Question

"Question36
If A = 60° and B = 30° , verify that : sin (A - B) = sinA 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 = 60° , B = 30°

Now,
sin(A - B) = sinA cos B - cos A sin B.

Substituting A = 60 , B = 30°

sin(60 - 30 ) = sin60cos30- cos 60sin 30



We know, Trigonometry ratios of particular angles : sin90 = 1 , sin30 = 1/2 , sin60 = √3/2 , cos30 = √3/2 , cos60 = 1/2

sin30 = √3/2 ( √3/2 ) - 1/2 ( 1/2 )

1/2 = 3/4 - 1/4

1 /2= 2/4

1/2 = 1/2

Both Sides of the equation are equal.

Hence, We proved and verified that sin(A - B) = sinA cos B - cos A sin B. holds good for A = 60, B = 30°

Answer 2

Hey mate !!

Here's your answer !!

Given :

A = 60° , B = 30°

To verify :

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

Proof :

Let's substitute the values of A and B directly in the equation that is required for verification.

= Sin ( 60 - 30 ) = Sin 60 . Cos 30 - Cos 60 . Sin 30

= Sin 30 = Sin 60 . Cos 30 - Cos 60 . Sin 30   -----( Eqn. 1 )

We know that,

Sin 30 = 1 / 2

Sin 60 = √ 3 / 2

Cos 30 = √ 3 / 2

Cos 60 = 1 / 2

Now let's substitute the values in Equation 1.

=>  1 / 2 = √ 3 / 2 * √ 3 / 2 - 1 / 2 * 1 / 2

=> 1 / 2 = ( √ 3 / 2 )² - ( 1 / 2 )²

=> 1 / 2 = ( 3 / 4 - 1 / 4 )

=> 1 / 2 = ( 2 / 4 )

=> 1 / 2 = 1 / 2

Hence LHS = RHS

Hence verified !!

Hope my answer helped you mate !!

Cheers !!