problems, always isolate the highest power term in the inductive assumption step.

Q1 f) \((1 + x)^n \geq 1 + nx\) whenever \(x\) is positive and \(n\) is a positive integer.
Answer: Let \(P(n)\) be \((1 + x)^n \geq 1 + nx\).
Step 1: For \(n = 1\):
LHS = \((1 + x)^1 = 1 + x\)
RHS = \(1 + 1(x) = 1 + x\)
Since LHS = RHS, \(P(1)\) is true.
Step 2: Assume \(P(k)\) is true for some positive integer \(k\):
\((1 + x)^k \geq 1 + kx\)
Step 3: Prove for \(n = k + 1\):
Multiply both sides of the inequality by \((1 + x)\) (since \(1 + x > 0\)):
\((1 + x)^{k + 1} \geq (1 + kx)(1 + x)\)
\((1 + x)^{k + 1} \geq 1 + x + kx + kx^2\)
\((1 + x)^{k + 1} \geq 1 + (k + 1)x + kx^2\)
Since \(kx^2 \geq 0\) for positive \(k\) and \(x\), we have:
\(1 + (k + 1)x + kx^2 \geq 1 + (k + 1)x\)
Therefore, \((1 + x)^{k + 1} \geq 1 + (k + 1)x\), which proves the statement for \(n = k + 1\).
In simple words: Multiply the assumed inequality by positive \((1 + x)\) and drop the positive extra term \(kx^2\) to get the required inequality.

Exam Tip: For inequality induction proofs, make sure to state that multiplying by positive quantities preserves the inequality sign.

Q1 g) \(4^n + 15n - 1\) is divisible by 9, for \(n \in N\).
Answer: Let \(P(n)\) be the statement that \(4^n + 15n - 1\) is divisible by 9, so \(4^n + 15n - 1 = 9m\) for some integer \(m\).
Step 1: For \(n = 1\):
\(4^1 + 15(1) - 1 = 4 + 15 - 1 = 18\), which is divisible by 9. So, \(P(1)\) is true.
Step 2: Assume \(P(k)\) is true:
\(4^k + 15k - 1 = 9m \implies 4^k = 9m - 15k + 1\)
Step 3: Prove for \(n = k + 1\):
\(4^{k + 1} + 15(k + 1) - 1 = 4^k \cdot 4 + 15k + 15 - 1\)
Substitute \(4^k = 9m - 15k + 1\):
\(= 4(9m - 15k + 1) + 15k + 14\)
\(= 36m - 60k + 4 + 15k + 14\)
\(= 36m - 45k + 18\)
\(= 9(4m - 5k + 2)\)
Since \(9(4m - 5k + 2)\) is a multiple of 9, it is divisible by 9. Hence, \(P(k + 1)\) is true.
In simple words: Substitute \(4^k\) from the inductive hypothesis and group terms to factor out 9.

Exam Tip: Always factor out the required divisor (like 9) explicitly in the final step to clearly show divisibility.

