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Punjab Board Class 11 Mathematics

Exercise 10.3 Solutions — Trigonometric Identities

Complete worked solutions for Exercise 10.3 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 10.3 Solutions

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

Question 1: Find the values of \(\sin 2\alpha\), \(\cos 2\alpha\) and \(\tan 2\alpha\), when: (i) \(\sin\alpha = \frac{3}{5}\), (ii) \(\cos\alpha = \frac{4}{5}\), where \(0 < \alpha < \frac{\pi}{2}\)

Pehle quadrant mein sab ratios positive hain. Pehle missing ratio (cos ya sin) Pythagorean identity se nikalenge, phir double angle formulas lagayenge.

(i) \(\sin\alpha = \frac{3}{5}\), so \(\cos\alpha = \sqrt{1-\frac{9}{25}} = \frac{4}{5}\).

\[ \sin 2\alpha = 2\sin\alpha\cos\alpha = 2\cdot\frac{3}{5}\cdot\frac{4}{5} = \frac{24}{25} \]
\[ \cos 2\alpha = 1 - 2\sin^2\alpha = 1 - 2\left(\frac{9}{25}\right) = \frac{7}{25}, \qquad \tan 2\alpha = \frac{24/25}{7/25} = \frac{24}{7} \]

(ii) \(\cos\alpha = \frac{4}{5}\), so \(\sin\alpha = \sqrt{1-\frac{16}{25}} = \frac{3}{5}\).

\[ \sin 2\alpha = 2\cdot\frac{3}{5}\cdot\frac{4}{5} = \frac{24}{25}, \qquad \cos 2\alpha = 2\cos^2\alpha - 1 = 2\left(\frac{16}{25}\right) - 1 = \frac{7}{25}, \qquad \tan 2\alpha = \frac{24}{7} \]
Answer: In both cases \(\sin 2\alpha = \frac{24}{25}\), \(\cos 2\alpha = \frac{7}{25}\), \(\tan 2\alpha = \frac{24}{7}\).

Question 2: Prove the following identities

Zyada complex side se start karenge aur double angle, half angle ya triple angle formulas laga kar simplify karenge.

(i) \(\cot\alpha - \tan\alpha = 2\cot 2\alpha\)

\[ \text{L.H.S} = \frac{\cos\alpha}{\sin\alpha} - \frac{\sin\alpha}{\cos\alpha} = \frac{\cos^2\alpha - \sin^2\alpha}{\sin\alpha\cos\alpha} = \frac{\cos 2\alpha}{\frac{1}{2}\sin 2\alpha} = 2\cot 2\alpha = \text{R.H.S} \]

(ii) \(\frac{\sin 2\alpha}{1+\cos 2\alpha} = \tan\alpha\)

\[ \text{L.H.S} = \frac{2\sin\alpha\cos\alpha}{1 + (2\cos^2\alpha - 1)} = \frac{2\sin\alpha\cos\alpha}{2\cos^2\alpha} = \frac{\sin\alpha}{\cos\alpha} = \tan\alpha = \text{R.H.S} \]

(iii) \(\frac{1-\cos\alpha}{\sin\alpha} = \tan\frac{\alpha}{2}\)

Using \(1-\cos\alpha = 2\sin^2\frac{\alpha}{2}\) and \(\sin\alpha = 2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}\):

\[ \text{L.H.S} = \frac{2\sin^2\frac{\alpha}{2}}{2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}} = \tan\frac{\alpha}{2} = \text{R.H.S} \]

(iv) \(\frac{\cos\alpha-\sin\alpha}{\cos\alpha+\sin\alpha} = \sec 2\alpha - \tan 2\alpha\)

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

(v) \(\sqrt{\frac{1+\sin\alpha}{1-\sin\alpha}} = \frac{\sin\frac{\alpha}{2}+\cos\frac{\alpha}{2}}{\sin\frac{\alpha}{2}-\cos\frac{\alpha}{2}}\)

Using \(1+\sin\alpha = \left(\sin\frac{\alpha}{2}+\cos\frac{\alpha}{2}\right)^2\) and \(1-\sin\alpha = \left(\sin\frac{\alpha}{2}-\cos\frac{\alpha}{2}\right)^2\):

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

(vi) \(\frac{\csc\theta + 2\csc 2\theta}{\sec\theta} = \cot\frac{\theta}{2}\)

\[ \text{L.H.S} = \cos\theta\left(\frac{1}{\sin\theta} + \frac{2}{\sin 2\theta}\right) = \cos\theta\left(\frac{1}{\sin\theta} + \frac{1}{\sin\theta\cos\theta}\right) = \frac{1+\cos\theta}{\sin\theta} \]
\[ = \frac{2\cos^2\frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}} = \cot\frac{\theta}{2} = \text{R.H.S} \]

