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 \(\left(\arccos \left(-\frac{\sqrt{3}}{2}\right)\right)=\sin 150^{\circ}\)
Answer 1:
\(\text { (1) } 210^{\circ}=\frac{7 \pi}{12}\)
\(\text { (2) }-90^{\circ}=-\frac{\pi}{2}\)
\(\text { (3) } \sin \left(\arccos \left(-\frac{\sqrt{3}}{2}\right)\right)=\frac{1}{2}\)
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 – \(\left[-\frac{\pi}{2} \cdot \frac{\pi}{2}\right]\) for arcsin, arccsc, arctan and arccot, while for arccos and arcsec tange is \([0, p]\).
Considering this we solve above as follows:
Answer 2:
To solve this type of problem we are to remember the range of inverse trigonometric function as shown in above figure
