NCERT Solutions: Class 11 Maths Chapter 3 - Trigonometric Functions

Use radian measure $=\dfrac{\pi}{180}\times$ degree measure. Thus $25^\circ=\dfrac{25\pi}{180}=\dfrac{5\pi}{36}$; $-47^\circ30'=-47.5^\circ=-\dfrac{95}{2}^\circ$, giving $-\dfrac{95\pi}{360}=-\dfrac{19\pi}{72}$; $240^\circ=\dfrac{240\pi}{180}=\dfrac{4\pi}{3}$; and $520^\circ=\dfrac{520\pi}{180}=\dfrac{26\pi}{9}$.

Use degree measure $=\dfrac{180}{\pi}\times$ radian measure. With $\pi=\dfrac{22}{7}$, $\dfrac{11}{16}$ radian gives $\dfrac{11}{16}\times\dfrac{180\times7}{22}=39.375^\circ$. Also $-4$ radians gives $-4\times\dfrac{180\times7}{22}=-\dfrac{2520}{11}^\circ$. For multiples of $\pi$, $\dfrac{5\pi}{3}=300^\circ$ and $\dfrac{7\pi}{6}=210^\circ$.

In one second, the wheel makes $\dfrac{360}{60}=6$ revolutions. One revolution is $2\pi$ radians, so the angle turned is $6\times2\pi=12\pi$ radians.

For arc length $l=r\theta$, $\theta=\dfrac{l}{r}=\dfrac{22}{100}=\dfrac{11}{50}$ radians. In degrees, $\theta=\dfrac{11}{50}\times\dfrac{180\times7}{22}=\dfrac{63}{5}=12.6^\circ=12^\circ36'$.

The radius is $20$ cm. If the chord subtends central angle $\theta$, then $20=2(20)\sin\dfrac{\theta}{2}$, so $\sin\dfrac{\theta}{2}=\dfrac12$. For the minor arc, $\theta=\dfrac{\pi}{3}$. Hence arc length $l=r\theta=20\times\dfrac{\pi}{3}=\dfrac{20\pi}{3}$ cm.

For the same arc length $l$, $l=r\theta$, so the radius is inversely proportional to the angle in radians. Therefore $r_1:r_2=\theta_2:\theta_1=75:60=5:4$.

Using $l=r\theta$ with $r=75$ cm, $\theta=\dfrac{l}{75}$. Thus for 10 cm, $\theta=\dfrac{10}{75}=\dfrac{2}{15}$; for 15 cm, $\theta=\dfrac{15}{75}=\dfrac15$; for 21 cm, $\theta=\dfrac{21}{75}=\dfrac{7}{25}$.

Since $\cos x=-\dfrac12$ and $x$ is in quadrant III, $\sin x$ is negative. From $\sin^2x+\cos^2x=1$, $\sin^2x=1-\dfrac14=\dfrac34$, so $\sin x=-\dfrac{\sqrt3}{2}$. The other functions follow by quotient and reciprocal definitions.

In quadrant II, sine is positive and cosine is negative. From $\cos^2x=1-\sin^2x=1-\dfrac9{25}=\dfrac{16}{25}$, $\cos x=-\dfrac45$. Then use $\tan x=\dfrac{\sin x}{\cos x}$ and reciprocals.

$\cot x=\dfrac43$ gives $\tan x=\dfrac34$. In quadrant III, sine and cosine are both negative while tangent and cotangent are positive. Using the $3$-$4$-$5$ triangle ratios gives $\sin x=-\dfrac35$ and $\cos x=-\dfrac45$.

$\sec x=\dfrac{13}{5}$ gives $\cos x=\dfrac5{13}$. In quadrant IV, sine is negative. From $\sin^2x=1-\dfrac{25}{169}=\dfrac{144}{169}$, $\sin x=-\dfrac{12}{13}$, and the remaining functions follow.

