Question

"Question31
If A = B = 45° , Verify that : sin (A+B)= sin A cos B + cos A sin B
Chapter6,T-Ratios of particular angles Exercise -6 ,Page number 289"

Answer 1

Hey there! Thanka for the question!

Given,
A = B = 45°

Now,
sin(A + B) = sinA cosB + cosA sinB

Substituting A = 45 , B = 45 °

sin( 45 + 45 ) = sin45 cos45 + cos45 sin45

sin90 = sin45 cos45 + cos45 sin45

We know, Trigonometry ratios of particular angles : sin90 = 1 , cos45 = 1/√2 , sin45 = 1/√2

1 = 1/√2 ( 1/√2 ) + 1/√2 ( 1/√2)

1 = 1/2 + 1/2

1 = 1 .

Both Sides of the equation are equal.

Hence, We proved and verified that sin (A+B)= sin A cos B + cos A sin B holds good for A = B = 45°

Answer 2

Hey mate !!

Here's your answer !!

Given :

1. A = B = 45°

To verify :

Sin ( A + B ) = Sin A . Cos B + Cos A . Sin B

Proof :

It is already given that A and B = 45°.

Hence we have to just substitute them in the equations given to verify them.

So after substituting we get,

Sin ( 45° + 45° ) = Sin 45° . Cos 45° + Cos 45° . Sin 45°

=> Sin ( 90° ) = Sin 45° . Cos 45° + Cos 45° . Sin 45°   -----( Eqn. 1 )

We know that,

Sin 90° = 1

Sin 45° = 1 / √ 2

Cos 45° = 1 / √ 2

Hence we must substitute the above values in Eqn. ( 1 )

So after substituting we get,

= 1 = 1 / √ 2 * 1 / √ 2 + 1 / √ 2 * 1 / √ 2

= 1 = 1 / 2 + 1 / 2

=> 1 = 1

=> LHS = RHS

Hence verified !!

Hope it helps you mate !!

Cheers !!