Question

fast plz...... trigonometric values in fractional form........no spams plz...

Answer 1

I skipped some small calculations, hope you can understand •_^

^_^ Trigonometry ^_^

1) We know that,
$\tan(2 \theta) = \frac{2 \tan( \theta) }{1 - {tan}^{2} ( \theta)}$

2) Now, 15°

We also know that,
$\tan(30 \degree) = \frac{1}{ \sqrt{3} }$
Using above Identity of Double angles :
$\tan(2 \times 15 \degree) = \frac{2tan 15 \degree }{1 - {tan}^{2}( 15\degree) }$
Let tan(15°) = x
$= > \frac{1}{ \sqrt{3} } = \frac{2x}{1 - {x}^{2} } \\ = > {x}^{2} + 2 \sqrt{3} x - 1 = 0 \\ = > x = \frac{ - 2 \sqrt{3} + 4 }{2} = 2 - \sqrt{3}$
Negative value of quadratic formula is rejected ,since tan(a) is positive when 'a' is acute angle.

Therefore, Value of tan(15°) = 2-√3

Since, We got value of one trigonometric angle and we can easily find other trigonometric ratios.
^_^ Try to find other trigonometric ratio as your own.

You will get,
$\sin(15 \degree) = \frac{ \sqrt{3} - 1 }{2 \sqrt{2} } \\ \cos(15 \degree) = \frac{ \sqrt{3} + 1}{2 \sqrt{2} }$

3)
$\boxed {22.5 \degree}$
We know that,
$\tan(45 \degree) = \frac{2 \tan(22.5 \degree) }{ 1 - {tan}^{2}22.5 \degree }$
Let tan(22.5°) = x,
$= > 1 = \frac{2x}{1 - {x}^{2} } \\ = > 2x = 1 - {x}^{2} \\ = > {x}^{2} + 2x - 1 = 0 \\ = > x = \frac{ - 2 + 2 \sqrt{2} }{2} = \sqrt{2} - 1$

Negative value of quadratic formula is rejected ,since tan(a) is positive when 'a' is acute angle.
^.^
Now, we can easily find other trigonometric ratios.

We get,
$\tan(22.5 \degree) = \sqrt{2} - 1 \\ \sin(22.5 \degree) = \frac{ \sqrt{2} - 1 }{ \sqrt{4 - 2 \sqrt{2} } } \\ \cos(22.5 \degree) = \frac{1}{ \sqrt{4 - 2 \sqrt{2} } }$

4)
$\boxed {7.5 \degree}$
We also know that,
$\tan(15 \degree) = \frac{2 \tan(7.5 \degree) }{ 1 - {tan}^{2}7.5 \degree }$

Let tan(7.5°) = x :
$2 - \sqrt{3} = \frac{2x}{1 - {x}^{2} } \\ = > {x}^{2} (2 - \sqrt{3} ) + 2x - (2 - \sqrt{3} ) = 0 \\ = > x = \frac{2 \sqrt{2 - \sqrt{3} } - 1}{(2 - \sqrt{3} )}$
Negative value of quadratic formula is rejected ,since tan(a) is positive when 'a' is acute angle.

Hence,
$\tan(7.5 \degree) = \frac{2 \sqrt{2 - \sqrt{3} } - 1 }{(2 - \sqrt{3} )}$
We Can easily find other trigonometric ratios.
Like :
$\sin(7.5 \degree) = \frac{2 \sqrt{2 - \sqrt{3} } - 1}{16 - 8 \sqrt{3} - 4 \sqrt{2 - \sqrt{3} } } \\ \cos(7.5 \degree) = \frac{2 - \sqrt{3} }{16 - 8 \sqrt{3} - 4 \sqrt{2 - \sqrt{3} } }$

Hope You Understand