In quadrant II, tangent is negative, sine is positive and cosine is negative. The ratio $|\tan x|=\dfrac5{12}$ corresponds to a $5$-$12$-$13$ triangle, so $\sin x=\dfrac5{13}$ and $\cos x=-\dfrac{12}{13}$.

$765^\circ=720^\circ+45^\circ$, and sine has period $360^\circ$. Hence $\sin765^\circ=\sin45^\circ=\dfrac{\sqrt2}{2}$.

$-1410^\circ+1440^\circ=30^\circ$. Thus $\sin(-1410^\circ)=\sin30^\circ=\dfrac12$, so $\cosec(-1410^\circ)=2$.

$\dfrac{19\pi}{3}=6\pi+\dfrac{\pi}{3}$. Since tangent has period $\pi$, $\tan\dfrac{19\pi}{3}=\tan\dfrac{\pi}{3}=\sqrt3$.

$-\dfrac{11\pi}{3}+4\pi=\dfrac{\pi}{3}$. Since sine has period $2\pi$, the value is $\sin\dfrac{\pi}{3}=\dfrac{\sqrt3}{2}$.

$-\dfrac{15\pi}{4}+4\pi=\dfrac{\pi}{4}$. Since cotangent has period $\pi$, the value is $\cot\dfrac{\pi}{4}=1$.

LHS $=\left(\dfrac12\right)^2+\left(\dfrac12\right)^2-1^2=\dfrac14+\dfrac14-1=-\dfrac12$, which equals RHS.

Since $\sin\dfrac{\pi}{6}=\dfrac12$, $\cosec\dfrac{7\pi}{6}=-2$ and $\cos\dfrac{\pi}{3}=\dfrac12$, LHS $=2\cdot\dfrac14+4\cdot\dfrac14=\dfrac12+1=\dfrac32$.

$\cot\dfrac{\pi}{6}=\sqrt3$, $\cosec\dfrac{5\pi}{6}=2$, and $\tan\dfrac{\pi}{6}=\dfrac1{\sqrt3}$. Thus LHS $=3+2+3\cdot\dfrac13=6$.

$\sin^2\dfrac{3\pi}{4}=\dfrac12$, $\cos^2\dfrac{\pi}{4}=\dfrac12$, and $\sec^2\dfrac{\pi}{3}=4$. Therefore LHS $=2\cdot\dfrac12+2\cdot\dfrac12+2\cdot4=1+1+8=10$.

$\sin75^\circ=\sin(45^\circ+30^\circ)=\sin45^\circ\cos30^\circ+\cos45^\circ\sin30^\circ=\dfrac{\sqrt6+\sqrt2}{4}$. Also $\tan15^\circ=\tan(45^\circ-30^\circ)=\dfrac{1-1/\sqrt3}{1+1/\sqrt3}=2-\sqrt3$.

Using $\cos A\cos B-\sin A\sin B=\cos(A+B)$, LHS $=\cos\left(\dfrac{\pi}{2}-x-y\right)=\sin(x+y)$.

$\tan\left(\dfrac{\pi}{4}+x\right)=\dfrac{1+\tan x}{1-\tan x}$ and $\tan\left(\dfrac{\pi}{4}-x\right)=\dfrac{1-\tan x}{1+\tan x}$. Dividing gives $\left(\dfrac{1+\tan x}{1-\tan x}\right)^2$.

Use $\cos(\pi+x)=-\cos x$, $\cos(-x)=\cos x$, $\sin(\pi-x)=\sin x$ and $\cos\left(\dfrac{\pi}{2}+x\right)=-\sin x$. Then LHS $=\dfrac{-\cos^2x}{-\sin^2x}=\cot^2x$.

