Question

prove that sin square (52×1/2) °-sin square (22×1/2)°=√3+1/4√2 ​

Answer 1

$\huge \bf \red{ǫᴜᴇsᴛɪᴏɴ}$

Prove that sin square (52×1/2) °-sin square (22×1/2)°=√3+1/4√2 .

$\huge \bf \red{ᴀɴsᴡᴇʀ}$

$sin2(52.5)−sin2(22.5)= \frac{3 + \sqrt{3} }{4 \sqrt{2} } \\ \\ </p><p>$

Step-by-step explanation:

➤ɢɪᴠᴇɴ :-

$Expression  = \sin^2 (52 \frac{1}{2}^\circ)-\sin^2 (22 \frac{1}{2}^\circ)sin2(5221∘)−sin2(2221∘)$

➤ To Find

Evaluate the expression :-

$We know, \sin^2 A-\sin^2 B=\sin(A+B)\sin(A-B)sin2A−sin2B=sin(A+B)sin(A−B)</p><p></p><p>\sin^2 (52.5)-\sin^2 (22.5)=\sin(52.5+22.5)\sin(52.5-22.5)sin2(52.5)−sin2(22.5)=sin(52.5+22.5)sin(52.5−22.5)</p><p></p><p>\sin^2 (52.5)-\sin^2 (22.5)=\sin(75)\sin(30)sin2(52.5)−sin2(22.5)=sin(75)sin(30)</p><p></p><p>\sin^2 (52.5)-\sin^2 (22.5)=\frac{1}{2}(\sin(45+30))sin2(52.5)−sin2(22.5)=21(sin(45+30))</p><p></p><p>Applying identity, \sin(A+B)=\sin A\cos B+\cos A\sin Bsin(A+B)=sinAcosB+cosAsinB</p><p></p><p>\sin^2 (52.5)-\sin^2 (22.5)=\frac{1}{2}(\sin 45\cos 30+\cos 45\sin 30)sin2(52.5)−sin2(22.5)=21(sin45cos30+cos45sin30)</p><p></p><p>Substitute the values of the function,</p><p></p><p>\sin^2 (52.5)-\sin^2 (22.5)=\frac{1}{2}((\frac{1}{\sqrt2})(\frac{\sqrt3}{2})+(\frac{1}{\sqrt2})(\frac{1}{2}))sin2(52.5)−sin2(22.5)=21((21)(23)+(21)(21))</p><p></p><p>\sin^2 (52.5)-\sin^2 (22.5)=\frac{1}{2}(\frac{\sqrt3}{2\sqrt2})+\frac{1}{2\sqrt2})sin2(52.5)−sin2(22.5)=21(223)+221)</p><p></p><p>\sin^2 (52.5)-\sin^2 (22.5)=\frac{1}{2}(\frac{\sqrt3+1}{2\sqrt2})sin2(52.5)−sin2(22.5)=21(223+1)</p><p></p><p>\sin^2 (52.5)-\sin^2 (22.5)=\frac{3+\sqrt3}{4\sqrt2}sin2(52.5)−sin2(22.5)=423+3</p><p></p><p>Therefore, \cos^2 (52.5)-\sin^2 (22.5)=\frac{3-\sqrt3}{4\sqrt2}cos2(52.5)−sin2(22.5)=423−3</p><p></p><p>$

Answer 2

Answer:

Here's ur answer

Step-by-step explanation:

Hope it helps u