# Exact Value of tan 7½°

How to find the exact value of tan 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, tan 7½° = $$\frac{sin 7½°}{cos 7½°}$$

= $$\frac{2 sin^{2} 7½°}{2 cos 7½° sin 7½°}$$

= $$\frac{1 - cos 15°}{sin 15°}$$

= $$\frac{1 - \frac{√3 + 1}{2√2}}{\frac{√3 - 1}{2√2}}$$

= $$\frac{2√2 - √3 - 1}{√3 - 1}$$

= $$\frac{(2√2 - √3 - 1)(√3 + 1)}{(√3 - 1)(√3 + 1)}$$

= $$\frac{2√6 - 3 - √3 + 2√2 - √3 - 11}{2}$$

= √6 - √3 + √2 - 2

Therefore, tan 7½° = √6 - √3 + √2 - 2

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