Examples (Revised) - Chapter 3 - Trigonometric Functions - Ncert Solutions class 11 - Maths

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Chapter 3: Trigonometric Functions | NCERT Solutions for Class 11 Maths

Example 1

Convert $40^{\circ} 20^{\prime}$ into radian measure.
Solution

We know that $180^{\circ}=\pi$ radian.
Hence $\quad 40^{\circ} 20^{\prime}=40 \frac{1}{3}$ degree $=\frac{\pi}{180} \times \frac{121}{3}$ radian $=\frac{121 \pi}{540}$ radian.
Therefore $\quad 40^{\circ} 20^{\prime}=\frac{121 \pi}{540}$ radian.
Example 2

Convert 6 radians into degree measure.
Solution

We know that $\pi$ radian $=180^{\circ}$.
Hence
$
\begin{aligned}
6 \text { radians } & =\frac{180}{\pi} \times 6 \text { degree }=\frac{1080 \times 7}{22} \text { degree } \\
& =343 \frac{7}{11} \text { degree }=343^{\circ}+\frac{7 \times 60}{11} \text { minute } \quad\left[\text { as } 1^{\circ}=60^{\prime}\right] \\
& =343^{\circ}+38^{\prime}+\frac{2}{11} \text { minute } \quad\left[\text { as } 1^{\prime}=60^{\prime \prime}\right] \\
& =343^{\circ}+38^{\prime}+10.9^{\prime \prime}=343^{\circ} 38^{\prime} 11^{\prime \prime} \text { approximately. }
\end{aligned}
$

Hence $\quad 6$ radians $=343^{\circ} 38^{\prime} 11^{\prime \prime}$ approximately.
Example 3

Find the radius of the circle in which a central angle of $60^{\circ}$ intercepts an arc of length $37.4 \mathrm{~cm}$ (use $\pi=\frac{22}{7}$ ).

Solution

Here $l=37.4 \mathrm{~cm}$ and $\theta=60^{\circ}=\frac{60 \pi}{180}$ radian $=\frac{\pi}{3}$
Hence,
$
\begin{aligned}
\text { by } r & =\frac{l}{\theta}, \text { we have } \\
r & =\frac{37.4 \times 3}{\pi}=\frac{37.4 \times 3 \times 7}{22}=35.7 \mathrm{~cm}
\end{aligned}
$

Example 4

The minute hand of a watch is $1.5 \mathrm{~cm}$ long. How far does its tip move in 40 minutes? (Use $\pi=3.14$ ).
Solution

In 60 minutes, the minute hand of a watch completes one revolution. Therefore, in 40 minutes, the minute hand turns through $\frac{2}{3}$ of a revolution. Therefore, $\theta=\frac{2}{3} \times 360^{\circ}$ or $\frac{4 \pi}{3}$ radian. Hence, the required distance travelled is given by
$
l=r \theta=1.5 \times \frac{4 \pi}{3} \mathrm{~cm}=2 \pi \mathrm{cm}=2 \times 3.14 \mathrm{~cm}=6.28 \mathrm{~cm} .
$

Example 5

If the arcs of the same lengths in two circles subtend angles $65^{\circ}$ and $110^{\circ}$ at the centre, find the ratio of their radii.
Solution

Let $r_1$ and $r_2$ be the radii of the two circles. Given that
$
\theta_1=65^{\circ}=\frac{\pi}{180} \times 65=\frac{13 \pi}{36} \text { radian }
$
and
$
\theta_2=110^{\circ}=\frac{\pi}{180} \times 110=\frac{22 \pi}{36} \text { radian }
$

Let $l$ be the length of each of the arc. Then $l=r_1 \theta_1=r_2 \theta_2$, which gives

$\begin{array}{ll} 
& \frac{13 \pi}{36} \times r_1=\frac{22 \pi}{36} \times r_2 \text {, i.e., } \frac{r_1}{r_2}=\frac{22}{13} \\
\text { Hence } \quad r_1: r_2=22: 13 .
\end{array}$

