How to find the exact value of tan 142½° using the value of sin 15° and cos 15°?
Solution:
For all values of the angle A and B we know that, tan (A + B) = \(\frac{tan A + tan B}{1 - tan A tan B}\),
sin A = 2 sin \(\frac{A}{2}\) cos \(\frac{A}{2}\)
and
cos A = cos\(^{2}\) \(\frac{A}{2}\) – sin\(^{2}\) \(\frac{A}{2}\)
Now tan 142½°
= tan (90 + 52½°)
= - cot 52½°
= \(\frac{-1}{tan 52½°}\)
= \(\frac{-1}{tan (45° + 7½°}\)
= - \(\frac{1 - tan 7½°}{1 + tan 7½°}\)
= - \(\frac{cos 7½° - sin 7½°}{cos 7½° + sin 7½°}\)
= - \(\frac{(cos 7½° - sin 7½°)(cos 7½° - sin 7½°)}{(cos 7½° + sin 7½°)(cos 7½° - sin 7½°)}\)
= - \(\frac{(cos 7½° - sin 7½°)^{2}}{cos^{2} 7½° - sin^{2} 7½°}\)
= - \(\frac{1 - 2 sin 7½° cos 7½°}{cos^{2} 7½° - sin^{2} 7½°}\)
= - \(\frac{1 - sin 15°}{cos 15°}\)
= - \(\frac{1 - sin (45° - 30°)}{cos (45° - 30°)}\)
= - \(\frac{1 - \frac{\sqrt{3} - 1}{2\sqrt{2}}}{\frac{\sqrt{3} + 1}{2\sqrt{2}}}\)
= - \(\frac{2√2 - √3 + 1}{√3 + 1}\)
= - \(\frac{(2√2 - √3 + 1)}{(√3 + 1)}\) \(\frac{(√3 - 1)}{(√3 - 1)}\)
= - \(\frac{(2√2 - √3 + 1)(√3 - 1)}{3 - 1}\)
= - \(\frac{(2√2(√3 - 1) - (√3 - 1)^{2}}{2}\)
= -[√2(√3 - 1) – (2 - √3)]
= -√6 + √2 + 2 - √3
= 2 + √2 - √3 - √6
11 and 12 Grade Math
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