Solveeit Logo
Login

Question

Question: \[\cos^{2}48{^\circ} - \sin^{2}12{^\circ} =\]...

cos⁡248∘−sin⁡212∘=\cos^{2}48{^\circ} - \sin^{2}12{^\circ} =

A

5−14\frac{\sqrt{5} - 1}{4}

B

5+18\frac{\sqrt{5} + 1}{8}

C

3−14\frac{\sqrt{3} - 1}{4}

D

3+122\frac{\sqrt{3} + 1}{2\sqrt{2}}

Answer

5+18\frac{\sqrt{5} + 1}{8}

Explanation

Solution

cos⁡2A−sin⁡2B=cos⁡(A+B).cos⁡(A−B)\cos^{2}A - \sin^{2}B = \cos(A + B).\cos(A - B)

∴cos⁡248o−sin⁡212o=cos⁡60o.cos⁡36o\therefore\cos^{2}48^{o} - \sin^{2}12^{o} = \cos{}60^{o}.\cos{}36^{o}

=12(5+14)=5+18.= \frac{1}{2}\left( \frac{\sqrt{5} + 1}{4} \right) = \frac{\sqrt{5} + 1}{8}.