Answer on Question #73819 – Math – Trigonometry
Question
Evaluate sin [ cos − 1 ( − 2 7 ) ] \sin \left[\cos^{-1}\left(\frac{-2}{7}\right)\right] sin [ cos − 1 ( 7 − 2 ) ]
a . 53 7 b . − 53 7 c . 3 5 7 d . − 3 5 7 e . None of these a.\frac{\sqrt{53}}{7} b.\frac{-\sqrt{53}}{7} c.\frac{3\sqrt{5}}{7} d.\frac{-3\sqrt{5}}{7} e.\text{None of these} a . 7 53 b . 7 − 53 c . 7 3 5 d . 7 − 3 5 e . None of these
Solution
Let φ = cos − 1 ( − 2 7 ) \varphi = \cos^{-1}\left(\frac{-2}{7}\right) φ = cos − 1 ( 7 − 2 ) then cos ( φ ) = − 2 7 \cos(\varphi) = \frac{-2}{7} cos ( φ ) = 7 − 2 , hence π 2 ≤ φ ≤ π \frac{\pi}{2} \leq \varphi \leq \pi 2 π ≤ φ ≤ π .
Find sin ( φ ) : sin ( φ ) = ± 1 − cos 2 ( φ ) = ± 3 5 7 \sin(\varphi): \sin(\varphi) = \pm \sqrt{1 - \cos^2(\varphi)} = \pm \frac{3\sqrt{5}}{7} sin ( φ ) : sin ( φ ) = ± 1 − cos 2 ( φ ) = ± 7 3 5
Because of π / 2 ≤ cos − 1 ( − 2 7 ) ≤ π \pi/2 \leq \cos^{-1}\left(\frac{-2}{7}\right) \leq \pi π /2 ≤ cos − 1 ( 7 − 2 ) ≤ π , then sin [ cos − 1 ( − 2 7 ) ] = 3 5 7 \sin \left[\cos^{-1}\left(\frac{-2}{7}\right)\right] = \frac{3\sqrt{5}}{7} sin [ cos − 1 ( 7 − 2 ) ] = 7 3 5
Answer: c . 3 5 7 c.\frac{3\sqrt{5}}{7} c . 7 3 5
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