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Punjab Board Class 11 Mathematics

Exercise 10.4 Solutions — Trigonometric Identities

Complete worked solutions for Exercise 10.4 from Chapter 10 (Trigonometric Identities), Punjab Board Class 11 Mathematics. Each question shows every step and names the formula used.

Mathematics 11 (Punjab Board)

Exercise 10.4 Solutions

Worked solutions for Punjab Board Class 11 Mathematics, Chapter 10: Trigonometric Identities.

Question 1: Express the following products as sums or differences: (i) \(2\sin 3\theta\cos\theta\), (ii) \(2\cos 5\theta\sin 3\theta\), (iii) \(\sin 5\theta\cos 2\theta\), (iv) \(2\sin 7\theta\sin 2\theta\), (v) \(\cos(x+y)\sin(x-y)\)

Product-to-sum formulas lagayenge: \(2\sin A\cos B = \sin(A+B)+\sin(A-B)\), \(2\cos A\sin B = \sin(A+B)-\sin(A-B)\), \(-2\sin A\sin B = \cos(A+B)-\cos(A-B)\).

(i) \(2\sin 3\theta\cos\theta = \sin(3\theta+\theta) + \sin(3\theta-\theta) = \sin 4\theta + \sin 2\theta\)

(ii) \(2\cos 5\theta\sin 3\theta = \sin(5\theta+3\theta) - \sin(5\theta-3\theta) = \sin 8\theta - \sin 2\theta\)

(iii) \(\sin 5\theta\cos 2\theta = \frac{1}{2}\big[\sin(5\theta+2\theta) + \sin(5\theta-2\theta)\big] = \frac{1}{2}(\sin 7\theta + \sin 3\theta)\)

(iv) \(2\sin 7\theta\sin 2\theta = \cos(7\theta-2\theta) - \cos(7\theta+2\theta) = \cos 5\theta - \cos 9\theta\)

(v) \(\cos(x+y)\sin(x-y) = \frac{1}{2}\big[2\sin(x-y)\cos(x+y)\big]\)

\[ = \frac{1}{2}\big[\sin(x-y+x+y) + \sin(x-y-x-y)\big] = \frac{1}{2}\big[\sin 2x - \sin 2y\big] \]
Answer: (i) \(\sin 4\theta + \sin 2\theta\), (ii) \(\sin 8\theta - \sin 2\theta\), (iii) \(\frac{1}{2}(\sin 7\theta + \sin 3\theta)\), (iv) \(\cos 5\theta - \cos 9\theta\), (v) \(\frac{1}{2}(\sin 2x - \sin 2y)\).

Question 2: Express the following sums or differences as products: (i) \(\sin 5\theta + \sin 3\theta\), (ii) \(\sin 8\theta - \sin 4\theta\), (iii) \(\cos 6\theta + \cos 3\theta\), (iv) \(\cos 7\theta - \cos\theta\), (v) \(\cos 12^\circ + \cos 48^\circ\)

Sum-to-product formulas lagayenge. Yaad rakhen ke \(\cos P - \cos Q\) mein minus sign aata hai.

(i) \(\sin 5\theta + \sin 3\theta = 2\sin\left(\frac{5\theta+3\theta}{2}\right)\cos\left(\frac{5\theta-3\theta}{2}\right) = 2\sin 4\theta\cos\theta\)

(ii) \(\sin 8\theta - \sin 4\theta = 2\cos\left(\frac{8\theta+4\theta}{2}\right)\sin\left(\frac{8\theta-4\theta}{2}\right) = 2\cos 6\theta\sin 2\theta\)

(iii) \(\cos 6\theta + \cos 3\theta = 2\cos\left(\frac{9\theta}{2}\right)\cos\left(\frac{3\theta}{2}\right)\)

(iv) \(\cos 7\theta - \cos\theta = -2\sin\left(\frac{8\theta}{2}\right)\sin\left(\frac{6\theta}{2}\right) = -2\sin 4\theta\sin 3\theta\)

(v) \(\cos 12^\circ + \cos 48^\circ = 2\cos 30^\circ\cos(-18^\circ) = 2\cdot\frac{\sqrt{3}}{2}\cdot\cos 18^\circ = \sqrt{3}\cos 18^\circ\)

Answer: (i) \(2\sin 4\theta\cos\theta\), (ii) \(2\cos 6\theta\sin 2\theta\), (iii) \(2\cos\frac{9\theta}{2}\cos\frac{3\theta}{2}\), (iv) \(-2\sin 4\theta\sin 3\theta\), (v) \(\sqrt{3}\cos 18^\circ\).

