NCERT Solutions Class 11 Maths Chapter 3 Exercise 3.3

⚡ Master Trigonometric Identities

This exercise is the heart of trigonometry! You will need to master the Sum, Difference, and Multiple Angle formulas.

Here is your Cheat Sheet before we begin:

🔑 Key Formulas to Remember:

• Sum & Difference:

- \(\sin(x \pm y) = \sin x \cos y \pm \cos x \sin y\)

- \(\cos(x \pm y) = \cos x \cos y \mp \sin x \sin y\) (Sign flips!)

• CD Formulas (Crucial for Q10-Q21):

- \(\sin C + \sin D = 2 \sin \frac{C+D}{2} \cos \frac{C-D}{2}\)

- \(\cos C - \cos D = -2 \sin \frac{C+D}{2} \sin \frac{C-D}{2}\) (Watch the negative sign!)

• Double Angle (\(\cos 2x\)):

- \(\cos^2 x - \sin^2 x\)

- \(2\cos^2 x - 1\) (Use to eliminate -1)

- \(1 - 2\sin^2 x\) (Use to eliminate +1)

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Detailed Solutions

🔴 Question 1: Prove that \(\sin^2 \frac{\pi}{6} + \cos^2 \frac{\pi}{3} - \tan^2 \frac{\pi}{4} = -\frac{1}{2}\).

🟢 Solution:

Substitute the standard values directly:

- \(\sin \frac{\pi}{6} = \frac{1}{2}\)

- \(\cos \frac{\pi}{3} = \frac{1}{2}\)

- \(\tan \frac{\pi}{4} = 1\)

LHS:

\[ \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 - (1)^2 \]

\[ = \frac{1}{4} + \frac{1}{4} - 1 \]

\[ = \frac{2}{4} - 1 = \frac{1}{2} - 1 = -\frac{1}{2} \]

LHS = RHS. Proved. ✅

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🔴 Question 5: Find the value of: (i) \(\sin 75^{\circ}\) (ii) \(\tan 15^{\circ}\).

🟢 Solution:

(i) \(\sin 75^{\circ}\)

We can split \(75^{\circ}\) into two standard angles: \(45^{\circ} + 30^{\circ}\).

Use formula: \(\sin(x+y) = \sin x \cos y + \cos x \sin y\).

\[ \sin(45^{\circ} + 30^{\circ}) = \sin 45^{\circ} \cos 30^{\circ} + \cos 45^{\circ} \sin 30^{\circ} \]

\[ = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} \]

\[ = \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}} = \frac{\sqrt{3}+1}{2\sqrt{2}} \]

(ii) \(\tan 15^{\circ}\)

Split \(15^{\circ}\) into \(45^{\circ} - 30^{\circ}\).

Use formula: \(\tan(x-y) = \frac{\tan x - \tan y}{1 + \tan x \tan y}\).

\[ \tan(45^{\circ} - 30^{\circ}) = \frac{\tan 45^{\circ} - \tan 30^{\circ}}{1 + \tan 45^{\circ} \tan 30^{\circ}} \]

\[ = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \cdot \frac{1}{\sqrt{3}}} = \frac{\frac{\sqrt{3}-1}{\sqrt{3}}}{\frac{\sqrt{3}+1}{\sqrt{3}}} \]

\[ = \frac{\sqrt{3}-1}{\sqrt{3}+1} \]

(Rationalize the denominator by multiplying \(\sqrt{3}-1\))

\[ = \frac{(\sqrt{3}-1)^2}{3-1} = \frac{3 + 1 - 2\sqrt{3}}{2} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3} \]

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🔴 Question 6: Prove that \(\cos(\frac{\pi}{4}-x)\cos(\frac{\pi}{4}-y) - \sin(\frac{\pi}{4}-x)\sin(\frac{\pi}{4}-y) = \sin(x+y)\).

🟢 Solution:

Observe the pattern on LHS. It looks like \(\cos A \cos B - \sin A \sin B\).

We know: \(\cos(A+B) = \cos A \cos B - \sin A \sin B\).

Let \(A = \frac{\pi}{4}-x\) and \(B = \frac{\pi}{4}-y\).

LHS:

\[ = \cos \left[ (\frac{\pi}{4}-x) + (\frac{\pi}{4}-y) \right] \]

\[ = \cos \left[ \frac{\pi}{2} - (x+y) \right] \]

Using identity \(\cos(90^{\circ} - \theta) = \sin \theta\):

\[ = \sin(x+y) \]

LHS = RHS. Proved. ✅

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🔴 Question 7: Prove that \(\frac{\tan(\frac{\pi}{4}+x)}{\tan(\frac{\pi}{4}-x)} = \left(\frac{1+\tan x}{1-\tan x}\right)^2\).

