Question

Prove that, sin54°+ cos 72° = 3/
4
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Answer 1

Answer:

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Step-by-step explanation:

Cos(72) = Cos(90-18)=Sin(18) = $\frac{\sqrt[]{5-1}}{4}$   ----(a)

Sin(54) = Sin(90-36) = Cos(36) =  $\frac{\sqrt[]{5+1}}{4}$  ----(b)

$a^{2} + b^{2}$

$\frac{5+1+2 \sqrt[]{5} }{16} + \frac{5+1-2 \sqrt[]{5} }{16}$

$\frac{12}{16}= \frac{3}{4}$