Question 3: Prove the following identities: (i) \(\frac{\sin 3x - \sin x}{\cos x - \cos 3x} = \cot 2x\), (ii) \(\frac{\sin 8x + \sin 2x}{\cos 8x + \cos 2x} = \tan 5x\), (iii) \(\frac{\sin A - \sin B}{\sin A + \sin B} = \cot\frac{A+B}{2}\tan\frac{A-B}{2}\), (iv) \(\frac{\sin 80^\circ + \sin 40^\circ}{\cos 80^\circ + \cos 40^\circ} = \sqrt{3}\)

Numerator aur denominator par sum-to-product formulas lagayenge, phir common factors cancel karenge.

(i)

\[ \text{L.H.S} = \frac{2\cos 2x\sin x}{-2\sin 2x(-\sin x)} = \frac{2\cos 2x\sin x}{2\sin 2x\sin x} = \frac{\cos 2x}{\sin 2x} = \cot 2x = \text{R.H.S} \]

(ii)

\[ \text{L.H.S} = \frac{2\sin 5x\cos 3x}{2\cos 5x\cos 3x} = \frac{\sin 5x}{\cos 5x} = \tan 5x = \text{R.H.S} \]

(iii)

\[ \text{L.H.S} = \frac{2\cos\frac{A+B}{2}\sin\frac{A-B}{2}}{2\sin\frac{A+B}{2}\cos\frac{A-B}{2}} = \cot\frac{A+B}{2}\tan\frac{A-B}{2} = \text{R.H.S} \]

(iv)

\[ \text{L.H.S} = \frac{2\sin 60^\circ\cos 20^\circ}{2\cos 60^\circ\cos 20^\circ} = \frac{\sin 60^\circ}{\cos 60^\circ} = \tan 60^\circ = \sqrt{3} = \text{R.H.S} \]
Answer: All four identities are proved.

Question 4: (i) Find the value of \(\cos 15^\circ + \cos 105^\circ + \cos 195^\circ + \cos 160^\circ + \cos 285^\circ\). Prove that: (ii) \(\frac{\sin 2\theta + \sin 4\theta + \sin 6\theta + \sin 8\theta}{\cos 2\theta + \cos 4\theta + \cos 6\theta + \cos 8\theta} = \tan 5\theta\), (iii) \(\cos^2\left(\frac{\pi}{4}-\frac{\alpha}{2}\right) - \cos^2\left(\frac{\pi}{4}+\frac{\alpha}{2}\right) = \sin\alpha\), (iv) \(\sin\left(\frac{\pi}{4}-\theta\right)\sin\left(\frac{\pi}{4}+\theta\right) = \frac{1}{2}\cos 2\theta\), (v) \(\frac{\sin\theta + \sin 3\theta + \sin 5\theta + \sin 7\theta}{\cos\theta + \cos 3\theta + \cos 5\theta + \cos 7\theta} = \tan 4\theta\)

Terms ko aise pair karenge ke sum-to-product formulas lag sakein. Part (i) mein do alag pairings se bhi same jawab aata hai — yehi cross-check hai.

(i) Grouping \((\cos 15^\circ + \cos 285^\circ) + (\cos 105^\circ + \cos 195^\circ) + \cos 160^\circ\):

\[ = 2\cos 150^\circ\cos(-135^\circ) + 2\cos 150^\circ\cos(-45^\circ) + \cos 160^\circ \]
\[ = 2\left(-\frac{\sqrt{3}}{2}\right)\left(-\frac{1}{\sqrt{2}}\right) + 2\left(-\frac{\sqrt{3}}{2}\right)\left(\frac{1}{\sqrt{2}}\right) + \cos 160^\circ = \frac{\sqrt{3}}{\sqrt{2}} - \frac{\sqrt{3}}{\sqrt{2}} + \cos 160^\circ = \cos 160^\circ \]

(ii) Grouping \((\sin 2\theta + \sin 8\theta) + (\sin 4\theta + \sin 6\theta)\):

\[ \text{L.H.S} = \frac{2\sin 5\theta\cos 3\theta + 2\sin 5\theta\cos\theta}{2\cos 5\theta\cos 3\theta + 2\cos 5\theta\cos\theta} = \frac{2\sin 5\theta(\cos 3\theta + \cos\theta)}{2\cos 5\theta(\cos 3\theta + \cos\theta)} = \tan 5\theta = \text{R.H.S} \]

(iii) Using \(a^2-b^2 = (a-b)(a+b)\):

\[ \text{L.H.S} = \left[\cos\left(\frac{\pi}{4}-\frac{\alpha}{2}\right) - \cos\left(\frac{\pi}{4}+\frac{\alpha}{2}\right)\right]\left[\cos\left(\frac{\pi}{4}-\frac{\alpha}{2}\right) + \cos\left(\frac{\pi}{4}+\frac{\alpha}{2}\right)\right] \]
\[ = \left[-2\sin\frac{\pi}{4}\sin\left(-\frac{\alpha}{2}\right)\right]\left[2\cos\frac{\pi}{4}\cos\frac{\alpha}{2}\right] = \left[\sqrt{2}\sin\frac{\alpha}{2}\right]\left[\sqrt{2}\cos\frac{\alpha}{2}\right] = \sin\alpha = \text{R.H.S} \]

