How to prove the identity?
(sin(3x))/sin(x) +(cos(3x)/cos(x))=4(1-2sin^2(x)

Question

Josiah Anderson · Accepted Answer

Please see the proof below.

We need #sin(a+b)=sinacosb+sinbcosa# #sin2x=2sinxcosx# #cos2x=1-2sin^2x# Consequently, #LHS=sin(3x)/sinx+cos(3x)/cosx# Applying a common denominator #=(sin3xcosx+cos3xsinx)/(sinxcosx)# #=(sin(3x+x))/(sin(2x)/2)# #=(2sin(4x))/sin(2x)# #=(4cancelsin(2x)cos(2x))/(cancelsin(2x))# #=4cos(2x)# #=4(1-2sin^2x)# #=RHS# #QED#

Delilah Aguilar · Answer

undefined To prove the identity (sin(3x))/sin(x) + (cos(3x)/cos(x)) = 4(1 - 2sin^2(x)), you can start by expressing sin(3x) and cos(3x) using trigonometric identities:

sin(3x) = 3sin(x) - 4sin^3(x)
cos(3x) = 4cos^3(x) - 3cos(x)

Then, substitute these expressions into the left-hand side of the given identity:

(sin(3x))/sin(x) + (cos(3x)/cos(x))
= (3sin(x) - 4sin^3(x))/sin(x) + (4cos^3(x) - 3cos(x))/cos(x)

Now, simplify each term:

= 3 - 4sin^2(x) + 4cos^2(x) - 3sin^2(x)
= 3 - 3sin^2(x) + 4cos^2(x) - 4sin^2(x)

Use the Pythagorean identity sin^2(x) + cos^2(x) = 1:

= 3 - 3(1 - cos^2(x)) + 4cos^2(x) - 4(1 - cos^2(x))
= 3 - 3 + 3cos^2(x) + 4cos^2(x) - 4 + 4cos^2(x)

Combine like terms:

= 7cos^2(x) - 4

Now, recall the Pythagorean identity again:

cos^2(x) = 1 - sin^2(x)

Substitute this into the expression:

= 7(1 - sin^2(x)) - 4
= 7 - 7sin^2(x) - 4

= 3 - 7sin^2(x)

Now, factor out 1 - 2sin^2(x):

= 4(1 - 2sin^2(x))

This matches the right-hand side of the given identity, proving its validity.

How to prove the identity? (sin(3x))/sin(x) +(cos(3x)/cos(x))=4(1-2sin^2(x)