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