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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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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