$\cos\left(\dfrac{3\pi}{2}+x\right)=\sin x$, $\cos(2\pi+x)=\cos x$, $\cot\left(\dfrac{3\pi}{2}-x\right)=\tan x$, and $\cot(2\pi+x)=\cot x$. Thus LHS $=\sin x\cos x(\tan x+\cot x)=\sin x\cos x\left(\dfrac{\sin x}{\cos x}+\dfrac{\cos x}{\sin x}\right)=1$.

Using $\cos(A-B)=\cos A\cos B+\sin A\sin B$, LHS $=\cos((n+1)x-(n+2)x)=\cos(-x)=\cos x$.

Use $\cos A-\cos B=-2\sin\dfrac{A+B}{2}\sin\dfrac{A-B}{2}$. Here $A=\dfrac{3\pi}{4}+x$ and $B=\dfrac{3\pi}{4}-x$, so LHS $=-2\sin\dfrac{3\pi}{4}\sin x=-\sqrt2\sin x$.

$\sin^2A-\sin^2B=(\sin A-\sin B)(\sin A+\sin B)=\sin(A+B)\sin(A-B)$. Taking $A=6x$, $B=4x$ gives $\sin10x\sin2x$.

$(\cos2x-\cos6x)(\cos2x+\cos6x)=\{2\sin4x\sin2x\}\{2\cos4x\cos2x\}=\sin8x\sin4x$.

$\sin2x+\sin6x=2\sin4x\cos2x$. Hence LHS $=2\sin4x\cos2x+2\sin4x=2\sin4x(1+\cos2x)=2\sin4x(2\cos^2x)=4\cos^2x\sin4x$.

LHS $=\cot4x\{2\sin4x\cos x\}=2\cos4x\cos x$. RHS $=\cot x\{2\cos4x\sin x\}=2\cos4x\cos x$. Hence both sides are equal.

$\cos9x-\cos5x=-2\sin7x\sin2x$ and $\sin17x-\sin3x=2\cos10x\sin7x$. Dividing gives $-\dfrac{\sin2x}{\cos10x}$.

Using sum-to-product, numerator $=2\sin4x\cos x$ and denominator $=2\cos4x\cos x$. Their ratio is $\tan4x$.

$\sin x-\sin y=2\cos\dfrac{x+y}{2}\sin\dfrac{x-y}{2}$ and $\cos x+\cos y=2\cos\dfrac{x+y}{2}\cos\dfrac{x-y}{2}$. Dividing gives $\tan\dfrac{x-y}{2}$.

$\sin x+\sin3x=2\sin2x\cos x$ and $\cos x+\cos3x=2\cos2x\cos x$. Hence the ratio is $\tan2x$.

$\sin x-\sin3x=2\cos2x\sin(-x)=-2\sin x\cos2x$. Also $\sin^2x-\cos^2x=-\cos2x$. Dividing gives $2\sin x$.

$\cos4x+\cos2x=2\cos3x\cos x$, so the numerator is $\cos3x(2\cos x+1)$. Also $\sin4x+\sin2x=2\sin3x\cos x$, so the denominator is $\sin3x(2\cos x+1)$. The ratio is $\cot3x$.

Using $\cot(A+B)=\dfrac{\cot A\cot B-1}{\cot A+\cot B}$ with $A=x$ and $B=2x$, $\cot3x(\cot x+\cot2x)=\cot x\cot2x-1$. Rearranging gives $\cot x\cot2x-\cot2x\cot3x-\cot3x\cot x=1$.

Let $t=\tan x$. Then $\tan2x=\dfrac{2t}{1-t^2}$. Hence $\tan4x=\dfrac{2\tan2x}{1-\tan^22x}=\dfrac{\frac{4t}{1-t^2}}{1-\frac{4t^2}{(1-t^2)^2}}=\dfrac{4t(1-t^2)}{(1-t^2)^2-4t^2}=\dfrac{4t(1-t^2)}{1-6t^2+t^4}$.

$\cos4x=1-2\sin^22x=1-2(2\sin x\cos x)^2=1-8\sin^2x\cos^2x$.

