Question

Match the following trigonometric functions with their corresponding values.

• A. $\frac{1}{\sqrt{2}}$
• B. $\frac{\sqrt{3}}{2}$
• C. $\frac{1}{\sqrt{3}}$
• D. $-\frac{\sqrt{3}}{2}$

Answer 1

Here, we have to find the values of all these trigonometric ratios.

$a)\cos {210}^{\circ }$
Now, ${210}^{\circ }$ can be written as: ${210}^{\circ }={270}^{\circ }-{60}^{\circ }$ and it lies in the $III$ quadrant, where cos of any angle is negative.
$\Rightarrow \cos {210}^{\circ }=\cos ({270}^{\circ }-{60}^{\circ })=-\sin {60}^{\circ }$

$\Rightarrow \cos {210}^{\circ }=-\sin {60}^{\circ }=-\frac{\sqrt{3}}{2}$

$b)\sin {135}^{\circ }$
Now, ${135}^{\circ }$ can be written as: ${135}^{\circ }={90}^{\circ }+{45}^{\circ }$ and it lies in the $II$ quadrant, where sin of any angle is positive.
$\Rightarrow \sin {135}^{\circ }=\sin ({90}^{\circ }+{45}^{\circ })=\cos {45}^{\circ }$

$\Rightarrow \sin {135}^{\circ }=\cos {45}^{\circ }=\frac{1}{\sqrt{2}}$

$c)\tan \frac{17\pi }{6}$
Now, $\frac{17\pi }{6}$ can be written as: $\frac{17\pi }{6}=\frac{5\pi }{2}+\frac{\pi }{3}$ and it lies in the $II$ quadrant, where $\tan$ of any angle is negative.
$\Rightarrow \tan \frac{17\pi }{6}=\tan (\frac{5\pi }{2}+\frac{\pi }{3})=-\cot \frac{\pi }{3}$

$\Rightarrow \tan \frac{17\pi }{6}=-\cot \frac{\pi }{3}=-\frac{1}{\sqrt{3}}$

$d)\cos \frac{23\pi }{3}$
Now, $\frac{23\pi }{3}$ can be written as: $\frac{23\pi }{3}=\frac{7\pi }{2}+\frac{\pi }{3}$ and it lies in the $IV$ quadrant, where cos of any angle is positive.
$\Rightarrow \cos \frac{23\pi }{3}=\cos (\frac{7\pi }{2}+\frac{\pi }{3})=\sin {60}^{\circ }$

$\Rightarrow \cos \frac{23\pi }{3}=\sin \frac{\pi }{3}=\frac{\sqrt{3}}{2}$