Worked Example of Trig Identity Proof

Observe the following identities:

   cos(2θ)  =  cos2(θ) - sin2(θ)  =  1 - 2sin2(θ)

   sin(2θ)  =  2sin(θ)cos(θ)

   cos(4θ)  =  cos2(2θ) - sin2(2θ)  =  1 - 2sin2(2θ)

   sin(4θ)  =  2sin(2θ)cos(2θ)

Substituting the sine and cosine for 2θ formulas into the 4θ formulas you have

   cos(4θ)  =  1 - 8sin2(θ)cos2(θ)

   sin(4θ)  =  4sin(θ)cos(θ)(1 - 2sin2(θ))

Using all this you can turn the quotient on the left side of the original equation into an expression of nothing but sines and cosines of θ. The numerator becomes:

  1 - 2sin2(θ)  -  (1 - 8sin2(θ)cos2(θ))  =  -2sin(θ)2(1 - 4cos2(θ))

The denominator becomes:

   2sin(θ)cos(θ)  +  4sin(θ)cos(θ)(1 - 2sin2(θ))  =

   2sin(θ)cos(θ)(1 + 2(1 - 2sin2(θ))  =

   2sin(θ)cos(θ)(3 - 4sin2(θ)

Clearly when we put this all together, you get a common factor of  2sin(θ)  in the numerator and denominator that you can cancel. So the new equation is

-sin(θ)(1 - 4cos2(θ)) = tan(θ) cos(θ)(3 - 4sin2(θ))

Now remember that  sin²(θ) = 1 - cos²(θ).  Make this substitution in the denominator and you have

-sin(θ)(1 - 4cos2(θ)) = tan(θ) cos(θ)(3 - (4 - 4cos2(θ)))

From here it's simple algebra to prove that  3 - (4 - 4cos²(θ)) = -(1 - 4cos²(θ)),  which gives you another cancellation and leaves

sin(θ) = tan(θ) cos(θ)

which you already knew.