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





11 and 12 Grade Math

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