We will learn to find the exact value of tan 27 degrees using the formula of submultiple angles.
How to find the exact value of tan 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° = \(\frac{1}{4}(\sqrt{5 + \sqrt{5}} - \sqrt{3 - \sqrt{5}})\)…………….….(iv)
Again, 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° = \(\frac{1}{4}(\sqrt{5 + \sqrt{5}} + \sqrt{3 - \sqrt{5}})\)…………….….(v)
Now dividing (iv) by (v) we get,
tan 27° = \(\frac{\sqrt{5 + \sqrt{5}} - \sqrt{3 - \sqrt{5}}}{\sqrt{5 + \sqrt{5}} + \sqrt{3 - \sqrt{5}}}\)
11 and 12 Grade Math
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