Exact Value of cos 27°

We will learn to find the exact value of cos 27 degrees using the formula of submultiple angles.


How to find the exact value of cos 27°?

Solution: 

We have, (sin 27° + cos 27°)\(^{2}\) = sin\(^{2}\) 27° + cos\(^{2}\) 27° + 2 sin 27° cos 27°

⇒ (sin 27° + cos 27°)\(^{2}\) = 1 + sin 2 ∙ 27°

⇒ (sin 27° + cos 27°)\(^{2}\) = 1 + sin 54° 

⇒ (sin 27° + cos 27°)\(^{2}\) = 1 + sin (90° - 36°)

⇒ (sin 27° + cos 27°)\(^{2}\) = 1 + cos 36° 

⇒ (sin 27° + cos 27°)\(^{2}\) = 1+ \(\frac{√5 +1}{4}\)

⇒ (sin 27° + cos 27°)\(^{2}\) = \(\frac{1}{4}\) ( 5 + √ 5)

Therefore,  sin 27° + cos 27° = \(\frac{1}{2}\sqrt{5 + \sqrt{5}}\) …………….….(i)

[Since, sin 27° > 0 and cos 27° > 0)

Similarly, we have, 

(sin 27° - cos 27°)\(^{2}\) = 1 - cos 36°

⇒ (sin 27° - cos 27°)\(^{2}\) = 1 - \(\frac{√5 + 1}{4}\)

⇒ (sin 27° - cos 27°)\(^{2}\) = \(\frac{1}{4}\) (3 - √5  )

Therefore, sin 27° - cos 27° = ± \(\frac{1}{2}\sqrt{3 - \sqrt{5}}\) …………….….(ii)

Now, sin 27° - cos 27° = √2 (\(\frac{1}{√2}\) sin 27˚ - \(\frac{1}{√2}\) cos 27°)

                               = √2 (cos 45° sin 27° - sin 45° cos 27°)

                               = √2 sin (27° - 45°)

                               = -√2 sin 18° < 0

Therefore, from (ii) we get,

sin 27° - cos 27° = -\(\frac{1}{2}\sqrt{3 - \sqrt{5}}\) …………….….(iii)

Now, subtracting (iii) and (i) we get,

2 cos 27° = \(\frac{1}{2}\sqrt{5 + \sqrt{5}}\) + \(\frac{1}{2}\sqrt{3 - \sqrt{5}}\)

⇒ cos 27° = \(\frac{1}{4}(\sqrt{5 + \sqrt{5}} + \sqrt{3 - \sqrt{5}})\)

Therefore, cos 27° = \(\frac{1}{4}(\sqrt{5 + \sqrt{5}} + \sqrt{3 - \sqrt{5}})\)      

● Submultiple Angles





11 and 12 Grade Math

From Exact Value of cos 27° 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.



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.

Share this page: What’s this?