Question

find cos^2 72-sin^2 54 =?

Answer 1

cos 72 = sin(90-72) = sin 18 = $\frac{ \sqrt{5} -1}{4}$

sin 54 = cos(90-54) = cos 36 = $\frac{ \sqrt{5} +1}{4}$

$cos^{2} 72 - sin^{2}54 \\ \\ =(\frac{ \sqrt{5}-1 }{4})^{2} - (\frac{ \sqrt{5}+1 }{4})^{2} \\ \\ =\frac{1}{16}[( \sqrt{5}-1 )^{2} - ( \sqrt{5}+1 )^{2} ] \\ \\ =\frac{1}{16} [-4* \sqrt{5}*1 ] \\ \\ =\frac{- \sqrt{5} }{4}$

Answer 2

HELLO DEAR,

cos²72 = cos²(90 - 18) = sin²18 = (√5 - 1)/4
sin²54 = sin²(90 - 36) = cos²36 = (√5 + 1)/4

now, cos²72 - sin²54

$\bold{\implies \{\frac{\sqrt{5} - 1}{4}\}^2 - \{\frac{\sqrt{5} + 1}{4}\}^2 }$

$\bold{\implies \frac{1}{16} [(\sqrt{5} - 1)^2 - (\sqrt{5} + 1)^2]}$

$\bold{\implies \frac{1}{16}[\{\sqrt{5} - 1 - \sqrt{5} - 1\}\{\sqrt{5} - 1 + \sqrt{5} + 1\}]}$

$\bold{\implies \frac{1}{16}[\{-2*(2\sqrt{5})\}]}$

$\bold{\implies \frac{-\sqrt{5}}{4}}$

I HOPE IT'S HELP YOU DEAR,
THANKS