(iv) Using \(2\sin A\sin B = \cos(A-B) - \cos(A+B)\):

\[ \text{L.H.S} = \frac{1}{2}\left[\cos(-2\theta) - \cos\frac{\pi}{2}\right] = \frac{1}{2}\cos 2\theta = \text{R.H.S} \]

(v) Grouping \((\sin\theta + \sin 7\theta) + (\sin 3\theta + \sin 5\theta)\):

\[ \text{L.H.S} = \frac{2\sin 4\theta(\cos 3\theta + \cos\theta)}{2\cos 4\theta(\cos 3\theta + \cos\theta)} = \tan 4\theta = \text{R.H.S} \]

Common mistake: Part (i) mein \(\cos 160^\circ\) ka term akela reh jata hai — is ka koi partner nahi banta. (Note: as written the sum equals \(\cos 160^\circ\), not 0; the printed RHS of 0 would need \(\cos 165^\circ\) in place of \(\cos 160^\circ\).)

Answer: (i) \(= \cos 160^\circ\); (ii)–(v) proved.

Question 5: Prove that: (i) \(\cos 20^\circ\cos 40^\circ\cos 60^\circ\cos 80^\circ = \frac{1}{16}\), (ii) \(\sin\frac{\pi}{9}\sin\frac{2\pi}{9}\sin\frac{3\pi}{9}\sin\frac{4\pi}{9} = \frac{3}{16}\), (iii) \(\sin 10^\circ\sin 30^\circ\sin 50^\circ\sin 70^\circ = \frac{1}{16}\)

Product mein \(2\sin A\) se multiply aur divide karenge taake \(\sin 2A\) banta jaye — ye "telescoping" trick in sawalon ki jaan hai.

(i) Since \(\cos 60^\circ = \frac{1}{2}\), multiply and divide by \(2\sin 20^\circ\):

\[ \text{L.H.S} = \frac{1}{2}\cdot\frac{2\sin 20^\circ\cos 20^\circ\cos 40^\circ\cos 80^\circ}{2\sin 20^\circ} = \frac{1}{2}\cdot\frac{\sin 40^\circ\cos 40^\circ\cos 80^\circ}{2\sin 20^\circ} \]
\[ = \frac{1}{2}\cdot\frac{\sin 80^\circ\cos 80^\circ}{4\sin 20^\circ} = \frac{1}{2}\cdot\frac{\sin 160^\circ}{8\sin 20^\circ} = \frac{1}{2}\cdot\frac{\sin 20^\circ}{8\sin 20^\circ} = \frac{1}{16} = \text{R.H.S} \]

(ii) \(\sin\frac{\pi}{9}\sin\frac{2\pi}{9}\sin\frac{3\pi}{9}\sin\frac{4\pi}{9} = \sin 20^\circ\sin 40^\circ\sin 60^\circ\sin 80^\circ = \frac{\sqrt{3}}{2}\sin 20^\circ\sin 40^\circ\sin 80^\circ\).

Now \(4\sin 20^\circ\sin 40^\circ\sin 80^\circ = 2(\cos 20^\circ - \cos 60^\circ)\sin 80^\circ = 2\cos 20^\circ\sin 80^\circ - \sin 80^\circ\):

\[ = (\sin 100^\circ + \sin 60^\circ) - \sin 80^\circ = \sin 80^\circ + \frac{\sqrt{3}}{2} - \sin 80^\circ = \frac{\sqrt{3}}{2} \]
\[ \text{L.H.S} = \frac{\sqrt{3}}{2}\cdot\frac{\sqrt{3}}{8} = \frac{3}{16} = \text{R.H.S} \]

(iii) \(\sin 10^\circ\sin 30^\circ\sin 50^\circ\sin 70^\circ = \frac{1}{2}\sin 10^\circ\sin 50^\circ\sin 70^\circ\).

Using \(\sin 50^\circ\sin 70^\circ = \frac{1}{2}\big[\cos(-20^\circ) - \cos 120^\circ\big] = \frac{1}{2}\left(\cos 20^\circ + \frac{1}{2}\right)\):

\[ \text{L.H.S} = \frac{1}{4}\sin 10^\circ\cos 20^\circ + \frac{1}{8}\sin 10^\circ = \frac{1}{8}\big[\sin 30^\circ - \sin 10^\circ\big] + \frac{1}{8}\sin 10^\circ \]
\[ = \frac{1}{8}\cdot\frac{1}{2} = \frac{1}{16} = \text{R.H.S} \]
Answer: All three products are proved.