Q2. Prove that \(\frac{\cos 8A \cos 5A - \cos 12A \cos 9A}{\sin 8A \cos 5A + \cos 12A \sin 9A} = \tan 4A\)
Answer: Consider the numerator: \(\cos 8A \cos 5A - \cos 12A \cos 9A\)
Multiply numerator and denominator by 2 to use product-to-sum formulas:
\(2 \cos 8A \cos 5A = \cos(8A + 5A) + \cos(8A - 5A) = \cos 13A + \cos 3A\)
\(2 \cos 12A \cos 9A = \cos(12A + 9A) + \cos(12A - 9A) = \cos 21A + \cos 3A\)
Numerator = \(\frac{1}{2} [(\cos 13A + \cos 3A) - (\cos 21A + \cos 3A)] = \frac{1}{2} (\cos 13A - \cos 21A)\)
Using sum-to-product formula \(\cos C - \cos D = 2 \sin\left(\frac{C + D}{2}\right) \sin\left(\frac{D - C}{2}\right)\):
Numerator = \(\frac{1}{2} \left[ 2 \sin\left(\frac{13A + 21A}{2}\right) \sin\left(\frac{21A - 13A}{2}\right) \right] = \sin 17A \sin 4A\)
Now consider the denominator: \(\sin 8A \cos 5A + \cos 12A \sin 9A\)
Using \(2 \sin A \cos B = \sin(A + B) + \sin(A - B)\):
\(2 \sin 8A \cos 5A = \sin 13A + \sin 3A\)
\(2 \cos 12A \sin 9A = \sin(12A + 9A) - \sin(12A - 9A) = \sin 21A - \sin 3A\)
Denominator = \(\frac{1}{2} [(\sin 13A + \sin 3A) + (\sin 21A - \sin 3A)] = \frac{1}{2} (\sin 13A + \sin 21A)\)
Using \(\sin C + \sin D = 2 \sin\left(\frac{C + D}{2}\right) \cos\left(\frac{C - D}{2}\right)\):
Denominator = \(\frac{1}{2} \left[ 2 \sin\left(\frac{13A + 21A}{2}\right) \cos\left(\frac{13A - 21A}{2}\right) \right] = \sin 17A \cos(-4A) = \sin 17A \cos 4A\)
Therefore, \(\text{LHS} = \frac{\sin 17A \sin 4A}{\sin 17A \cos 4A} = \frac{\sin 4A}{\cos 4A} = \tan 4A = \text{RHS}\).
In simple words: Convert products of cosines and sines into sums using standard trigonometric formulas, then simplify terms to get \(\tan 4A\).

Exam Tip: Be careful with signs when applying product-to-sum and sum-to-product formulas.

Q3. Prove that \(\tan 3x \tan 2x \tan x = \tan 3x - \tan 2x - \tan x\)
Answer: Start with the angle identity: \(3x = 2x + x\)
Taking tangent on both sides:
\(\tan(3x) = \tan(2x + x)\)
Using the tangent addition formula \(\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}\):
\(\tan 3x = \frac{\tan 2x + \tan x}{1 - \tan 2x \tan x}\)
Cross-multiplying:
\(\tan 3x (1 - \tan 2x \tan x) = \tan 2x + \tan x\)
\(\tan 3x - \tan 3x \tan 2x \tan x = \tan 2x + \tan x\)
Rearranging terms to isolate the product term:
\(\tan 3x \tan 2x \tan x = \tan 3x - \tan 2x - \tan x\)
Hence proved.
In simple words: Use the relation \(3x = 2x + x\) and expand using the tan formula to rearrange into the desired identity.

Exam Tip: This is a standard trigonometric identity; always start from the relation involving the sum of angles.

Q4. If \(A = \{1, 2, 5\}\), \(B = \{1, 2, 3, 4\}\) and \(C = \{5, 6, 2\}\), Verify that
a) \(A \times (B \cap C) = (A \times B) \cap (A \times C)\)
Answer: Given sets: \(A = \{1, 2, 5\}\), \(B = \{1, 2, 3, 4\}\), \(C = \{5, 6, 2\}\).
Step 1: Find \(B \cap C\):
\(B \cap C = \{1, 2, 3, 4\} \cap \{5, 6, 2\} = \{2\}\)
Step 2: Find LHS \(= A \times (B \cap C)\):
\(A \times (B \cap C) = \{1, 2, 5\} \times \{2\} = \{(1, 2), (2, 2), (5, 2)\}\)
Step 3: Find \(A \times B\):
\(A \times B = \{(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (2,4), (5,1), (5,2), (5,3), (5,4)\}\)
Step 4: Find \(A \times C\):
\(A \times C = \{(1,5), (1,6), (1,2), (2,5), (2,6), (2,2), (5,5), (5,6), (5,2)\}\)
Step 5: Find RHS \(= (A \times B) \cap (A \times C)\):
The common ordered pairs in \(A \times B\) and \(A \times C\) are \(\{(1, 2), (2, 2), (5, 2)\}\).
Since LHS = RHS, the property is verified.
In simple words: Calculate intersection of sets first, then form Cartesian products and compare both sides.

