How do you evaluate $cos (A r c sin (\frac{5}{13}))$ $Cos(Arc sin (5/13))$ ?

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Answer 1 · 2016-06-23T17:26:32

$\pm \frac{12}{13}$ $+-12/13$

Let $a = a r c sin (\frac{5}{13})$ $a = arc sin (5/13)$ . Then, $sin a = \frac{5}{13} > 0$ $sin a = 5/13 > 0$ . a is in either 1st

quadrant or in the 2nd. Accordingly, the given expression

$cos a = \pm \sqrt{1 - {sin}^{2} a} = \pm \sqrt{1 - \frac{5^{2}}{13^{2}}} = \pm \frac{12}{13}$ $cos a=+-sqrt(1-sin^2 a)=+-sqrt(1-5^2/13^2)=+-12/13$ ,

The negative value is from $sin (π - a) = sin a = \frac{5}{13} and cos (π - a) = - cos a$ $sin (pi-a)=sin a = 5/13 and cos(pi-a)=-cos a$ .

This is off, if arc sin (5/13) is restricted to be the principal value in

$[- \frac{π}{2}, \frac{π}{2}]$ $[-pi/2, pi/2]$ .

How do you evaluate cos(Arcsin(513))Cos(Arc sin (5/13))?