# Find the value of?

## (1) $\arccos \left(- \frac{\sqrt{3}}{2}\right) - \arcsin \left(- \frac{\sqrt{3}}{2}\right) - \arccos \left(\frac{1}{2}\right) + \arcsin \left(\frac{\sqrt{3}}{2}\right)$ (2) $\arcsin \left(- \frac{1}{2}\right) + \arcsin \left(- \frac{\sqrt{3}}{2}\right)$ (3) $\sin \left(\arccos \left(- \frac{\sqrt{3}}{2}\right)\right) = \sin {150}^{o}$

Dec 15, 2016

(1) ${210}^{o} = \frac{7 \pi}{12}$

(2) $- {90}^{o} = - \frac{\pi}{2}$

(3) $\sin \left(\arccos \left(- \frac{\sqrt{3}}{2}\right)\right) = \frac{1}{2}$

#### Explanation:

The trigonometric ratios of standard angles are given in

However, before we use this let us remember that range for inverse trigonometric functions are - $\left[- \frac{\pi}{2.} \frac{\pi}{2}\right]$ for arcsin, arccsc, arctan and arccot, while for arccos and arcsec tange is $\left[0 , p\right]$.

Considering this we solve above as follows:

(1) $\arccos \left(- \frac{\sqrt{3}}{2}\right) - \arcsin \left(- \frac{\sqrt{3}}{2}\right) - \arccos \left(\frac{1}{2}\right) + \arcsin \left(\frac{\sqrt{3}}{2}\right)$

= ${150}^{o} - \left(- {60}^{o}\right) - {60}^{o} + {60}^{o}$

= ${150}^{o} + {60}^{o} - {60}^{o} + {60}^{o} = {210}^{o}$ or $\frac{7 \pi}{12}$

(2) $\arcsin \left(- \frac{1}{2}\right) + \arcsin \left(- \frac{\sqrt{3}}{2}\right)$

= $- {30}^{o} - {60}^{o} = - {90}^{o} = - \frac{\pi}{2}$

(3) $\sin \left(\arccos \left(- \frac{\sqrt{3}}{2}\right)\right) = \sin {150}^{o} = \frac{1}{2}$

Dec 15, 2016

To solve this type of problem we are to remember the range of inverse trigonometric function as shown in above figure

1)
$\arccos \left(- \frac{\sqrt{3}}{2}\right) - \arcsin \left(- \frac{\sqrt{2}}{2}\right) - \arccos \left(\frac{1}{2}\right) + \arcsin \left(\frac{\sqrt{3}}{2}\right)$

$= \arccos \left(- \cos \left(\frac{\pi}{6}\right)\right) - \arcsin \left(- \sin \left(\frac{\pi}{4}\right)\right) - \arccos \left(\cos \left(\frac{\pi}{3}\right)\right) + \arcsin \left(\sin \left(\frac{\pi}{3}\right)\right)$

$= \arccos \left(\cos \left(\pi - \frac{\pi}{6}\right)\right) - \arcsin \sin \left(- \frac{\pi}{4}\right) - \arccos \left(\cos \left(\frac{\pi}{3}\right)\right) + \arcsin \left(\sin \left(\frac{\pi}{3}\right)\right)$

$= \frac{5 \pi}{6} + \frac{\pi}{4} - \frac{\pi}{3} + \frac{\pi}{3}$

$= \frac{10 \pi + 3 \pi}{12} = \frac{13 \pi}{12}$

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2).$\arcsin \left(- \frac{1}{2}\right) + \arcsin \left(- \frac{\sqrt{3}}{2}\right)$

$= \arcsin \left(\sin \left(- \frac{\pi}{6}\right)\right) + \arcsin \left(\sin \left(- \frac{\pi}{3}\right)\right)$

$= - \frac{\pi}{6} - \frac{\pi}{3} = - \frac{\pi}{2}$

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3).sin(arccos(-√3/2))

=sin(arccos(cos(pi-pi/6))

$= \sin \left(\pi - \frac{\pi}{6}\right) = \sin \left(\frac{\pi}{6}\right) = \frac{1}{2}$