Example 6

If $\cos x=-\frac{3}{5}, x$ lies in the third quadrant, find the values of other five trigonometric functions.
Solution Since $\cos x=-\frac{3}{5}$, we have $\sec x=-\frac{5}{3}$
Now $\sin ^2 x+\cos ^2 x=1$, i.e., $\sin ^2 x=1-\cos ^2 x$
or
$
\sin ^2 x=1-\frac{9}{25}=\frac{16}{25}
$

Hence $\quad \sin x= \pm \frac{4}{5}$
Since $x$ lies in third quadrant, $\sin x$ is negative. Therefore
$
\sin x=-\frac{4}{5}
$
which also gives
$
\operatorname{cosec} x=-\frac{5}{4}
$

Further, we have
$
\tan x=\frac{\sin x}{\cos x}=\frac{4}{3} \text { and } \cot x=\frac{\cos x}{\sin x}=\frac{3}{4} .
$

Example 7

If $\cot x=-\frac{5}{12}, x$ lies in second quadrant, find the values of other five trigonometric functions.

Solution

Since $\cot x=-\frac{5}{12}$, we have $\tan x=-\frac{12}{5}$

Now
$
\sec ^2 x=1+\tan ^2 x=1+\frac{144}{25}=\frac{169}{25}
$

Hence
$
\sec x= \pm \frac{13}{5}
$

Since $x$ lies in second quadrant, $\sec x$ will be negative. Therefore
$
\sec x=-\frac{13}{5}
$
which also gives
$
\cos x=-\frac{5}{13}
$

Further, we have
$
\sin x=\tan x \cos x=\left(-\frac{12}{5}\right) \times\left(-\frac{5}{13}\right)=\frac{12}{13}
$
and
$
\operatorname{cosec} x=\frac{1}{\sin x}=\frac{13}{12}
$

Example 8

Find the value of $\sin \frac{31 \pi}{3}$.
Solution

We know that values of $\sin x$ repeats after an interval of $2 \pi$. Therefore
$
\sin \frac{31 \pi}{3}=\sin \left(10 \pi+\frac{\pi}{3}\right)=\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2} \text {. }
$

Example 9

Find the value of $\cos \left(-1710^{\circ}\right)$.
Solution

We know that values of $\cos x$ repeats after an interval of $2 \pi$ or $360^{\circ}$.
Therefore, $\cos \left(-1710^{\circ}\right)=\cos \left(-1710^{\circ}+5 \times 360^{\circ}\right)$
$
=\cos \left(-1710^{\circ}+1800^{\circ}\right)=\cos 90^{\circ}=0 \text {. }
$

Example 10

Prove that
$
3 \sin \frac{\pi}{6} \sec \frac{\pi}{3}-4 \sin \frac{5 \pi}{6} \cot \frac{\pi}{4}=1
$

Solution

We have
$
\begin{aligned}
\text { L.H.S. } & =3 \sin \frac{\pi}{6} \sec \frac{\pi}{3}-4 \sin \frac{5 \pi}{6} \cot \frac{\pi}{4} \\
& =3 \times \frac{1}{2} \times 2-4 \sin \left(\pi-\frac{\pi}{6}\right) \times 1=3-4 \sin \frac{\pi}{6} \\
& =3-4 \times \frac{1}{2}=1=\text { R.H.S. }
\end{aligned}
$

Example 11

Find the value of $\sin 15^{\circ}$.
Solution

We have
$
\begin{aligned}
\sin 15^{\circ} & =\sin \left(45^{\circ}-30^{\circ}\right) \\
& =\sin 45^{\circ} \cos 30^{\circ}-\cos 45^{\circ} \sin 30^{\circ} \\
& =\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}} \times \frac{1}{2}=\frac{\sqrt{3}-1}{2 \sqrt{2}} .
\end{aligned}
$

Example 12

Find the value of $\tan \frac{13 \pi}{12}$.

