Question

2sin^2(45°-A)=1+sin2A prove this

Answer 1

Step-by-step explanation:

2 sin square( 45°-A)

(therefore;

by trigonometry ratios

sin45°=1)

=2sin2(1-A)

=2sin(1-A)

=1+sin2A

hence proved

I HOPE THIS WILL HELP YOU

Answer 2

Trigonometry

Before we solve the problem, we must know some formulae:

$\quad cos2\theta=1-2\:sin^{2}\theta$

$\quad cos(90^{\circ}-\theta)=sin\theta$

We have to prove:

$\quad 2\:sin^{2}(45^{\circ}-A)=1-sin2A$

Proof.

Now, $2\:sin^{2}(45^{\circ}-A)$

$=1-cos\{2(45^{\circ}-A)\}$

$=1-cos(90^{\circ}-2A)$

$=1-sin2A$

$\therefore \color{blue}2\:sin^{2}(45^{\circ}-A)=1-sin2A$

This completes the proof.

Another approach:

Now, $2\:sin^{2}(45^{\circ}-A)$

$=2\:(sin45^{\circ}\:cosA-cos45^{\circ}\:sinA)^{2}$

$=2\:\left(\frac{1}{\sqrt{2}}\:cosA-\frac{1}{\sqrt{2}}\:sinA\right)^{2}$

$=2*\frac{1}{2}*(cosA-sinA)^{2}$

$=cos^{2}A+sin^{2}A-2\:sinA\:cosA$

$=1-sin2A$

$\Rightarrow \color{blue}2\:sin^{2}(45^{\circ}-A)=1-sin2A$

This completes the proof.