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UPSC NDA Math Model Paper 8

Question : 115 of 120

Marks: +1, -0

What is the value of cos 36°?

Solution:

Let us consider B = 18°

So, 5B = 90°

⇒ 2B + 3B = 90˚

⇒ 2B = 90˚ - 3B

By taking sine on both sides, we get

Sin 2B = sin (90˚ - 3B) = cos 3B

\Rightarrow 2 \sin B \cos B = 4 \cos^{3} B - 3 \cos B

\Rightarrow 2 \sin B \cos B - 4 \cos^{3} B + 3 \cos B = 0

\Rightarrow \cos B (2 \sin B - 4 \cos^{2} B + 3) = 0

Dividing both sides by

\cos B = \cos 18^{\circ} \neq 0

, we get

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

\Rightarrow 4 \sin^{2} B + 2 \sin B - 1 = 0

This is a quadratic equation in Sine,

So,

\sin B = \frac{-2 \pm \sqrt{4+16}}{2 \times 4} = \frac{-1 \pm \sqrt{5}}{4}

Now since 18° is first quadrant so sin 18° is positive

Therefore,

\sin B = \frac{-1+\sqrt{5}}{4}

Now,

\cos 36^{\circ} = \cos (2 \times 18^{\circ})

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

\Rightarrow \cos 36^{\circ} = 1 - 2\left(\frac{\sqrt{5}-1}{4}\right)^{2}

= \frac{16-2 \times (5+1-2\sqrt{5})}{16} = \frac{4+4\sqrt{5}}{16}

\cos 36^{\circ} = \frac{\sqrt{5}+1}{4}

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