We will learn to find the exact value of cos 36 degrees using the formula of multiple angles.
How to find exact value of cos 54°?
Solution:
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° = \(\frac{\sqrt{10 - 2\sqrt{5}}}{4}\)
Now cos 54° = cos (90° - 36°) = sin 36° = \(\frac{\sqrt{10 - 2\sqrt{5}}}{4}\)
Therefore, cos 54° = \(\frac{\sqrt{10 - 2\sqrt{5}}}{4}\)
11 and 12 Grade Math
From Exact Value of cos 54° to HOME PAGE
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.
Nov 23, 24 03:45 PM
Nov 23, 24 03:14 PM
Nov 23, 24 02:51 PM
Nov 23, 24 12:22 AM
Nov 22, 24 12:34 AM
New! Comments
Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.