Question

Evaluate cos² 72° - sin² 54°

Answer 1

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

Answer 2

Answer:

Step-by-step explanation:

Formula used:

sinA=cos(90-A)

cosA=sin(90-A)

cos² 72° - sin² 54°

$={sin}^2(18)-{cos}^2(36)$

$={(\frac{\sqrt{5}-1}{4})}^2-{(\frac{\sqrt{5}+1}{4})}^2\\=(\frac{5+1-2\sqrt{5}}{16})-(\frac{5+1+2\sqrt{5}}{16})\\=(\frac{6-2\sqrt{5}}{16})-(\frac{6+2\sqrt{5}}{16})\\=\frac{6-2\sqrt{5}-6-2\sqrt{5}}{16}$

$=\frac{-4\sqrt{5}}{16}\\=\frac{</strong><strong>-</strong><strong>\sqrt{5}}{4}$