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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