Question

Guys please answer this...I will mark as brainliest....plz dont spam anything...plz...tomorrow exam is there...plz​

Answer 1

Answer:

hey bro the picture is looking blur

Answer 2

Step-by-step explanation:

we have,

$\cos(2 \alpha ) = 1 - 2 \sin^{2} ( \alpha )$

$= > 2 \sin^{2} ( \alpha ) = 1 - \cos( 2\alpha )$

$= > \sin( \alpha ) = \sqrt{ \frac{1 - \cos(2 \alpha ) }{2} } \: ......(i)$

Now, we know that,

$\cos( \alpha - \beta ) = \cos( \alpha ) \cos( \beta ) + \sin( \alpha ) \sin( \beta )$

So,

$\cos( \frac{\pi}{4} - \frac{\pi}{6} ) = \cos( \frac{\pi}{4} ) \cos( \frac{\pi}{6} ) + \sin( \frac{\pi}{4} ) \sin(\frac{\pi}{6})$

$= > \cos( \frac{\pi}{12} ) = \frac{1}{ \sqrt{2} } . \frac{ \sqrt{3} }{2} + \frac{1}{ \sqrt{2} }. \frac{1}{2}$

$= > \cos( \frac{\pi}{12} ) = \frac{ \sqrt{3} + 1 }{2 \sqrt{2} }$

From (i), putting α= π/24, we get,

$\sin( \frac{\pi}{24} ) = \sqrt{ \frac{1 - \cos( \frac{\pi}{12} ) }{2} }$

$= > \sin( \frac{\pi}{24} ) = \sqrt{ \frac{1 - \frac{ \sqrt{3} + 1}{2 \sqrt{2} } }{2} }$

$= > \sin( \frac{\pi}{24} ) = \sqrt{ \frac{2 \sqrt{2} - \sqrt{3} - 1}{4 \sqrt{2} } }$

Here, π/24= (15/2)°