Trigonometric Functions
NCERT Class 11 Maths • Miscellaneous Exercise1. Trigonometric Proofs
Q1: Prove $2\cos\frac{\pi}{13}\cos\frac{9\pi}{13} + \cos\frac{3\pi}{13} + \cos\frac{5\pi}{13} = 0$
Identity: $2\cos A \cos B = \cos(A+B) + \cos(A-B)$
Consider the first term: $2\cos\frac{\pi}{13}\cos\frac{9\pi}{13}$
$= \cos(\frac{9\pi}{13} + \frac{\pi}{13}) + \cos(\frac{9\pi}{13} – \frac{\pi}{13})$
$= \cos\frac{10\pi}{13} + \cos\frac{8\pi}{13}$
$= \cos\frac{10\pi}{13} + \cos\frac{8\pi}{13}$
Substitute back into LHS: $\cos\frac{10\pi}{13} + \cos\frac{8\pi}{13} + \cos\frac{3\pi}{13} + \cos\frac{5\pi}{13}$
Convert angles: $\frac{10\pi}{13} = \pi – \frac{3\pi}{13}$ and $\frac{8\pi}{13} = \pi – \frac{5\pi}{13}$
$\cos(\pi – \theta) = -\cos \theta$. So:
$= -\cos\frac{3\pi}{13} – \cos\frac{5\pi}{13} + \cos\frac{3\pi}{13} + \cos\frac{5\pi}{13}$
$= -\cos\frac{3\pi}{13} – \cos\frac{5\pi}{13} + \cos\frac{3\pi}{13} + \cos\frac{5\pi}{13}$
$= 0 = \text{RHS}$
Q2: Prove $(\sin 3x + \sin x)\sin x + (\cos 3x – \cos x)\cos x = 0$
Expand terms:
$= \sin 3x \sin x + \sin^2 x + \cos 3x \cos x – \cos^2 x$
$= \sin 3x \sin x + \sin^2 x + \cos 3x \cos x – \cos^2 x$
Group terms:
$= (\cos 3x \cos x + \sin 3x \sin x) – (\cos^2 x – \sin^2 x)$
$= (\cos 3x \cos x + \sin 3x \sin x) – (\cos^2 x – \sin^2 x)$
$\cos(A-B) = \cos A \cos B + \sin A \sin B$
$= \cos(3x – x) – \cos 2x$
$= \cos 2x – \cos 2x$
$= \cos 2x – \cos 2x$
$= 0 = \text{RHS}$
Q3: Prove $(\cos x + \cos y)^2 + (\sin x – \sin y)^2 = 4\cos^2\frac{x+y}{2}$
Expand squares:
$= (\cos^2 x + \cos^2 y + 2\cos x \cos y) + (\sin^2 x + \sin^2 y – 2\sin x \sin y)$
$= (\cos^2 x + \cos^2 y + 2\cos x \cos y) + (\sin^2 x + \sin^2 y – 2\sin x \sin y)$
Group $(\sin^2 \theta + \cos^2 \theta = 1)$:
$= (\cos^2 x + \sin^2 x) + (\cos^2 y + \sin^2 y) + 2(\cos x \cos y – \sin x \sin y)$
$= 1 + 1 + 2\cos(x+y) = 2 + 2\cos(x+y)$
$= (\cos^2 x + \sin^2 x) + (\cos^2 y + \sin^2 y) + 2(\cos x \cos y – \sin x \sin y)$
$= 1 + 1 + 2\cos(x+y) = 2 + 2\cos(x+y)$
$= 2(1 + \cos(x+y))$
$1 + \cos A = 2\cos^2\frac{A}{2}$
$= 2(2\cos^2\frac{x+y}{2}) = 4\cos^2\frac{x+y}{2} = \text{RHS}$
Q4: Prove $(\cos x – \cos y)^2 + (\sin x – \sin y)^2 = 4\sin^2\frac{x-y}{2}$
Expand:
$= \cos^2 x + \cos^2 y – 2\cos x \cos y + \sin^2 x + \sin^2 y – 2\sin x \sin y$
$= \cos^2 x + \cos^2 y – 2\cos x \cos y + \sin^2 x + \sin^2 y – 2\sin x \sin y$
Use identity $\sin^2\theta + \cos^2\theta = 1$:
$= 2 – 2(\cos x \cos y + \sin x \sin y)$
$= 2 – 2\cos(x-y) = 2(1 – \cos(x-y))$
$= 2 – 2(\cos x \cos y + \sin x \sin y)$
$= 2 – 2\cos(x-y) = 2(1 – \cos(x-y))$
$1 – \cos A = 2\sin^2\frac{A}{2}$
$= 2(2\sin^2\frac{x-y}{2}) = 4\sin^2\frac{x-y}{2} = \text{RHS}$
Q5: Prove $\sin x + \sin 3x + \sin 5x + \sin 7x = 4\cos x \cos 2x \sin 4x$
Group $(\sin 7x + \sin x)$ and $(\sin 5x + \sin 3x)$.