$\cos6x=\cos(3\cdot2x)=4\cos^32x-3\cos2x$. Put $\cos2x=2\cos^2x-1$. Then $4(2c^2-1)^3-3(2c^2-1)=32c^6-48c^4+18c^2-1$, where $c=\cos x$.

Pair the terms as $\cos\dfrac{3\pi}{13}+\cos\dfrac{9\pi}{13}=2\cos\dfrac{6\pi}{13}\cos\left(-\dfrac{3\pi}{13}\right)$ and use $\cos\dfrac{6\pi}{13}=\sin\dfrac{\pi}{26}$ with complementary-angle reductions; equivalently, these angles are roots arranged symmetrically from $13$th roots of $-1$, and their real parts sum to 0.

LHS $=\sin3x\sin x+\cos3x\cos x+\sin^2x-\cos^2x=\cos2x-\cos2x=0$.

Expanding gives $\cos^2x+\sin^2x+\cos^2y+\sin^2y+2\cos x\cos y-2\sin x\sin y=2+2\cos(x+y)=4\cos^2\dfrac{x+y}{2}$.

Expanding gives $2-2(\cos x\cos y+\sin x\sin y)=2-2\cos(x-y)=4\sin^2\dfrac{x-y}{2}$.

$\sin x+\sin7x=2\sin4x\cos3x$ and $\sin3x+\sin5x=2\sin4x\cos x$. Thus LHS $=2\sin4x(\cos3x+\cos x)=2\sin4x(2\cos2x\cos x)$.

The numerator is $2\sin6x\cos x+2\sin6x\cos3x=2\sin6x(\cos x+\cos3x)$. The denominator is $2\cos6x\cos x+2\cos6x\cos3x=2\cos6x(\cos x+\cos3x)$. The ratio is $\tan6x$.

LHS $=(\sin3x-\sin x)+\sin2x=2\cos2x\sin x+2\sin x\cos x=2\sin x(\cos2x+\cos x)$. Since $\cos2x+\cos x=2\cos\dfrac{3x}{2}\cos\dfrac{x}{2}$, the result follows.

Since $x$ is in quadrant II, $x/2$ is in quadrant I. From $\tan x=-\dfrac43$, take $\sin x=\dfrac45$ and $\cos x=-\dfrac35$. Then $\sin\dfrac{x}{2}=\sqrt{\dfrac{1-\cos x}{2}}=\sqrt{\dfrac{1+3/5}{2}}=\dfrac{2}{\sqrt5}$ and $\cos\dfrac{x}{2}=\sqrt{\dfrac{1+\cos x}{2}}=\dfrac{1}{\sqrt5}$.

Since $x$ is in quadrant III, $x/2$ is in quadrant II, so sine is positive and cosine is negative. Thus $\sin\dfrac{x}{2}=\sqrt{\dfrac{1-\cos x}{2}}=\sqrt{\dfrac{1+1/3}{2}}=\sqrt{\dfrac23}$ and $\cos\dfrac{x}{2}=-\sqrt{\dfrac{1+\cos x}{2}}=-\sqrt{\dfrac{1-1/3}{2}}=-\dfrac1{\sqrt3}$.

In quadrant II, $\cos x=-\sqrt{1-\sin^2x}=-\dfrac{\sqrt{15}}{4}$, and $x/2$ lies in quadrant I. Therefore $\sin\dfrac{x}{2}=\sqrt{\dfrac{1-\cos x}{2}}=\sqrt{\dfrac{4+\sqrt{15}}{8}}$ and $\cos\dfrac{x}{2}=\sqrt{\dfrac{1+\cos x}{2}}=\sqrt{\dfrac{4-\sqrt{15}}{8}}$. Also $\tan\dfrac{x}{2}=\dfrac{\sin x}{1+\cos x}=\dfrac{1}{4-\sqrt{15}}=4+\sqrt{15}$.