Resolved: "Verify the identity. sin2t2=tan⁡t−sin⁡t2tan⁡t"

floymdiT

Answered question

2021-09-14

Verify the identity.

\sin^{2} \frac{t}{2} = \frac{\tan t - \sin t}{2 \tan t}

rogreenhoxa8

Skilled2021-09-15Added 109 answers

Trigonometry:

\tan t = \frac{2 \tan \frac{t}{2}}{1 - \tan^{2} \frac{t}{2}}

\sin t = \frac{2 \tan \frac{t}{2}}{1 + \tan^{2} \frac{t}{2}}

\frac{\tan t - \sin t}{2 \tan t} = \frac{\frac{2 \tan \frac{t}{2}}{1 - \tan^{2} \frac{t}{2}} - \frac{2 \tan \frac{t}{2}}{1 + \tan^{2} \frac{t}{2}}}{2 \frac{2 \tan \frac{t}{2}}{1 - \tan^{2} \frac{t}{2}}}

= \frac{2 \tan \frac{t}{2} + 2 \tan^{3} \frac{t}{2} - 2 \tan \frac{t}{2} + 2 \tan^{3} \frac{t}{2}}{(1 - \tan^{2} \frac{t}{2}) (1 + \tan^{2} \frac{t}{2})} \cdot \frac{1 - \tan^{2} \frac{t}{2}}{4 \tan \frac{t}{2}}

= \frac{4 \tan^{3} \frac{t}{2}}{4 \tan \frac{t}{2} \cdot (1 + \tan^{2} \frac{t}{2})}

= \frac{\tan^{2} \frac{t}{2}}{\sec^{2} \frac{t}{2}}

= \frac{\sin^{2} \frac{t}{2}}{\cos^{2} \frac{t}{2}} \cos^{2} \frac{t}{2}

= \sin^{2} \frac{t}{2}

Proved

Jeffrey Jordon

Expert2021-12-10Added 2605 answers

Answer is given below (on video)

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