Question

The value of tan
7.1/2 is equal to​

Answer 1

$\tan(7.1/2)$

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°)

$\frac{7}{root2}. \frac{root3}{2} + \frac{1}{root2} . \frac{1}{2}$

$\frac{root3}{2root2}. \frac{1}{2root2}$

$\frac{root3 - 1}{2root2}$

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°

$\frac{1}{root2} . \frac{root3}{2} + \frac{1}{root2} . \frac{1}{2}$

$\frac{root3}{2root2} + \frac{1}{2root2}$

$\frac{root3 + 1}{2root2}$

now tan √2 =

$\frac{ \sin(7 \times \frac{1}{2} ) }{ \cos(7 \times \frac{1}{2} ) }$

$\frac{2 ( {sin}^{2}7 \times \frac{1}{2} ) }{2 \cos(7 \times \frac{1}{2} ) \sin(7 \times \frac{1}{2} ) }$

$\frac{1 - \cos(15) }{ \sin(15) }$

$\frac{1 - \binom{root3 + 1 }{2root2} }{ \binom{root3 - 1}{2root2} }$

$\frac{(2root2 - root3 - 1)( - root3 + 1)}{root3 - 1}$

$\frac{2root6 - 3 - root3 \ + 2root2 - root3 - 11}{2}$

= √6 - √3 + √2 - 2

Therefore, tan 7½° = √6 - √3 + √2 - 2

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