Question

sin18: OHL then cos36°​

Answer 1

Answer:

Hi your answer is as follows

Step-by-step explanation:

let θ = 18°

5θ = 90°

2θ+3θ = 90°

2θ = 90 - 3θ

applying sine on both sides,

sin(2θ) = sin (90-3θ)    ------------------------ (sin 90 - θ = cosθ)

sin(2θ) = -cos3θ           -------------(sin2θ = 2sinθcosθ) ; (cos3θ = $4cos^3 - 3cos$)

2sinθcosθ = $-4cos^3 + 3cos$       ==========> (minus (-) is outside)

2sinθ = $-4cos^2 + 3$

2 sinθ = -4 + 4$sin^2$θ + 3          -------------------------($cos^2 = 1-sin^2$)

$4 sin^2 -2sin -1 =0$

on using shridaracharya's formula,          -------------------- [$\frac{-b+-\sqrt{b^2 - 4ac} }{2a}$]

θ = $\frac{\sqrt{5} -1}{4}$

$\frac{-1-\sqrt{5} }{4}$ is rejected as angles in 1st quadrant are positive

sin 18° = $\frac{\sqrt{5} -1}{4}$

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As θ  = 18°

we can write cos(36°) as cos 2(18°)

which is in the form of cos 2θ  = $1-sin^2$θ

Therefore,

cos2θ  =                           ---------------- (cos2θ = $1-2sin^2$)

$1- 2*[\frac{\sqrt{5}-1 }{4} ]^2\\= \frac{\sqrt{5}+1}{4}$    

cos 36° = $\frac{\sqrt{5} +1}{4}$

Hope it helps,

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