Circular trig. function: 1-cos^2(arcsin(sin(x-pi/2)))

[OhMrsDarcy]

1-cos^2(arcsin(sin(x-pi/2)))

The problems asks to express this as a single circular trigonometric function. I don't even have the slightest clue where to begin, or even what arcsin means. Any help would be appreciated.

 

[royhaas]

The arcsin function is another name for the inverse function of the sine, i.e, $\displaystyle arcxin(x) = sin^{-1}(x)$.

 

[soroban]

Hello, OhMrsDarcy!

Not only are you expected to be familiar with Inverse Trig Functions,
. . you should know all the earlier identities, too!

Express as a single trig function:

. . $\displaystyle 1\,-\,\cos^2\left(\arcsin\left[\sin\left(x\,-\,\frac{\pi}{2}\right)\right]\right)$

Since $\displaystyle \arcsin$ and $\displaystyle \sin$ are inverse functions: $\displaystyle \:\arcsin(\sin\theta)\,=\,\theta$
. . (assuming principal values only).

So we have: $\displaystyle \:1\,-\,\cos^2\left(x\,-\,\frac{\pi}{2}\right)$


Since $\displaystyle \sin^2\theta\,+\,\cos^2\theta\:=\:1\;\;\Rightarrow\;\;1\,-\,\cos^2\theta\:=\:\sin^2\theta$

. . the problem becomes: $\displaystyle \:\sin^2\left(x\,-\,\frac{\pi}{2}\right) \:= \:\left[\sin\left(x\,-\,\frac{\pi}{2}\right)\right]^2$


Since $\displaystyle \sin(-\theta)\,=\,-\sin\theta$, we have: $\displaystyle \:\left[-\sin\left(\frac{\pi}{2}\,-\,x\right)\right]^2$


Since $\displaystyle \sin\left(\frac{\pi}{2}\,-\,\theta\right)\:=\:\cos\theta$, we have: \(\displaystyle \:\left[-\cos x\right]^2 \:=\:\L\cos^2x\;\;\) . . . There!

 

[OhMrsDarcy]

I had gotten to the sin^2(x-pi/2) but beyond that i was cluless. Thank you so much! This was Uber helpful!