Question
Evaluate the following:
$\tan\frac{1}{2}\Big(\cos^{-1}\frac{\sqrt5}{3}\Big)$
$\tan\frac{1}{2}\Big(\cos^{-1}\frac{\sqrt5}{3}\Big)$
✓
Answer
Let $\frac{1}{2}\sin^{-1}\frac{3}{4}=\text{x}$
$ \sin^{-1}\frac{3}{4}=2\text{x}$
$ \sin2\text{x}=\frac{3}{4}$
$\cos2\text{x}=\frac{\sqrt7}{4}$
$\tan\Big(\frac{1}{2}\sin^{-1}\frac{3}{4}\Big)$
$=\tan\text{x}$
$ =\sqrt{\frac{1-\cos2\text{x}}{1+\cos2\text{x}}}$
$=\sqrt{\frac{1-\frac{\sqrt7}{4}}{1+\frac{\sqrt7}{4}}}$
$=\sqrt{\frac{4-\sqrt7}{4+\sqrt7}}$
$=\sqrt{\frac{\big(4-\sqrt7\big)\big(4-\sqrt7\big)}{\big(4+\sqrt7\big)\big(4-\sqrt7\big)}}$
$=\sqrt{\frac{\big(4-\sqrt7\big)^2}{9}}$
$ =\frac{4-\sqrt7}{3}$
$ \sin^{-1}\frac{3}{4}=2\text{x}$
$ \sin2\text{x}=\frac{3}{4}$
$\cos2\text{x}=\frac{\sqrt7}{4}$
$\tan\Big(\frac{1}{2}\sin^{-1}\frac{3}{4}\Big)$
$=\tan\text{x}$
$ =\sqrt{\frac{1-\cos2\text{x}}{1+\cos2\text{x}}}$
$=\sqrt{\frac{1-\frac{\sqrt7}{4}}{1+\frac{\sqrt7}{4}}}$
$=\sqrt{\frac{4-\sqrt7}{4+\sqrt7}}$
$=\sqrt{\frac{\big(4-\sqrt7\big)\big(4-\sqrt7\big)}{\big(4+\sqrt7\big)\big(4-\sqrt7\big)}}$
$=\sqrt{\frac{\big(4-\sqrt7\big)^2}{9}}$
$ =\frac{4-\sqrt7}{3}$
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