Solution

We have
$
\begin{aligned}
\tan \frac{13 \pi}{12} & =\tan \left(\pi+\frac{\pi}{12}\right)=\tan \frac{\pi}{12}=\tan \left(\frac{\pi}{4}-\frac{\pi}{6}\right) \\
& =\frac{\tan \frac{\pi}{4}-\tan \frac{\pi}{6}}{1+\tan \frac{\pi}{4} \tan \frac{\pi}{6}}=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}=\frac{\sqrt{3}-1}{\sqrt{3}+1}=2-\sqrt{3}
\end{aligned}
$

Example 13

Prove that
$
\frac{\sin (x+y)}{\sin (x-y)}=\frac{\tan x+\tan y}{\tan x-\tan y} .
$

Solution

We have
$
\text { L.H.S. }=\frac{\sin (x+y)}{\sin (x-y)}=\frac{\sin x \cos y+\cos x \sin y}{\sin x \cos y-\cos x \sin y}
$

Dividing the numerator and denominator by $\cos x \cos y$, we get
$
\frac{\sin (x+y)}{\sin (x-y)}=\frac{\tan x+\tan y}{\tan x-\tan y} .
$

Example 14

Show that
$
\tan 3 x \tan 2 x \tan x=\tan 3 x-\tan 2 x-\tan x
$

Solution

We know that $3 x=2 x+x$
Therefore, $\tan 3 x=\tan (2 x+x)$
or
$
\tan 3 x=\frac{\tan 2 x+\tan x}{1-\tan 2 x \tan x}
$
or
$
\tan 3 x-\tan 3 x \tan 2 x \tan x=\tan 2 x+\tan x
$

or $\quad \tan 3 x-\tan 2 x-\tan x=\tan 3 x \tan 2 x \tan x$
or $\quad \tan 3 x \tan 2 x \tan x=\tan 3 x-\tan 2 x-\tan x$.
Example 15

Prove that
$
\cos \left(\frac{\pi}{4}+x\right)+\cos \left(\frac{\pi}{4}-x\right)=\sqrt{2} \cos x
$

Solution

Using the Identity 20(i), we have

L.H.S. $\quad=\cos \left(\frac{\pi}{4}+x\right)+\cos \left(\frac{\pi}{4}-x\right)$
$=2 \cos \left(\frac{\frac{\pi}{4}+x+\frac{\pi}{4}-x}{2}\right) \cos \left(\frac{\frac{\pi}{4}+x-\left(\frac{\pi}{4}-x\right)}{2}\right)$
$=2 \cos \frac{\pi}{4} \cos x=2 \times \frac{1}{\sqrt{2}} \cos x=\sqrt{2} \cos x=$ R.H.S.

Example 16

Prove that $\frac{\cos 7 x+\cos 5 x}{\sin 7 x-\sin 5 x}=\cot x$
Solution

Using the Identities 20 (i) and 20 (iv), we get
L.H.S. $=\frac{2 \cos \frac{7 x+5 x}{2} \cos \frac{7 x-5 x}{2}}{2 \cos \frac{7 x+5 x}{2} \sin \frac{7 x-5 x}{2}}=\frac{\cos x}{\sin x}=\cot x=$ R.H.S.

Example 17

Prove that $=\frac{\sin 5 x-2 \sin 3 x+\sin x}{\cos 5 x-\cos x}=\tan x$
Solution

We have
L.H.S. $=\frac{\sin 5 x-2 \sin 3 x+\sin x}{\cos 5 x-\cos x}=\frac{\sin 5 x+\sin x-2 \sin 3 x}{\cos 5 x-\cos x}$

$\begin{aligned}
& =\frac{2 \sin 3 x \cos 2 x-2 \sin 3 x}{-2 \sin 3 x \sin 2 x}=-\frac{\sin 3 x(\cos 2 x-1)}{\sin 3 x \sin 2 x} \\
& =\frac{1-\cos 2 x}{\sin 2 x}=\frac{2 \sin ^2 x}{2 \sin x \cos x}=\tan x=\text { R.H.S. }
\end{aligned}$