Question 6: Prove that \(\frac{\sin 3\theta}{1 + 2\cos 2\theta} = \sin\theta\); deduce the value of \(\sin 18^\circ\)

\(\sin 3\theta\) ko expand kar ke numerator mein \(\sin\theta\) common lenge — denominator wali expression cancel ho jayegi. Phir \(\theta = 18^\circ\) rakh kar \(\sin 18^\circ\) nikalenge.

\[ \text{L.H.S} = \frac{3\sin\theta - 4\sin^3\theta}{1 + 2(1-2\sin^2\theta)} = \frac{\sin\theta(3-4\sin^2\theta)}{3-4\sin^2\theta} = \sin\theta = \text{R.H.S} \]

Putting \(\theta = 18^\circ\): \(\sin 54^\circ = \sin 18^\circ(1 + 2\cos 36^\circ)\). Since \(\sin 54^\circ = \cos 36^\circ = \frac{\sqrt{5}+1}{4}\):

\[ \sin 18^\circ = \frac{\frac{\sqrt{5}+1}{4}}{1 + \frac{\sqrt{5}+1}{2}} = \frac{\sqrt{5}+1}{2(3+\sqrt{5})} = \frac{(\sqrt{5}+1)(3-\sqrt{5})}{8} = \frac{\sqrt{5}-1}{4} \]
Answer: The identity is proved, and \(\sin 18^\circ = \frac{\sqrt{5}-1}{4}\).

Question 7: Prove that \(\tan 75^\circ - \tan 15^\circ = 2\sqrt{3}\)

Dono ki exact values (jo Exercise 10.2 mein nikali thin) substitute karenge.

\[ \text{L.H.S} = (2+\sqrt{3}) - (2-\sqrt{3}) = 2\sqrt{3} = \text{R.H.S} \]
Answer: The identity is proved.

Question 8: Prove that \(\cos 15^\circ - \sin 15^\circ = \frac{1}{\sqrt{2}}\)

Dono taraf square karenge — left side mein \(\sin 30^\circ\) ban jayega. Aakhir mein sign check karenge (cos15° > sin15°).

\[ (\cos 15^\circ - \sin 15^\circ)^2 = \cos^2 15^\circ + \sin^2 15^\circ - 2\sin 15^\circ\cos 15^\circ = 1 - \sin 30^\circ = 1 - \frac{1}{2} = \frac{1}{2} \]

Since \(\cos 15^\circ > \sin 15^\circ\), taking the positive square root: \(\cos 15^\circ - \sin 15^\circ = \frac{1}{\sqrt{2}}\). Proved.

Answer: The identity is proved.

Question 9: Prove that \(\frac{\sin^2\alpha - \sin^2\beta}{\sin\alpha\cos\alpha - \sin\beta\cos\beta} = \tan(\alpha+\beta)\)

Numerator mein \(\sin^2\alpha - \sin^2\beta = \sin(\alpha+\beta)\sin(\alpha-\beta)\) aur denominator mein \(\sin 2\alpha - \sin 2\beta\) ki sum-to-product form lagayenge.

\[ \text{L.H.S} = \frac{\sin(\alpha+\beta)\sin(\alpha-\beta)}{\frac{1}{2}(\sin 2\alpha - \sin 2\beta)} = \frac{\sin(\alpha+\beta)\sin(\alpha-\beta)}{\cos(\alpha+\beta)\sin(\alpha-\beta)} = \tan(\alpha+\beta) = \text{R.H.S} \]
Answer: The identity is proved.

Question 10: Prove that \(\sin\alpha + \sin\beta + \sin\gamma - \sin(\alpha+\beta+\gamma) = 4\sin\frac{\beta+\gamma}{2}\sin\frac{\gamma+\alpha}{2}\sin\frac{\alpha+\beta}{2}\)

Pehle \(\sin\alpha + \sin\beta\) ko product mein badlenge, phir \(\sin\gamma - \sin(\alpha+\beta+\gamma)\) ko. Dono mein \(\sin\frac{\alpha+\beta}{2}\) common niklega.

\[ \text{L.H.S} = 2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2} - 2\cos\left(\gamma + \frac{\alpha+\beta}{2}\right)\sin\frac{\alpha+\beta}{2} \]
\[ = 2\sin\frac{\alpha+\beta}{2}\left[\cos\frac{\alpha-\beta}{2} - \cos\left(\gamma + \frac{\alpha+\beta}{2}\right)\right] \]
\[ = 2\sin\frac{\alpha+\beta}{2}\left[-2\sin\frac{\alpha+\gamma}{2}\sin\frac{-\beta-\gamma}{2}\right] = 4\sin\frac{\alpha+\beta}{2}\sin\frac{\beta+\gamma}{2}\sin\frac{\gamma+\alpha}{2} = \text{R.H.S} \]
Answer: The identity is proved.