Question

Prove the following trigonometric identities.

if `T_n = sin^n theta + cos^n theta`, prove that `(T_3 - T_5)/T_1 = (T_5 - T_7)/T_3`

Answer 1

In the given question, we are given `T_n = sin^n theta + cos^n theta`

We need to prove `(T_3 - T_5)/T_1 = (T_5 - T_7)/T_3`

Here L.H.S is

`(T_3 - T_5)/T_1 = ((sin^3 theta = cos^3 theta) - (sin^5 theta + cos^5 theta))/((sin theta + cos theta))`

Now, solving the L.H.S, we get

`((sin^3 theta + cos^3 theta)- (sin^5 theta + cos^5 theta))/((sin theta + cos theta)) = (sin^3 theta - sin^5 theta + cos^3 theta - cos^ 5 theta)/(sin theta + cos theta)`

` = (sin^3 theta (1 - sin^2 theta) + cos^3 theta (1 - cos^2 theta))/((sin theta  + cos theta))`

Further Using the property `sin^2 theta + cos^2 theta = 1` we get

`cos^2 theta = 1 - sin^2 theta`

`sin^2 theta = 1 - cos^2 theta`

So,

`(sin^3 theta(1 - sin^2 theta) + cos^3 theta (1 - cos^2 theta))/(sin theta + cos theta)  = (sin^3 theta cos^2 theta + cos^3 theta sin^2 theta)/(sin theta + cos theta)`

`= (sin^2 theta cos^2 theta (sin theta + cos theta))/(sin theta + cos theta)`

`= sin^2 theta cos^2 theta`

Now, solving the R.H.S, we get

`(T_5 - T_7)/T_3 = ((sin^5 theta + cois ^5)(sin^7 theta + cos^2 theta))/(sin^3 theta + cos^3 theta)`

So,

`((sin^5 theta + cos^5 theta) - (sin^7 theta + cos^7 theta))/(sin^3 theta + cos^3 theta) = (sin^5 theta - sin^7 theta + cos^5 theta - cos^7 theta)/(sin^3 theta + cos^3 theta)`

`= (sin^5 theta (1  - sin^2 theta) + cos^5 theta (1 + cos^2 theta))/ (sin^3 theta + cos^3 theta)`

`= sin^2 theta cos^2 theta`

Hence proved