Exam Tip: Write all ordered pairs clearly in set-builder or roster form to avoid missing elements.

Q4 b) \((A - B) \times C = (A \times C) - (B \times C)\)
Answer: Given sets: \(A = \{1, 2, 5\}\), \(B = \{1, 2, 3, 4\}\), \(C = \{5, 6, 2\}\).
Step 1: Find \(A - B\):
\(A - B = \{1, 2, 5\} - \{1, 2, 3, 4\} = \{5\}\)
Step 2: Find LHS \(= (A - B) \times C\):
\(\{5\} \times \{5, 6, 2\} = \{(5, 5), (5, 6), (5, 2)\}\)
Step 3: Find \(A \times C\):
\(A \times C = \{(1,5), (1,6), (1,2), (2,5), (2,6), (2,2), (5,5), (5,6), (5,2)\}\)
Step 4: Find \(B \times C\):
\(B \times C = \{(1,5), (1,6), (1,2), (2,5), (2,6), (2,2), (3,5), (3,6), (3,2), (4,5), (4,6), (4,2)\}\)
Step 5: Find RHS \(= (A \times C) - (B \times C)\):
Removing all elements of \(B \times C\) from \(A \times C\) leaves: \(\{(5, 5), (5, 6), (5, 2)\}\).
Since LHS = RHS, the property is verified.
In simple words: Subtract set B from set A first, then take Cartesian product with C and compare with set difference of Cartesian products.

Exam Tip: Cartesian product is distributive over set difference and intersection.

Q6. What is the value of \(\cos\left(\frac{\pi}{4} - x\right) \cos\left(\frac{\pi}{4} - y\right) - \sin\left(\frac{\pi}{4} - x\right) \sin\left(\frac{\pi}{4} - y\right)\)
Answer: Let \(A = \frac{\pi}{4} - x\) and \(B = \frac{\pi}{4} - y\).
The expression is in the form \(\cos A \cos B - \sin A \sin B\), which is the cosine addition formula: \(\cos(A + B)\).
Thus, expression = \(\cos\left( \left(\frac{\pi}{4} - x\right) + \left(\frac{\pi}{4} - y\right) \right)\)
\(= \cos\left( \frac{\pi}{2} - (x + y) \right)\)
Using the trigonometric identity \(\cos\left(\frac{\pi}{2} - \theta\right) = \sin\theta\):
\(= \sin(x + y)\)
In simple words: Apply the cosine sum formula \(\cos(A + B) = \cos A \cos B - \sin A \sin B\) and simplify using complementary angle relations.

Exam Tip: Recognize standard trigonometric expansion formulas immediately to condense expressions.

Q7. What is the eccentricity of the curve \(4x^2 + y^2 = 100\)?
Answer: Given equation of the curve: \(4x^2 + y^2 = 100\)
Divide both sides by 100 to convert it into standard ellipse form:
\(\frac{4x^2}{100} + \frac{y^2}{100} = 1\)
\(\frac{x^2}{25} + \frac{y^2}{100} = 1\)
Here, \(a^2 = 25\) and \(b^2 = 100\). Since \(b > a\), this is a vertical ellipse where \(a^2 = b^2(1 - e^2)\).
\(25 = 100(1 - e^2)\)
\(\frac{25}{100} = 1 - e^2\)
\(\frac{1}{4} = 1 - e^2\)
\(e^2 = 1 - \frac{1}{4} = \frac{3}{4}\)
\(e = \frac{\sqrt{3}}{2}\)
In simple words: Convert the equation to standard ellipse form, identify the semi-major and semi-minor axes, and use the eccentricity formula \(a^2 = b^2(1 - e^2)\).

Exam Tip: Always check which axis is larger (\(a\) or \(b\)) before applying the eccentricity formula for an ellipse.

