Expert Answer to "Express as a trigonometric function of one angle...."

College
Calculus and Analysis
Integral Calculus
Parametric equations, polar coordinates, and vector-valued functions

Jaden Easton

Answered question

2021-06-07

Express as a trigonometric function of one angle.
a)

\cos 2 \sin (- 9) - \cos 9 \sin 2

Find the exact value of the expression.
b)

\sin (\arcsin \frac{\sqrt{3}}{2} + \arccos 0)

Answer & Explanation

Fatema Sutton

Skilled2021-06-08Added 88 answers

Step 1
$\cos 2 \sin (- 9) - \cos 9 \sin 2 = \cos 2 - \sin q - \cos 9 \sin 2$
$\sin (- θ) = - \sin θ$
$= - \cos 2 \sin 9 - \cos 9 \sin 2$
$= - (\sin 9 \cos 2 + \cos 9 \sin 2)$
$= - \sin (9 + 2)$
$= - \sin 11$
$∵ \sin (A + B)$
$= \sin A \cos B + \cos A \sin B$
$= \sin (- 11)$
$∵ \sin (- θ) = - \sin θ$
Step 2
$\sin [\arcsin \frac{\sqrt{3}}{2} + \arccos 0] = \sin [\sin^{- 1} (\frac{\sqrt{3}}{2}) + \cos^{- 1} (0)]$
$= \sin [60^{\circ} + 90^{\circ}] = \sin (150^{\circ})$
$= \sin (90^{\circ} + 60^{\circ})$
$∵ \sin 60^{\circ} = \frac{\sqrt{3}}{2}$
$= \cos 60^{\circ}$
$\cos 90^{\circ} = 0$
$= \frac{1}{2}$
$\cos 60^{\circ} = \frac{1}{2}$

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