Question

sin²(n+1)Theta - Sin²ntheta = sin(2n+1)Theta.Sin Theta​

Answer 1

To prove ----->

Sin² ( n + 1 )θ - Sin²nθ = Sin ( 2n + 1 )θ Sinθ

Proof -----> We know that,

Sin² A - Sin²B = Sin ( A + B ) Sin ( A - B )

Now taking LHS

= Sin² ( n + 1 ) θ - Sin²nθ

= Sin² ( nθ + θ ) - Sin²nθ

= Sin ( nθ + θ + nθ ) Sin ( nθ + θ - nθ )

= Sin ( 2nθ + θ ) Sin θ

= Sin ( 2n + 1 )θ Sinθ = RHS

Additional information ----->

1) Cos²Α - Sin²A = Cos ( A + B ) Cos ( A - B )

2) 2 SinA CosB = Sin ( A + B ) + Sin ( A - B )

3) 2 CosA SinB = Sin ( A + B ) - Sin ( A - B )

4) 2 CosA CosB = Cos ( A + B ) + Cos ( A - B )

5) 2 SinA SinB = Cos ( A - B ) - Cos ( A + B )

6) Sin 2A = 2 SinA CosA

7) Cos2A = 2 Cos²A - 1

= 1 - 2 Sin²A

= Cos²A - Sin²A

8) tan2A = 2tanA / ( 1 - tan²A )

9) Sin2A = 2tanA / ( 1 + tan²A )