🟢 Solution:

Expand numerator and denominator using \(\tan(A \pm B)\) formulas.

Numerator: \(\tan(\frac{\pi}{4}+x) = \frac{1+\tan x}{1-\tan x}\)

Denominator: \(\tan(\frac{\pi}{4}-x) = \frac{1-\tan x}{1+\tan x}\)

LHS:

\[ \frac{\frac{1+\tan x}{1-\tan x}}{\frac{1-\tan x}{1+\tan x}} = \frac{1+\tan x}{1-\tan x} \times \frac{1+\tan x}{1-\tan x} \]

\[ = \left( \frac{1+\tan x}{1-\tan x} \right)^2 \]

LHS = RHS. Proved. ✅

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🔴 Question 10: Prove that \(\sin(n+1)x \sin(n+2)x + \cos(n+1)x \cos(n+2)x = \cos x\).

🟢 Solution:

Rearrange LHS to look like standard formula: \(\cos A \cos B + \sin A \sin B\).

This matches the formula for \(\cos(A - B)\).

Let \(A = (n+2)x\) and \(B = (n+1)x\).

LHS:

\[ = \cos [ (n+2)x - (n+1)x ] \]

\[ = \cos [ nx + 2x - nx - x ] \]

\[ = \cos x \]

LHS = RHS. Proved. ✅

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🔴 Question 12: Prove that \(\sin^2 6x - \sin^2 4x = \sin 2x \sin 10x\).

🟢 Solution:

Use algebraic identity: \(a^2 - b^2 = (a-b)(a+b)\).

LHS:

\[ (\sin 6x - \sin 4x)(\sin 6x + \sin 4x) \]

Now apply CD formulas:

1. \(\sin C - \sin D = 2 \cos \frac{C+D}{2} \sin \frac{C-D}{2}\)

2. \(\sin C + \sin D = 2 \sin \frac{C+D}{2} \cos \frac{C-D}{2}\)

\[ = \left( 2 \cos 5x \sin x \right) \left( 2 \sin 5x \cos x \right) \]

Rearrange the terms to group angles:

\[ = (2 \sin x \cos x) (2 \sin 5x \cos 5x) \]

Use \(\sin 2\theta = 2 \sin \theta \cos \theta\):

\[ = \sin 2x \sin 10x \]

LHS = RHS. Proved. ✅

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🔴 Question 22: Prove \(\cot x \cot 2x - \cot 2x \cot 3x - \cot 3x \cot x = 1\).

💡 Expert Trick: Don't start from LHS! Start with the relation of angles.

🟢 Solution:

We know that \(3x = 2x + x\).

Apply \(\cot\) on both sides:

\[ \cot 3x = \cot(2x + x) \]

Use formula \(\cot(A+B) = \frac{\cot A \cot B - 1}{\cot B + \cot A}\):

\[ \cot 3x = \frac{\cot 2x \cot x - 1}{\cot 2x + \cot x} \]

Cross multiply:

\[ \cot 3x (\cot 2x + \cot x) = \cot 2x \cot x - 1 \]

\[ \cot 3x \cot 2x + \cot 3x \cot x = \cot 2x \cot x - 1 \]

Rearrange terms to match the question:

\[ 1 = \cot 2x \cot x - \cot 3x \cot 2x - \cot 3x \cot x \]

Proved. ✅

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🔴 Question 25: Prove that \(\cos 6x = 32 \cos^6 x - 48 \cos^4 x + 18 \cos^2 x - 1\).

🟢 Solution:

Write \(6x\) as \(2(3x)\). Use double angle formula first.

LHS: \(\cos 6x = \cos 2(3x) = 2 \cos^2 (3x) - 1\)

Now use triple angle formula: \(\cos 3x = 4 \cos^3 x - 3 \cos x\).

\[ = 2 (4 \cos^3 x - 3 \cos x)^2 - 1 \]

Expand using \((a-b)^2 = a^2 + b^2 - 2ab\):

\[ = 2 [ 16 \cos^6 x + 9 \cos^2 x - 24 \cos^4 x ] - 1 \]

\[ = 32 \cos^6 x + 18 \cos^2 x - 48 \cos^4 x - 1 \]

\[ = 32 \cos^6 x - 48 \cos^4 x + 18 \cos^2 x - 1 \]

LHS = RHS. Proved. ✅

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🧐 Common Mistakes to Avoid:

⚠️ The "Cos" Trap: Remember that \(\cos(A+B)\) has a MINUS sign in the middle. Students often put a plus sign by

mistake.

⚠️ Signs in CD Formulas: In \(\cos C - \cos D\), there is a negative sign at the front: \(-2 \sin... \sin...\). Don't forget

it!

⚠️ Opening Brackets: In Q25, when expanding \((a-b)^2\), remember to multiply the outside '2' with EVERY term inside.