Question

Prove the following:

$\cos 36 ° = \frac{\sqrt{5} + 1}{4}$

Accepted Answer

Let, $\cos 36 ° = \frac{\sqrt{5} + 1}{4}$

$\Rightarrow 5 θ = 90 °$

$∴ 2 θ + 3 θ = 90 °$

$\Rightarrow 2 θ = 90 ° - 3 θ$

$\Rightarrow \sin 2 θ = \sin (90 ° - 3 θ) = \cos 3 θ$

$\Rightarrow 2 sinθ cosθ = 4 \cos^{3} θ - 3 cosθ$ $[∵ \sin 2 A = 2 \sin A \cos A & \cos 3 A = 4 \cos^{3} A - 3 \cos A]$

$\Rightarrow 2 sinθcosθ - 4 \cos^{3} θ + 3 cosθ = 0$

$\Rightarrow cosθ (2 sinθ - 4 \cos^{2} θ + 3) = 0$

$\Rightarrow 2 sinθ - 4 \cos^{2} θ + 3 = 0$

$\Rightarrow 2 sinθ - 4 (1 - \sin^{2} θ) + 3 = 0$

$\Rightarrow 2 sinθ - 4 + 4 \sin^{2} θ + 3 = 0$

$\Rightarrow 4 \sin^{2} θ + 2 sinθ - 1 = 0$

Roots of a quadratic equation $a x^{2} + b x + c$ are $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ .

$\Rightarrow sinθ = \frac{- 2 \pm \sqrt{2^{2} - 4 \times 4 \times (- 1)}}{2 \times 4}$

$\Rightarrow sinθ = \frac{- 2 \pm \sqrt{20}}{8}$

$\Rightarrow sinθ = \frac{\sqrt{5} \pm 1}{4}$

$∴ \sin 18 ° = \frac{\sqrt{5} - 1}{4}$ $[∵ \sin 18 ° > 0]$

We know that, $\cos 2 θ = 1 - 2 \sin^{2} θ$

$\Rightarrow \cos 36 ° = 1 - 2 \sin^{2} 18 °$

$\Rightarrow \cos 36 ° = 1 - 2 {(\frac{\sqrt{5} - 1}{4})}^{2} = 1 - (\frac{5 + 1 - 2 \sqrt{5}}{8})$

$∴ \cos 36 ° = \frac{\sqrt{5} + 1}{4}$

Hence, proved.

Maharashtra Board Solutions for Chapter: Trigonometry - II, Exercise 6: MISCELLANEOUS EXERCISE - 3

Maharashtra Board Mathematics Solutions for Exercise - Maharashtra Board Solutions for Chapter: Trigonometry - II, Exercise 6: MISCELLANEOUS EXERCISE - 3

Questions from Maharashtra Board Solutions for Chapter: Trigonometry - II, Exercise 6: MISCELLANEOUS EXERCISE - 3 with Hints & Solutions

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