# Exact Value of tan 27°

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° = $$\frac{\sqrt{5 + \sqrt{5}} - \sqrt{3 - \sqrt{5}}}{\sqrt{5 + \sqrt{5}} + \sqrt{3 - \sqrt{5}}}$$

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