We will learn to find the exact value of sin 36 degrees using the formula of multiple angles.
How to find exact value of sin 36°?
Let A = 18°
Therefore, 5A = 90°
⇒ 2A + 3A = 90˚
⇒ 2θ = 90˚ - 3A
Taking sine on both sides, we get
sin 2A = sin (90˚ - 3A) = cos 3A
⇒ 2 sin A cos A = 4 cos\(^{3}\) A - 3 cos A
⇒ 2 sin A cos A - 4 cos\(^{3}\) A + 3 cos A = 0
⇒ cos A (2 sin A - 4 cos\(^{2}\) A + 3) = 0
Dividing both sides by cos A = cos 18˚ ≠ 0, we get
⇒ 2 sin θ - 4 (1 - sin\(^{2}\) A) + 3 = 0
⇒ 4 sin\(^{2}\) A + 2 sin A - 1 = 0, which is a quadratic in sin A
Therefore, sin θ = \(\frac{-2 \pm \sqrt{- 4 (4)(-1)}}{2(4)}\)
⇒ sin θ = \(\frac{-2 \pm \sqrt{4 + 16}}{8}\)
⇒ sin θ = \(\frac{-2 \pm 2 \sqrt{5}}{8}\)
⇒ sin θ = \(\frac{-1 \pm \sqrt{5}}{4}\)
Now sin 18° is positive, as 18° lies in first quadrant.
Therefore, sin 18° = sin A = \(\frac{-1 \pm \sqrt{5}}{4}\)
Now, cos 36° = cos 2 ∙ 18°
⇒ cos 36° = 1 - 2 sin\(^{2}\) 18°
⇒ cos 36° = 1 - 2\((\frac{\sqrt{5} - 1}{4})^{2}\)
⇒ cos 36° = \(\frac{16 - 2(5 + 1 - 2\sqrt{5})}{16}\)
⇒ cos 36° = \(\frac{1 + 4\sqrt{5}}{16}\)
⇒ cos 36° = \(\frac{\sqrt{5} + 1}{4}\)
Therefore, sin 36° = \(\sqrt{1 - cos^{2} 36°}\),[Taking sin 36° is positive, as 36° lies in first quadrant, sin 36° > 0]
⇒ sin 36° = \(\sqrt{1 - (\frac{\sqrt{5} + 1}{4})^{2}}\)
⇒ sin 36° = \(\sqrt{\frac{16 - (5 + 1 + 2\sqrt{5})}{16}}\)
⇒ sin 36° = \(\sqrt{\frac{10 - 2\sqrt{5}}{16}}\)
⇒ sin 36° = \(\frac{\sqrt{10 - 2\sqrt{5}}}{4}\)
Therefore, sin 36° = \(\frac{\sqrt{10 - 2\sqrt{5}}}{4}\)
11 and 12 Grade Math
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