How to find the exact value of cos 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, 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° = \(\frac{\sqrt{3} + 1}{2\sqrt{2}}\)
11 and 12 Grade Math
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