Question

hii friends please don't spam and give answer very fast​

Answer 1

${\red{\huge{\underline{\mathbb{Given:-}}}}}$

$\sin( \alpha ) = \frac{ \sqrt{3} }{2}$

$\cos( \beta = \frac{1}{ \sqrt{2} } )$

and both alpha and beta are in the first quadrant .

${\red{\huge{\underline{\mathbb{Answer:-}}}}}$

1) we know that

$\sin(2 \alpha ) = 2 \sin( \alpha ) \cos( \alpha )$

$= 2 \times \frac{ \sqrt{3} }{2} \times \frac{1}{ 2 }$

$= \frac{ \sqrt{3} }{2}$

2) we know that

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

then ,

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

$\cos( \frac{ \alpha }{2} ) = \sqrt{ \frac{ \cos( \alpha ) + 1 }{2} }$

________________

sin alpha = √3/2

then , cos alpha = 1/2

______________

$\cos( \frac{ \alpha }{2} ) = \sqrt{ \frac{ \frac{1}{2} + 1}{2} }$

$= \sqrt{ \frac{3}{4} } = \frac{ \sqrt{3} }{2}$

3)given

cosß = 1/√2

then ,

sinß = 1/√2

⇾tanß = sinß/cosß = 1

we know that

$\tan(2 \beta ) = \frac{2 \tan( \beta ) }{1 - \tan {}^{2} ( \beta ) }$

it is not defined value , i.e denominator is 0

Answer 2

${\Blue {\huge{\underline{\mathbb{Given:-}}}}}$

$\sin( \alpha ) = \frac{ \sqrt{3} }{2}$

$and \: \cos( \beta ) = \frac{1}{ \sqrt{2} }$

$\huge{\orange{\boxed{\boxed{\boxed{\pink{\underline{\underline{\mathfrak{\red{♡ĂnSwer♡}}}}}}}}}}$

1)

$\sin( \alpha ) = \frac{ \sqrt{3} }{2} \: then \: \cos( \alpha ) = \frac{1}{2}$

we know that

$\sin(2 \alpha ) = 2 \sin( \alpha ) \cos( \alpha )$

put the values

_________________

$\sin( 2\alpha ) = 2 \times \frac{ \sqrt{3} }{2} \times \frac{1}{ 2}$

$= \frac{ \sqrt{3} }{2}$

2) we know that

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

$then \: \cos( \alpha ) = 2 \cos {}^{2} ( \frac{ \alpha }{2} ) - 1$

therefore,

$\cos( \frac{ \alpha }{2} ) = \sqrt{ \frac{ \cos( \alpha ) + }{2} }$

$put \: the \cos( \alpha ) = \frac{1}{ 2}$

$\cos( \frac{ \alpha }{2} ) = \sqrt{ \frac{ \frac{1}{2} + 1 }{2} }$

$= \frac{ \sqrt{3} }{2}$

3) we know that

$\tan( 2\beta ) = \frac{ \tan( \beta ) }{1 - \tan {}^{2} ( \beta ) }$

$\cos( \beta ) = \frac{1}{ \sqrt{2} } \: then \: \sin( \beta ) = \frac{1}{ \sqrt{2} }$

$therefore \: \tan( \beta ) = 1$

put the tanß value in the given formula

u get not defined term

${\red{\huge{\underline{\mathbb{Follow:-}}}}}$