Exact Value of sin 27°

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


How to find the exact value of sin 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, adding (i) and (iii) we get,

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

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

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

 Submultiple Angles





11 and 12 Grade Math

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