Resolved: "Prove that sin2(π8+x2)−sin2(π8−x2)=12sin⁡x"

aortiH

Answered question

2021-09-14

Prove that

\sin^{2} (\frac{π}{8} + \frac{x}{2}) - \sin^{2} (\frac{π}{8} - \frac{x}{2}) = \frac{1}{\sqrt{2}} \sin x

FieniChoonin

Skilled2021-09-15Added 102 answers

We use Trigonometry identities , double angle formulas to prove this Question

\sin^{2} (\frac{π}{8} + \frac{x}{2}) - \sin^{2} (\frac{π}{8} - \frac{x}{2})

\Rightarrow [\sin (\frac{π}{8} + \frac{x}{2}) + \sin (\frac{π}{8} - \frac{x}{2}] [\sin (\frac{π}{8} + \frac{x}{2}) - \sin (\frac{π}{8} - \frac{x}{2})]

\Rightarrow [2 \sin (\frac{π}{8}) \cos (\frac{x}{2})] [2 \cos (\frac{π}{8}) \sin (\frac{x}{2})]

\Rightarrow (2 \sin (\frac{π}{8}) \cos (\frac{π}{8})) (2 \sin (\frac{π}{2}) \cos (\frac{π}{2}))

\Rightarrow \sin 2 (\frac{π}{8}) \sin 2 (\frac{π}{2})

∵ \sin 2 A = 2 \sin A \cos A

\Rightarrow \sin (\frac{π}{4}) \sin x

\Rightarrow \frac{1}{\sqrt{2}} \sin x

\sin^{2} (\frac{π}{8} + \frac{x}{2}) - \sin^{2} (\frac{π}{8} - \frac{x}{2}) = \frac{1}{\sqrt{2}} \sin x

Jeffrey Jordon

Expert2021-12-24Added 2605 answers

Answer is given below (on video)

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?