# Exact Value of cot 7½°

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

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