Question

It A = 60° and B = 30°, prove that:
sin (A+B) = sin A x Cas B + COSA X Sin B
​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

Step-by-step explanation:

Here are few methods to get the solution : -

                 

                Method 1 -

Given measures of A is 60° and B is 30°. If we observe, sum of both the angles is 90°, so angle B can be written as ( 90 - 60 )° or ( 90 - A ) { 90 - 60 = 30 }.

Now, measure of angle A is 60° and measure of B is ( 90 - 60 )°.

We have to prove that sin( A + B ) = sinAcosB + cosAsinB

Solving Right Hand Side :

= > sinAcos( 90 - A ) + cosAsin( 90 - A )

From the properties of trigonometry : -

• sin( 90 - θ ) = cosθ
• cos( 90 - θ ) = sinθ
• sin^2 θ + cos^2 θ = 1

= > sinAsinA + cosAcosA

= > sin^2 A + cos^2 A

= > 1                                { from above properties }

Solving Left Hand Side :

= > sin( A + B )

= > sin( 60° + 30° )

= > sin90°

= > 1                           { from trigonometric properties }

                     Method 2 -

= > sin60°cos30° + cos60°sin30°

From the properties of trigonometry : -

• sin60° = √3 / 2
• cos30° = √3 / 2
• cos60° = 1 / 2
• sin30° = 1 / 2

= > ( √3 / 2 )( √3 / 2 ) + ( 1 / 2 )( 1 / 2 )

= > ( 3 / 4 ) + ( 1 / 4 )

= > ( 3 + 1 ) / 4

= > 4 / 4

= > 1

LHS = RHS.

Hence proved.