(vii) \(1 + \tan\alpha\tan 2\alpha = \sec 2\alpha\)

\[ \text{L.H.S} = 1 + \tan\alpha\cdot\frac{2\tan\alpha}{1-\tan^2\alpha} = \frac{1+\tan^2\alpha}{1-\tan^2\alpha} = \frac{1}{\cos 2\alpha} = \sec 2\alpha = \text{R.H.S} \]

(viii) \(\frac{2\sin\theta\sin 2\theta}{\cos\theta + \cos 3\theta} = \tan 2\theta\tan\theta\)

Denominator: \(\cos\theta + \cos 3\theta = 2\cos 2\theta\cos\theta\).

\[ \text{L.H.S} = \frac{2\sin\theta(2\sin\theta\cos\theta)}{2\cos 2\theta\cos\theta} = \frac{2\sin^2\theta}{\cos 2\theta} \]
\[ \text{R.H.S} = \frac{\sin 2\theta}{\cos 2\theta}\cdot\frac{\sin\theta}{\cos\theta} = \frac{2\sin\theta\cos\theta\sin\theta}{\cos 2\theta\cos\theta} = \frac{2\sin^2\theta}{\cos 2\theta} = \text{L.H.S} \]

(ix) \(\frac{\sin 3\theta}{\sin\theta} - \frac{\cos 3\theta}{\cos\theta} = 2\)

\[ \text{L.H.S} = \frac{\sin 3\theta\cos\theta - \cos 3\theta\sin\theta}{\sin\theta\cos\theta} = \frac{\sin 2\theta}{\sin\theta\cos\theta} = \frac{2\sin\theta\cos\theta}{\sin\theta\cos\theta} = 2 = \text{R.H.S} \]

(x) \(\frac{\cos 3\theta}{\cos\theta} + \frac{\sin 3\theta}{\sin\theta} = 4\cos 2\theta\)

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

(xi) \(\frac{\tan\frac{\theta}{2} + \cot\frac{\theta}{2}}{\cot\frac{\theta}{2} - \tan\frac{\theta}{2}} = \sec\theta\)

\[ \text{L.H.S} = \frac{\frac{1}{\sin\frac{\theta}{2}\cos\frac{\theta}{2}}}{\frac{\cos^2\frac{\theta}{2}-\sin^2\frac{\theta}{2}}{\sin\frac{\theta}{2}\cos\frac{\theta}{2}}} = \frac{1}{\cos\theta} = \sec\theta = \text{R.H.S} \]

(xii) \(\frac{\sin 3\theta}{\cos\theta} + \frac{\cos 3\theta}{\sin\theta} = 2\cot 2\theta\)

\[ \text{L.H.S} = \frac{\sin 3\theta\sin\theta + \cos 3\theta\cos\theta}{\sin\theta\cos\theta} = \frac{\cos 2\theta}{\sin\theta\cos\theta} = \frac{2\cos 2\theta}{\sin 2\theta} = 2\cot 2\theta = \text{R.H.S} \]

(xiii) \(\frac{3+\cos 4\theta}{1-\cos 4\theta} = \frac{1}{2}(\tan^2\theta + \cot^2\theta)\)

\[ \text{R.H.S} = \frac{1}{2}\cdot\frac{\sin^4\theta+\cos^4\theta}{\sin^2\theta\cos^2\theta} = \frac{1}{2}\cdot\frac{1-2\sin^2\theta\cos^2\theta}{\sin^2\theta\cos^2\theta} \]

Since \(\sin^2\theta\cos^2\theta = \frac{1-\cos 4\theta}{8}\) and \(1-2\sin^2\theta\cos^2\theta = \frac{3+\cos 4\theta}{4}\):

\[ \text{R.H.S} = \frac{1}{2}\cdot\frac{(3+\cos 4\theta)/4}{(1-\cos 4\theta)/8} = \frac{3+\cos 4\theta}{1-\cos 4\theta} = \text{L.H.S} \]

(xiv) \(\frac{1+\sin 2\theta}{1-\sin 2\theta} = \tan^2\left(\frac{\pi}{4}+\theta\right)\)

Using \(1+\sin 2\theta = (\sin\theta+\cos\theta)^2\) and \(1-\sin 2\theta = (\sin\theta-\cos\theta)^2\):

\[ \text{L.H.S} = \left(\frac{\sin\theta+\cos\theta}{\sin\theta-\cos\theta}\right)^2, \qquad \text{R.H.S} = \left(\frac{1+\tan\theta}{1-\tan\theta}\right)^2 = \left(\frac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta}\right)^2 = \text{L.H.S} \]

(xv) \(\cos^2\frac{\pi}{8} + \cos^2\frac{3\pi}{8} + \cos^2\frac{5\pi}{8} + \cos^2\frac{7\pi}{8} = 2\)

Since \(\cos^2\frac{5\pi}{8} = \cos^2\frac{3\pi}{8}\) and \(\cos^2\frac{7\pi}{8} = \cos^2\frac{\pi}{8}\) (allied angles), and \(\cos^2\frac{3\pi}{8} = \sin^2\frac{\pi}{8}\):

\[ \text{L.H.S} = 2\cos^2\frac{\pi}{8} + 2\cos^2\frac{3\pi}{8} = 2\left(\cos^2\frac{\pi}{8} + \sin^2\frac{\pi}{8}\right) = 2 = \text{R.H.S} \]
Answer: All fifteen identities are proved.

