How to solve for Arcsin(cos 5pi/8)???
\[\sin^{-1}(\cos(\frac{5 \pi}{8}))\]
You need to convert the inner cos into sin, and you are done.. :)
\[ \cos(x) = \sin(\frac \pi 2 - x) \]
Yeah we can do this also.. :)
arcsin=-5pi
cos=5pi/8 usetriangle a2+b2=c2 solver for opp arcsin=opp/adj simplify to get arcsin=-5pi
\[ \sin^{-1}(\cos(\frac{5 \pi}{8})) = \sin^{-1}(\sin(\frac \pi 2 - \frac{5 \pi}{8})) = \sin^{-1}(\sin(\frac 4\pi 8 - \frac{5 \pi}{8})) = \\ \sin^{-1}(\sin(-\frac {\pi}{8})) = -\frac{\pi}{8} \]
If you do the other way pi/2=5pi/8=pi/8, so you get two answers
negative because arc -pi/8
so two solutions depend on the formula you use, so -5pi, or -pi/8
I just checked and -pi/8 was also the final answer in the problem set that I'm answering but thanks to everyone for your efforts
it just depends what formula you use, you will get different answers
@aum I have a question why did you subtract 5pi/8 from pi/2? is it because of the range of arcsin?
-5pi is not the correct answer. sin(-5pi) is not equal to cos(5pi/8)
because cos(x) can be written as sin(pi/2 - x)
formula sin(pi/2-5pi/8) use your fx and solver
what I did was use a right triangle we know cos adj/opp so solve for opp =8-5pi/8
Noticed a typo in an earlier reply: \[ \sin^{-1}(\cos(\frac{5 \pi}{8})) = \sin^{-1}(\sin(\frac \pi 2 - \frac{5 \pi}{8})) = \sin^{-1}(\sin(\frac {4\pi}{ 8} - \frac{5 \pi}{8})) = \\ \sin^{-1}(\sin(-\frac {\pi}{8})) = -\frac{\pi}{8} \]
sin opp/hyp 8-5pi/8 simplify get 5pi
aum is using the trig identites, so what ever way is easier for you, do it
he is using the cofunction identites
5pi/8 is an ANGLE. It is not the sides. So the adjacent is not 5pi and the hypotenuse is not 8. 5pi/8 = 112.5 degrees.
Yeah you are right, if you had sides then you could have done it that way
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