How to find the exact value of sin 7½° using the value of cos 15°?
Solution:
7½° lies in the first quadrant.
Therefore, sin 7½° is positive.
For all values of the angle A we know that, cos (α - β) = cos α cos β + sin α sin β.
Therefore, cos 15° = cos (45° - 30°)
cos 15° = cos 45° cos 30° + sin 45° sin 30°
= \(\frac{1}{√2}\)∙\(\frac{√3}{2}\) + \(\frac{1}{√2}\)∙\(\frac{1}{2}\)
= \(\frac{√3}{2√2}\) + \(\frac{1}{2√2}\)
= \(\frac{√3 + 1}{2√2}\)
Again for all values of the angle A we know that, cos A = 1 - 2 sin\(^{2}\)\(\frac{A}{2}\)
⇒
1 - cos A = 2 sin\(^{2}\) \(\frac{A}{2}\)
⇒
2 sin\(^{2}\) \(\frac{A}{2}\) = 1 - cos A
⇒
2 sin\(^{2}\) 7½˚ = 1 - cos 15°
⇒
sin\(^{2}\) 7½˚ = \(\frac{1 - cos 15°}{2}\)
⇒ sin\(^{2}\) 7½˚ = \(\frac{1 - \frac{√3 + 1}{2√2}}{2}\)
⇒
sin\(^{2}\) 7½˚ = \(\frac{2√2 - √3 - 1}{4√2}\)
⇒
sin 7½˚ = \(\sqrt{\frac{4 - √6 - √2}{8}}\), [Since sin 7½° is positive]
⇒ sin 7½˚ = \(\frac{\sqrt{4 - √6 - √2}}{2√2}\)
Therefore, sin 7½˚ = \(\frac{\sqrt{4 - √6 - √2}}{2√2}\)
11 and 12 Grade Math
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