$\sin A + \sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2}$
$= 2\sin 4x \cos 3x + 2\sin 4x \cos x$
Factor out $2\sin 4x$:
$= 2\sin 4x (\cos 3x + \cos x)$
$= 2\sin 4x (\cos 3x + \cos x)$
Apply $\cos A + \cos B = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2}$:
$= 2\sin 4x (2\cos 2x \cos x)$
$= 2\sin 4x (2\cos 2x \cos x)$
$= 4\cos x \cos 2x \sin 4x = \text{RHS}$
Q6: Prove $\frac{(\sin 7x + \sin 5x) + (\sin 9x + \sin 3x)}{(\cos 7x + \cos 5x) + (\cos 9x + \cos 3x)} = \tan 6x$
Numerator: $2\sin 6x \cos x + 2\sin 6x \cos 3x = 2\sin 6x(\cos x + \cos 3x)$
Denominator: $2\cos 6x \cos x + 2\cos 6x \cos 3x = 2\cos 6x(\cos x + \cos 3x)$
Divide:
$\frac{2\sin 6x(\cos x + \cos 3x)}{2\cos 6x(\cos x + \cos 3x)}$
$\frac{2\sin 6x(\cos x + \cos 3x)}{2\cos 6x(\cos x + \cos 3x)}$
$= \frac{\sin 6x}{\cos 6x} = \tan 6x = \text{RHS}$
Q7: Prove $\sin 3x + \sin 2x – \sin x = 4\sin x \cos\frac{x}{2}\cos\frac{3x}{2}$
Group $(\sin 3x – \sin x) + \sin 2x$.
$= 2\cos\frac{4x}{2}\sin\frac{2x}{2} + 2\sin x \cos x$
$= 2\cos 2x \sin x + 2\sin x \cos x$
$= 2\cos 2x \sin x + 2\sin x \cos x$
Factor $2\sin x$:
$= 2\sin x (\cos 2x + \cos x)$
$= 2\sin x (\cos 2x + \cos x)$
Apply $\cos A + \cos B$:
$= 2\sin x (2\cos\frac{3x}{2}\cos\frac{x}{2})$
$= 2\sin x (2\cos\frac{3x}{2}\cos\frac{x}{2})$
$= 4\sin x \cos\frac{x}{2}\cos\frac{3x}{2} = \text{RHS}$
2. Find $\sin\frac{x}{2}, \cos\frac{x}{2}, \tan\frac{x}{2}$
Q8: $\tan x = -\frac{4}{3}$, x in Quadrant II
x in Q2 $\rightarrow$ cos x is negative
$\sec^2 x = 1 + \tan^2 x = 1 + \frac{16}{9} = \frac{25}{9} \Rightarrow \sec x = -\frac{5}{3}$
$\cos x = -\frac{3}{5}$
$\cos x = -\frac{3}{5}$
Check Quadrant for $x/2$:
$\frac{\pi}{2} < x < \pi \Rightarrow \frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2}$ (Quadrant I)
All values will be positive.
$\frac{\pi}{2} < x < \pi \Rightarrow \frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2}$ (Quadrant I)
All values will be positive.
$\sin\frac{x}{2} = \sqrt{\frac{1-\cos x}{2}} = \sqrt{\frac{1 – (-3/5)}{2}} = \sqrt{\frac{8/5}{2}} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}}$
$\cos\frac{x}{2} = \sqrt{\frac{1+\cos x}{2}} = \sqrt{\frac{1 – 3/5}{2}} = \sqrt{\frac{2/5}{2}} = \frac{1}{\sqrt{5}}$
$\tan\frac{x}{2} = \frac{2/\sqrt{5}}{1/\sqrt{5}} = 2$
Q9: $\cos x = -\frac{1}{3}$, x in Quadrant III
x in Q3 $\rightarrow$ $x/2$ in Q2
$\pi < x < \frac{3\pi}{2} \Rightarrow \frac{\pi}{2} < \frac{x}{2} < \frac{3\pi}{4}$ (Quadrant II)
$\sin$ is positive, $\cos$ is negative.
$\sin$ is positive, $\cos$ is negative.
$\sin\frac{x}{2} = \sqrt{\frac{1-(-1/3)}{2}} = \sqrt{\frac{4/3}{2}} = \sqrt{\frac{2}{3}}$
$\cos\frac{x}{2} = -\sqrt{\frac{1+(-1/3)}{2}} = -\sqrt{\frac{2/3}{2}} = -\sqrt{\frac{1}{3}} = -\frac{1}{\sqrt{3}}$
$\tan\frac{x}{2} = \frac{\sqrt{2/3}}{-1/\sqrt{3}} = -\sqrt{2}$
Q10: $\sin x = \frac{1}{4}$, x in Quadrant II
x in Q2 $\rightarrow$ $x/2$ in Q1
Find $\cos x$: Since x is in Q2, $\cos x < 0$.
$\cos x = -\sqrt{1 – \sin^2 x} = -\sqrt{1 – \frac{1}{16}} = -\frac{\sqrt{15}}{4}$
$\cos x = -\sqrt{1 – \sin^2 x} = -\sqrt{1 – \frac{1}{16}} = -\frac{\sqrt{15}}{4}$
Quadrant Check:
$\frac{\pi}{2} < x < \pi \Rightarrow \frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2}$ (Quadrant I)
All values positive.
$\frac{\pi}{2} < x < \pi \Rightarrow \frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2}$ (Quadrant I)
All values positive.
$\sin\frac{x}{2} = \sqrt{\frac{1 – (-\frac{\sqrt{15}}{4})}{2}} = \sqrt{\frac{4+\sqrt{15}}{8}} = \frac{\sqrt{8+2\sqrt{15}}}{4}$
Simplify nested root: $8+2\sqrt{15} = (\sqrt{5}+\sqrt{3})^2$
$\sin\frac{x}{2} = \frac{\sqrt{5}+\sqrt{3}}{4}$
$\sin\frac{x}{2} = \frac{\sqrt{5}+\sqrt{3}}{4}$
$\cos\frac{x}{2} = \sqrt{\frac{1 + (-\frac{\sqrt{15}}{4})}{2}} = \frac{\sqrt{5}-\sqrt{3}}{4}$
$\tan\frac{x}{2} = \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}} = 4+\sqrt{15}$