Q8. What is the value of \(\cot\left(-\frac{15\pi}{4}\right)\)?
Answer: We know that \(\cot(-\theta) = -\cot\theta\).
\(\cot\left(-\frac{15\pi}{4}\right) = -\cot\left(\frac{15\pi}{4}\right)\)
Simplify the angle by dividing by \(2\pi\):
\(\frac{15\pi}{4} = 4\pi - \frac{\pi}{4}\)
Since \(\cot(2n\pi + \theta) = \cot\theta\) and \(\cot(4\pi - \theta) = -\cot\theta\):
\(\cot\left(4\pi - \frac{\pi}{4}\right) = -\cot\left(\frac{\pi}{4}\right) = -1\)
Therefore, \(-\cot\left(\frac{15\pi}{4}\right) = -(-1) = 1\).
In simple words: Use periodic properties of trigonometric functions to reduce large angles into the first rotation.

Exam Tip: Use periodicity of \(2\pi\) to bring any large angle into the range \([0, 2\pi]\).

Q9. Find \(\sin\frac{x}{2}, \cos\frac{x}{2}\) and \(\tan\frac{x}{2}\) if \(\tan x = -\frac{4}{3}\), \(x\) in quadrant II.
Answer: Since \(x\) lies in quadrant II, \(\frac{x}{2}\) lies in quadrant I (since \(\frac{\pi}{2} < x < \pi \implies \frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2}\)).
Thus, all trigonometric functions of \(\frac{x}{2}\) will be positive.
Given \(\tan x = -\frac{4}{3}\), we find \(\sec x\):
\(\sec^2 x = 1 + \tan^2 x = 1 + \left(-\frac{4}{3}\right)^2 = 1 + \frac{16}{9} = \frac{25}{9}\)
Since \(x\) is in quadrant II, \(\sec x = -\frac{5}{3}\), so \(\cos x = -\frac{3}{5}\).
Now, using half-angle formulas:
\(\sin^2\frac{x}{2} = \frac{1 - \cos x}{2} = \frac{1 - (-3/5)}{2} = \frac{8/5}{2} = \frac{4}{5}\)
Since \(\frac{x}{2}\) is in quadrant I, \(\sin\frac{x}{2} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5}\).
\(\cos^2\frac{x}{2} = \frac{1 + \cos x}{2} = \frac{1 + (-3/5)}{2} = \frac{2/5}{2} = \frac{1}{5}\)
Since \(\frac{x}{2}\) is in quadrant I, \(\cos\frac{x}{2} = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5}\).
\(\tan\frac{x}{2} = \frac{\sin(x/2)}{\cos(x/2)} = \frac{2/\sqrt{5}}{1/\sqrt{5}} = 2\).
In simple words: Determine the quadrant for \(\frac{x}{2}\), find \(\cos x\) from \(\tan x\), and use half-angle formulas.

Exam Tip: Always determine the quadrant of the half-angle \(\frac{x}{2}\) first to assign correct positive or negative signs.

Q10. Find the square root of \(-7 - 24i\)
Answer: Let \(\sqrt{-7 - 24i} = x + iy\).
Squaring both sides:
\(-7 - 24i = (x + iy)^2 = x^2 - y^2 + 2xyi\)
Equating real and imaginary parts:
\(x^2 - y^2 = -7\) ---(1)
\(2xy = -24 \implies xy = -12\) ---(2)
We know that \((x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2\)
\((x^2 + y^2)^2 = (-7)^2 + (-24)^2 = 49 + 576 = 625\)
\(x^2 + y^2 = 25\) ---(3) (since \(x^2 + y^2\) cannot be negative)
Adding (1) and (3):
\(2x^2 = 18 \implies x^2 = 9 \implies x = \pm 3\)
Subtracting (1) from (3):
\(2y^2 = 32 \implies y^2 = 16 \implies y = \pm 4\)
From (2), \(xy\) is negative, so \(x\) and \(y\) have opposite signs.
Thus, the square roots are \(3 - 4i\) and \(-3 + 4i\), or \(\pm(3 - 4i)\).
In simple words: Equate real and imaginary parts after squaring, use the modulus identity to find \(x^2 + y^2\), and solve simultaneous equations.

