Question

sin square 52 1/2- sin square 22 1/2

Answer 1

Please see the attachment

Answer 2

Answer:

$\sin^2 (52.5)-\sin^2 (22.5)=\frac{3+\sqrt3}{4\sqrt2}$

Step-by-step explanation:

Given : Expression $\sin^2 (52 \frac{1}{2}^\circ)-\sin^2 (22 \frac{1}{2}^\circ)$

To find : Evaluate the expression?

Solution :  

We know, $\sin^2 A-\sin^2 B=\sin(A+B)\sin(A-B)$

$\sin^2 (52.5)-\sin^2 (22.5)=\sin(52.5+22.5)\sin(52.5-22.5)$

$\sin^2 (52.5)-\sin^2 (22.5)=\sin(75)\sin(30)$

$\sin^2 (52.5)-\sin^2 (22.5)=\frac{1}{2}(\sin(45+30))$

Applying identity, $\sin(A+B)=\sin A\cos B+\cos A\sin B$

$\sin^2 (52.5)-\sin^2 (22.5)=\frac{1}{2}(\sin 45\cos 30+\cos 45\sin 30)$

Substitute the values of the function,

$\sin^2 (52.5)-\sin^2 (22.5)=\frac{1}{2}((\frac{1}{\sqrt2})(\frac{\sqrt3}{2})+(\frac{1}{\sqrt2})(\frac{1}{2}))$

$\sin^2 (52.5)-\sin^2 (22.5)=\frac{1}{2}(\frac{\sqrt3}{2\sqrt2})+\frac{1}{2\sqrt2})$

$\sin^2 (52.5)-\sin^2 (22.5)=\frac{1}{2}(\frac{\sqrt3+1}{2\sqrt2})$

$\sin^2 (52.5)-\sin^2 (22.5)=\frac{3+\sqrt3}{4\sqrt2}$

Therefore, $\cos^2 (52.5)-\sin^2 (22.5)=\frac{3-\sqrt3}{4\sqrt2}$