# Exact Value of tan 15°

How to find the exact value of tan 15° using the value of sin 30°?

Solution:

For all values of the angle A we know that, (sin $$\frac{A}{2}$$ + cos $$\frac{A}{2}$$)$$^{2}$$  = sin$$^{2}$$ $$\frac{A}{2}$$  + cos$$^{2}$$ $$\frac{A}{2}$$  + 2 sin $$\frac{A}{2}$$ cos $$\frac{A}{2}$$ = 1 + sin A

Therefore, sin $$\frac{A}{2}$$  + cos $$\frac{A}{2}$$ = ± √(1 + sin A), [taking square root on both the sides]

Now, let A = 30° then, $$\frac{A}{2}$$ = $$\frac{30°}{2}$$ = 15° and from the above equation we get,

sin 15° + cos 15° = ± √(1 + sin 30°)                       ….. (i)

Similarly, for all values of the angle A we know that, (sin $$\frac{A}{2}$$ - cos $$\frac{A}{2}$$)$$^{2}$$  = sin$$^{2}$$ $$\frac{A}{2}$$ + cos$$^{2}$$ $$\frac{A}{2}$$ - 2 sin $$\frac{A}{2}$$ cos $$\frac{A}{2}$$  = 1 - sin A

Therefore, sin $$\frac{A}{2}$$  - cos $$\frac{A}{2}$$  = ± √(1 - sin A), [taking square root on both the sides]

Now, let A = 30° then, $$\frac{A}{2}$$ = $$\frac{30°}{2}$$ = 15° and from the above equation we get,

sin 15° - cos 15° = ± √(1 - sin 30°)                  …… (ii)

Clearly, sin 15° > 0 and cos 15˚ > 0

Therefore, sin 15° + cos 15° > 0

Therefore, from (i) we get,

sin 15° + cos 15° = √(1 + sin 30°)                                  ..... (iii)

Again, sin 15° - cos 15° = √2 ($$\frac{1}{√2}$$ sin 15˚ - $$\frac{1}{√2}$$ cos 15˚)
or, sin 15° - cos 15° = √2 (cos 45° sin 15˚ - sin 45° cos 15°)

or, sin 15° - cos 15° = √2 sin (15˚ - 45˚)

or, sin 15° - cos 15° = √2 sin (- 30˚)

or, sin 15° - cos 15° = -√2 sin 30°

or, sin 15° - cos 15° = -√2 ∙ $$\frac{1}{2}$$

or, sin 15° - cos 15° = - $$\frac{√2}{2}$$

Thus, sin 15° - cos 15° < 0

Therefore, from (ii) we get, sin 15° - cos 15°= -√(1 - sin 30°)      ..... (iv)

Now, adding (iii) and (iv) we get,

2 sin 15° = $$\sqrt{1 + \frac{1}{2}} - \sqrt{1 - \frac{1}{2}}$$

2 sin 15° = $$\frac{\sqrt{3} - 1}{\sqrt{2}}$$

sin 15° = $$\frac{\sqrt{3} - 1}{2\sqrt{2}}$$

Therefore, sin 15° = $$\frac{\sqrt{3} - 1}{2\sqrt{2}}$$

Similarly, subtracting (iv) from (iii) we get,

2 cos 15° = $$\sqrt{1 + \frac{1}{2}} + \sqrt{1 - \frac{1}{2}}$$

2 cos 15° = $$\frac{\sqrt{3} + 1}{\sqrt{2}}$$

cos 15° = $$\frac{\sqrt{3} + 1}{2\sqrt{2}}$$

Therefore, cos 15° = $$\frac{\sqrt{3} + 1}{2\sqrt{2}}$$

Now, tan 15° = $$\frac{sin 15°}{cos 15°}$$

= $$\frac{\frac{\sqrt{3} - 1}{2\sqrt{2}}}{\frac{\sqrt{3} + 1}{2\sqrt{2}}}$$

= $$\frac{\sqrt{3} - 1}{\sqrt{3} + 1}$$

Thus, tan 15° = $$\frac{\sqrt{3} - 1}{\sqrt{3} + 1}$$

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