It A = 60° and B = 30°, prove that:
sin (A+B) = sin A x Cas B + COSA X Sin B
Answers
Answer:
Step-by-step explanation:
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.