Question

# Express as a trigonometric function of one angle. a) \cos2\sin(-9)-\cos9\sin2 Find the exact value of the expression. b) \sin\left(\arcsin\frac{\sqrt{3}}{2}+\arccos0\right)

Express as a trigonometric function of one angle.
a) $$\cos2\sin(-9)-\cos9\sin2$$
Find the exact value of the expression.
b) $$\sin\left(\arcsin\frac{\sqrt{3}}{2}+\arccos0\right)$$

2021-06-08

Step 1
$$\cos2\sin(-9)-\cos9\sin2=\cos2-\sin q-\cos9\sin2$$
$$\sin(-\theta)=-\sin\theta$$
$$=-\cos2\sin9-\cos9\sin2$$
$$=-(\sin9\cos2+\cos9\sin2)$$
$$=-\sin(9+2)$$
$$=-\sin11$$
$$\because\sin(A+B)$$
$$=\sin A\cos B+\cos A\sin B$$
$$=\sin(-11)$$
$$\because\sin(-\theta)=-\sin\theta$$
Step 2
$$\sin[ \arcsin\frac{\sqrt{3}}{2}+\arccos0]=\sin[\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)+\cos^{-1}(0)]$$
$$=\sin[60^{\circ}+90^{\circ}]=\sin(150^{\circ})$$
$$=\sin(90^{\circ}+60^{\circ})$$
$$\because\sin60^{\circ}=\frac{\sqrt{3}}{2}$$
$$=\cos60^{\circ}$$
$$\cos90^{\circ}=0$$
$$=\frac{1}{2}$$
$$\cos60^{\circ}=\frac{1}{2}$$