Question

Sin 12theta-sin 4theta = 2 cos 8theta . Sin 4theta ​

Answer 1

Answer:

2cos8θsin4θ

Step-by-step explanation:

To Prove :-

sin12θ - sin4θ = 2cos8θ.sin4θ

How To Prove :-

Here we are asked to prove that 'sin12θ - sin4θ = 2cos8θ.sin4θ' . So we need to take the L.H.S(Left hand side). We can observe that in L.H.S it is 'sin12θ - sin4θ'. We can observe that 'sin12θ - sin4θ' is in the form of 'sinA - sinB' [ ∴ We need to consider 'A' as 12θ and 'B' as 4θ ]. So we need to apply the formula 'sinA - sinB' and we need expand that and we need to prove that.

Formula Required :-

sinA - sinB = 2cos(A + B/2).sin(A - B/2)

Solution :-

Taking L.H.S :-

= sin12θ - sin4θ

$= 2cos\left(\dfrac{12\theta+4\theta}{2}\right)\times sin\left(\dfrac{12\theta-4\theta}{2}\right)$

$=2 cos\left(\dfrac{16\theta}{2}\right) sin\left(\dfrac{8\theta}{2}\right)$

= 2cos(8θ)sin(4θ)

= 2cos8θsin4θ

= R.H.S

hence Proved that 'sin12θ - sin4θ = 2cos8θsin4θ'.

Answer 2

Answer:

2cos8θsin4θ

Step-by-step explanation:

To Prove :-

sin12θ - sin4θ = 2cos8θ.sin4θ

How To Prove :-

Here we are asked to prove that 'sin12θ - sin4θ = 2cos8θ.sin4θ' . So we need to take the L.H.S(Left hand side). We can observe that in L.H.S it is 'sin12θ - sin4θ'. We can observe that 'sin12θ - sin4θ' is in the form of 'sinA - sinB' [ ∴ We need to consider 'A' as 12θ and 'B' as 4θ ]. So we need to apply the formula 'sinA - sinB' and we need expand that and we need to prove that.

ANSWER

Formula Required :-

sinA - sinB = 2cos(A + B/2).sin(A - B/2)

Solution :-

Taking L.H.S :-

= sin12θ - sin4θ

= 2cos\left(\dfrac{12\theta+4\theta}{2}\right)\times sin\left(\dfrac{12\theta-4\theta}{2}\right)

= 2cos\left(\dfrac{12\theta+4\theta}{2}\right)\times sin\left(\dfrac{12\theta-4\theta}{2}\right)

=2 cos\left(\dfrac{16\theta}{2}\right) sin\left(\dfrac{8\theta}{2}\right)

=2 cos\left(\dfrac{16\theta}{2}\right) sin\left(\dfrac{8\theta}{2}\right)

= 2cos(8θ)sin(4θ)

= 2cos8θsin4θ

= R.H.S

hence Proved that 'sin12θ - sin4θ = 2cos8θsin4θ'.