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Trigonometry> inverse-function...

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sin inverse root{2-root3)/4}

manoj jain , 12 Years ago

Grade 12

1 Answers

Jitender Singh

Last Activity: 11 Years ago

Ans:

$t = sin^{-1}\sqrt{\frac{2-\sqrt{3}}{4}}$

$sint = sinsin^{-1}\sqrt{\frac{2-\sqrt{3}}{4}}$

$sint = \sqrt{\frac{2-\sqrt{3}}{4}}$

$sin^{2}t = {\frac{2-\sqrt{3}}{4}}$

$\frac{1-cos2t}{2} = {\frac{2-\sqrt{3}}{4}}$

$2-2cos2t = {\frac{2-\sqrt{3}}{1}}$

$cos2t=\frac{\sqrt{3}}{2}$

$2t = \frac{\pi }{6}$

$t = \frac{\pi }{12}$

Thanks & Regards

Jitender Singh

IIT Delhi

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