Exam Tip: Remember to check the sign of the product \(xy\) from the imaginary part to assign correct signs to \(x\) and \(y\).

Q11. Prove that \(\cos^2 x + \cos^2\left(x + \frac{\pi}{3}\right) + \cos^2\left(x - \frac{\pi}{3}\right) = \frac{3}{2}\)
Answer: We know the identity \(2\cos^2\theta = 1 + \cos 2\theta\), which gives \(\cos^2\theta = \frac{1 + \cos 2\theta}{2}\).
Applying this to each term on the LHS:
LHS = \(\frac{1 + \cos 2x}{2} + \frac{1 + \cos\left(2x + \frac{2\pi}{3}\right)}{2} + \frac{1 + \cos\left(2x - \frac{2\pi}{3}\right)}{2}\)
\(= \frac{1}{2} \left[ 3 + \cos 2x + \cos\left(2x + \frac{2\pi}{3}\right) + \cos\left(2x - \frac{2\pi}{3}\right) \right]\)
Using the sum formula \(\cos(A + B) + \cos(A - B) = 2\cos A \cos B\):
\(\cos\left(2x + \frac{2\pi}{3}\right) + \cos\left(2x - \frac{2\pi}{3}\right) = 2 \cos(2x) \cos\left(\frac{2\pi}{3}\right)\)
Since \(\cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}\):
\(= 2 \cos(2x) \left(-\frac{1}{2}\right) = -\cos 2x\)
Substitute this back into the expression:
LHS = \(\frac{1}{2} [3 + \cos 2x - \cos 2x] = \frac{3}{2} = \text{RHS}\).
Hence proved.
In simple words: Convert squared cosine terms into double-angle cosines and use the sum-to-product formula to cancel terms.

Exam Tip: Use the half-angle formula for cosine squared to linearize powers in trigonometric proofs.

Q12. Show that \((A \cup B)' = A' \cap B'\) and \((A \cap B)' = A' \cup B'\)
Where \(U = \{1, 2, 3, 4, 5, 6, 7, 8\}\), \(A = \{2, 3, 4\}\) and \(B = \{3, 4, 5\}\)
Answer: Given universal set \(U = \{1, 2, 3, 4, 5, 6, 7, 8\}\), \(A = \{2, 3, 4\}\), \(B = \{3, 4, 5\}\).
Part 1: Show \((A \cup B)' = A' \cap B'\)
\(A \cup B = \{2, 3, 4\} \cup \{3, 4, 5\} = \{2, 3, 4, 5\}\)
LHS \(= (A \cup B)' = U - (A \cup B) = \{1, 2, 3, 4, 5, 6, 7, 8\} - \{2, 3, 4, 5\} = \{1, 6, 7, 8\}\)
Now find complements of \(A\) and \(B\):
\(A' = U - A = \{1, 5, 6, 7, 8\}\)
\(B' = U - B = \{1, 2, 6, 7, 8\}\)
RHS \(= A' \cap B' = \{1, 5, 6, 7, 8\} \cap \{1, 2, 6, 7, 8\} = \{1, 6, 7, 8\}\)
Since LHS = RHS, \((A \cup B)' = A' \cap B'\) is verified.
Part 2: Show \((A \cap B)' = A' \cup B'\)
\(A \cap B = \{2, 3, 4\} \cap \{3, 4, 5\} = \{3, 4\}\)
LHS \(= (A \cap B)' = U - (A \cap B) = \{1, 2, 3, 4, 5, 6, 7, 8\} - \{3, 4\} = \{1, 2, 5, 6, 7, 8\}\)
RHS \(= A' \cup B' = \{1, 5, 6, 7, 8\} \cup \{1, 2, 6, 7, 8\} = \{1, 2, 5, 6, 7, 8\}\)
Since LHS = RHS, \((A \cap B)' = A' \cup B'\) is verified (De Morgan's Laws).
In simple words: Compute unions, intersections, and set differences step-by-step using the given universal set to verify De Morgan's laws.

Exam Tip: These are De Morgan's Laws of set theory; ensure proper subtraction from the universal set \(U\) for complements.