^-1()-^-1()=x, then x=

Question

$\cot^{-1}(\sqrt{\cos\alpha})-	an^{-1}(\sqrt{\cos\alpha})=x$, then $\sin x=$

Accepted Answer

$$\cot^{-1}(\sqrt{\cos\alpha})-\tan^{-1}(\sqrt{\cos\alpha})=\tan^{-1}\left(\frac{1}{\sqrt{\cos\alpha}}\right)-\tan^{-1}(\sqrt{\cos\alpha})=\tan^{-1}\left(\frac{\frac{1}{\sqrt{\cos\alpha}}-\sqrt{\cos\alpha}}{1+1}\right)=\tan^{-1}\left(\frac{1-\cos\alpha}{2\sqrt{\cos\alpha}}\right).$$

So $\tan x=\dfrac{1-\cos\alpha}{2\sqrt{\cos\alpha}}$, giving (with perpendicular $=1-\cos\alpha$, base $=2\sqrt{\cos\alpha}$, hypotenuse $=1+\cos\alpha$):

$$\sin x=\frac{1-\cos\alpha}{1+\cos\alpha}=\frac{2\sin^2\frac{\alpha}{2}}{2\cos^2\frac{\alpha}{2}}=\tan^2\frac{\alpha}{2}.$$

Solutions · 1