Question

Prove that sin 75°=(√6 √2)/4

Answer 1

Answer:

We have to find Sin 75

Sin75 =Sin(45+30)

Always remember

sin(A+B) = SinA CosB + CosA SinB

take A =45, B=30

Sin(45+30) = Sin45 cos 30 + cos45 sin 30

Remember Sin 45 = cos 45 = 1/√2

cos30= √3/2

sin30= 1/2

sin(45+30)= (1/√2)( √3/2) + (1/√2)( 1/2)

= √3/2√2 + 1/2√2

= √3 +1)/2√2

Multiply by √2 in N and D

=√2( √3 +1)/2√2 √2

= √6 +√2)/4

Hence proved

#answerwithquality #BAL

Answer 2

Answer:

To prove : \sin 75=\frac{\sqrt6+2}{4}

Proof :

Taking LHS,

LHS=\sin 75

LHS=\sin (30+45)

Using identity, \sin(A+B)=\sin A\cos B+\cos A\sin B

LHS=\sin 30\cos 45+\cos 30\sin 45

Substitute the trigonometric values,

LHS=\frac{1}{2}\times \frac{1}{\sqrt2}+\frac{\sqrt3}{2}\times \frac{1}{\sqrt2}

LHS=\frac{1}{2\sqrt2}+\frac{\sqrt3}{2\sqrt2}

LHS=\frac{1+\sqrt3}{2\sqrt2}

Multiply and divide by \sqrt{2}

LHS=\frac{1+\sqrt3}{2\sqrt2}\times \frac{\sqrt2}{\sqrt2}

LHS=\frac{\sqrt2+\sqrt6}{2\times 2}

LHS=\frac{\sqrt2+\sqrt6}{4}

LHS=RHS

Hence proved.