# How can i show this equality in trigonometry?

Help me to prove that $$\arcsin(\sqrt{2}\sin(t))+\arcsin(\sqrt{\cos(2t)})=\pi/2$$ Starting here is a bit difficult. Thank you in advance.

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Do you have any restrictions on $t$? Or do you work with the sine function over the complex numbers? If you insert $t=\pi/2$ you have to calculate $\arcsin(\sqrt{2})$ which you can do over the complex numbers. – Michalis Jan 5 '13 at 17:08
No. It is printed in my trigonometry text and just used $\theta$ istead of $t$ and noting added here as i am looking that. :( – Ned Dabby Jan 5 '13 at 17:10
This is only valid where $\cos(2t)\geq 0$. – 1015 Jan 5 '13 at 17:20

Put

$$f(x):=\arcsin(\sqrt 2\sin t)+\arcsin(\sqrt{\cos 2t})\Longrightarrow$$

$$f'(x)=\frac{\sqrt 2\cos t}{\sqrt{1-2\sin^2 t}}-\frac{\sin 2t}{\sqrt{\cos 2t}}\frac{1}{\sqrt{1-\cos 2t}}=:I$$

But $\,1-2\sin^2t=\cos^2t-\sin^2t=\cos 2t\,$ , so that also $\,1-\cos 2t= 2\sin^2 t\,$ , and then:

$$I=\frac{1}{\sqrt {\cos 2t}}\left(\sqrt 2\cos t-\frac{\sin 2t}{\sqrt 2\sin t}\right)=\frac{1}{\sqrt{\cos 2t}}\left(\sqrt 2\cos t-\sqrt 2\cos t\right)=0$$

using the double angle formula for the sine function, and from here that $\,f(x)=K=\,$constant.

Now just evaluate $\,f(0)\,$ to find out what $\,K\,$ is....

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Hint: $\left(\sqrt{2}\sin t\right)^2+\left(\sqrt{\cos 2t}\right)^2=1$. This is a disguised form of the familiar double-angle identity $\cos 2t=2\cos^2 t-1=\cos^2 t-\sin^2 t=1-2\sin^2 t$.

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Let $$\arcsin(\sqrt 2\sin t)=y\implies \sin y=\sqrt 2\sin t$$

$$\cos2t=1-2\sin^2t=1-2\left(\frac{\sin y}{\sqrt2}\right)^2=1-\sin^2y=\cos^2y$$

So, $$\arcsin \sqrt {\cos2t}=\arcsin (\cos y)=\arcsin (\sin(\frac\pi2-y))$$ as $\cos x=\sin(\frac\pi2-x)$ and $\arcsin(\sin z)=z$

Hence, $$\arcsin \sqrt {\cos2t}=\frac\pi2-y=\frac\pi2-\arcsin(\sqrt 2\sin t)$$

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