Question

iv) sin [(n+1)A].sin((n+2)A]+cos[(n+1)A].
cos[(n+2)A]=cos A​

Answer 1

To prove-->

Sin(n+1)A Sin(n+2)A + Cos(n+1)A

Cos(n+2)A = CosA

Proof --->

We have a formula

Cos(x - y ) = Cosx Cosy - Sinx Siny

Now , LHS

=Sin(n+1)A Sin(n+2)A + Cos(n+1)A

Cos(n+2)A

Let ( n + 1 )A = x , ( n + 2 )A = y

= Sin x Sin y + Cosx Cos y

= Cos x Cos y + Sin x Sin y

= Cos ( x - y )

Putting values of x and y we get

= Cos { ( n + 1 )A - ( n + 2 ) A }

= Cos { ( n + 1 ) - ( n + 2 ) } A

= Cos ( n + 1 - n - 2 ) A

= Cos ( - 1 ) A

= Cos ( - A )

We know that , Cos( - θ ) = Cosθ

= Cos A = RHS

Additional information--->

1) Cos(A + B ) =CosA CosB - SinA SinB

2) Sin(A + B ) = SinA CosB + CosA SinB

3) Sin(A - B ) = SinA CosB - CosA SinB

4) Sin2A = 2 SinA CosA

5) Cos2A = Cos²A - Sin²A

= 2 Cos²A - 1

= 1 - 2 Sin²A