Question 3: Show that \(2\cos\theta = \sqrt{2 + \sqrt{2 + 2\cos 4\theta}}\)

R.H.S se start karenge aur andar se bahar ki taraf \(1+\cos 2A = 2\cos^2 A\) ki identity lagayenge taake square roots khatam ho jayein.

\[ \text{R.H.S} = \sqrt{2 + \sqrt{2(1+\cos 4\theta)}} = \sqrt{2 + \sqrt{2(2\cos^2 2\theta)}} = \sqrt{2 + 2\cos 2\theta} \]
\[ = \sqrt{2(1+\cos 2\theta)} = \sqrt{2(2\cos^2\theta)} = \sqrt{4\cos^2\theta} = 2\cos\theta = \text{L.H.S} \]
Answer: The identity is proved.

Question 4: Reduce \(\sin^4\theta\) to an expression involving only functions of multiples of \(\theta\), raised to the first power

Power reducing formula \(\sin^2\theta = \frac{1-\cos 2\theta}{2}\) ko do martaba apply karenge.

\[ \sin^4\theta = (\sin^2\theta)^2 = \left(\frac{1-\cos 2\theta}{2}\right)^2 = \frac{1}{4}\big(1 - 2\cos 2\theta + \cos^2 2\theta\big) \]

Using \(\cos^2 2\theta = \frac{1+\cos 4\theta}{2}\):

\[ = \frac{1}{4}\left[1 - 2\cos 2\theta + \frac{1+\cos 4\theta}{2}\right] = \frac{3 - 4\cos 2\theta + \cos 4\theta}{8} \]
Answer: \(\sin^4\theta = \frac{3 - 4\cos 2\theta + \cos 4\theta}{8}\).

Question 5: Find the exact values of \(\sin\) and \(\cos\) of (i) \(18^\circ\), (ii) \(36^\circ\), (iii) \(54^\circ\), (iv) \(72^\circ\)

Pehle \(\sin 18^\circ\) nikalenge: \(18^\circ\) ko \(x\) rakhen, to \(2x + 3x = 90^\circ\) se quadratic equation banti hai. Phir baqi angles is se derive karenge (\(\sin 54^\circ = \cos 36^\circ\) waghera).

Let \(x = 18^\circ\). Then \(2x + 3x = 90^\circ\), so \(\sin 2x = \sin(90^\circ - 3x) = \cos 3x\).

\[ 2\sin x\cos x = 4\cos^3 x - 3\cos x \implies 2\sin x = 4(1-\sin^2 x) - 3 \]
\[ 4\sin^2 x + 2\sin x - 1 = 0 \implies \sin x = \frac{-2 + \sqrt{20}}{8} = \frac{\sqrt{5}-1}{4} \]

(i) \(\sin 18^\circ = \frac{\sqrt{5}-1}{4}\), \(\cos 18^\circ = \sqrt{1-\sin^2 18^\circ} = \frac{\sqrt{10+2\sqrt{5}}}{4}\)

(ii) \(\sin 36^\circ = 2\sin 18^\circ\cos 18^\circ = \frac{\sqrt{10-2\sqrt{5}}}{4}\), \(\cos 36^\circ = 1-2\sin^2 18^\circ = \frac{\sqrt{5}+1}{4}\)

(iii) \(\sin 54^\circ = \cos 36^\circ = \frac{\sqrt{5}+1}{4}\), \(\cos 54^\circ = \sin 36^\circ = \frac{\sqrt{10-2\sqrt{5}}}{4}\)

(iv) \(\sin 72^\circ = \cos 18^\circ = \frac{\sqrt{10+2\sqrt{5}}}{4}\), \(\cos 72^\circ = \sin 18^\circ = \frac{\sqrt{5}-1}{4}\)

Answer: (i) \(\sin 18^\circ = \frac{\sqrt{5}-1}{4}\), \(\cos 18^\circ = \frac{\sqrt{10+2\sqrt{5}}}{4}\); (ii) \(\sin 36^\circ = \frac{\sqrt{10-2\sqrt{5}}}{4}\), \(\cos 36^\circ = \frac{\sqrt{5}+1}{4}\); (iii) \(\sin 54^\circ = \frac{\sqrt{5}+1}{4}\), \(\cos 54^\circ = \frac{\sqrt{10-2\sqrt{5}}}{4}\); (iv) \(\sin 72^\circ = \frac{\sqrt{10+2\sqrt{5}}}{4}\), \(\cos 72^\circ = \frac{\sqrt{5}-1}{4}\).