Question

if A=580° then prove that sinA/2= 1/2(- root over of 1+sinA - root over of 1- sinA)​

Answer 1

Answer:

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Answer 2

If $A=580^\circ$, then prove that $\sin \frac{A}{2}=\frac{1}{2}(-\sqrt{1+\sin A}-\sqrt{1-\sin A})$

The proof is given below:

Given

$\angle A=580^\circ$

To Prove that

$\sin \frac{A}{2}=\frac{1}{2}(-\sqrt{1+\sin A}-\sqrt{1-\sin A})$

RHS

$=\frac{1}{2}(-\sqrt{1+\sin A}-\sqrt{1-\sin A})$

$=-\frac{1}{2}(\sqrt{1+\sin A}+\sqrt{1-\sin A})$

$=-\frac{1}{2}(\sqrt{\sin^2\frac{A}{2}+\cos^2\frac{A}{2}+2\sin \frac{A}{2}\cos\frac{A}{2}}+\sqrt{\sin^2\frac{A}{2}+\cos^2\frac{A}{2}-2\sin \frac{A}{2}\cos\frac{A}{2}})$

$=-\frac{1}{2}(\sqrt{(\sin\frac{A}{2}+\cos\frac{A}{2})^2}+\sqrt{(\sin\frac{A}{2}+\cos\frac{A}{2})^2}$

$=-\frac{1}{2}(|\sin\frac{A}{2}+\cos\frac{A}{2}|+|\sin\frac{A}{2}-\cos\frac{A}{2}|)$    (Note: We know that $\sqrt{x^2}=|x|$ and |x| = x if x ≥ 0; |x| = -x if x < 0)

∵ $\angle A=580^\circ$

∴ $\angle A/2=290^\circ$

Therefore,

RHS

$=-\frac{1}{2}(|\sin 290^\circ+\cos 290^\circ|+|\sin 290^\circ-\cos 290^\circ|)$

$=-\frac{1}{2}(|\sin (360^\circ-70^\circ)+\cos (360^\circ-70^\circ)|+|\sin (360^\circ-70^\circ)-\cos (360^\circ-70^\circ)|)$

$=-\frac{1}{2}(|-\sin 70^\circ+\cos 70^\circ|+|-\sin 70^\circ-\cos 70^\circ|)$

$=-\frac{1}{2}(-(-\sin 70^\circ+\cos 70^\circ)-(-\sin 70^\circ-\cos 70^\circ))$

$=\frac{1}{2}(-\sin 70^\circ+\cos 70^\circ)+(-\sin 70^\circ-\cos 70^\circ))$

$=-\frac{1}{2}(2\sin 70^\circ)$

$=-\sin 70^\circ$

Now,

LHS

$=\sin\frac{A}{2}$

$=\sin290^\circ$

$=\sin(360^\circ-70^\circ)$

$=-\sin 70^\circ$

Therefore, LHS = RHS                                            (Proved)

Hope this answer is helpful.

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