Question

If T_n = sinⁿ θ + cosⁿ θ, prove that `(T_3 - T_5)/(T_1) = (T_5 - T_7)/(T_3)`

Answer 1

Given: Let T_n = sinⁿ θ + cosⁿ θ.

To Prove: `(T_3 - T_5)/(T_1) = (T_5 - T_7)/(T_3)`

Proof (Step-wise):

1. Put s = sin θ and c = cos θ.

So T_n = sⁿ + cⁿ.

2. Compute T₃ – T₅: 

T₃ – T₅ = (s³ + c³) – (s⁵ + c⁵) 

= s³(1 – s²) + c³(1 – c²) 

= s³c² + c³s²

= s²c²(s + c)

3. Note T₁ = s + c. 

Therefore `(T_3 - T_5)/T_1 = (s^2c^2 (s + c))/(s + c)`   ...(Provided s + c ≠ 0)

= s²c²

4. Compute T₅ – T₇: 

T₅ – T₇ = (s⁵ + c⁵) – (s⁷ + c⁷) 

= s⁵(1 – s²) + c⁵(1 – c²) 

= s⁵c² + c⁵s²

= s²c²(s³ + c³) 

= s²c²T₃

5. Therefore `(T_5 - T_7)/T_3 = (s^2 c^2T_3)/T_3`   ...(Provided T₃ ≠ 0)

= s²c² 

Both ratios equal s²c².

So `(T_3 - T_5)/T_1 = (T_5 - T_7)/T_3`, as required.