Question

Prove the following trigonometric identities.

if ${\displaystyle T_n={\sin }^n\theta +{\cos }^n\theta }$, prove that ${\displaystyle \frac{T_3-T_5}{T_1}=\frac{T_5-T_7}{T_3}}$

Answer 1

In the given question, we are given ${\displaystyle T_n={\sin }^n\theta +{\cos }^n\theta }$

We need to prove ${\displaystyle \frac{T_3-T_5}{T_1}=\frac{T_5-T_7}{T_3}}$

Here L.H.S is

${\displaystyle \frac{T_3-T_5}{T_1}=\frac{({\sin }^3\theta ={\cos }^3\theta )-({\sin }^5\theta +{\cos }^5\theta )}{(\sin \theta +\cos \theta )}}$

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

${\displaystyle \frac{({\sin }^3\theta +{\cos }^3\theta )-({\sin }^5\theta +{\cos }^5\theta )}{(\sin \theta +\cos \theta )}=\frac{{\sin }^3\theta -{\sin }^5\theta +{\cos }^3\theta -{\cos }^5\theta }{\sin \theta +\cos \theta }}$

${\displaystyle =\frac{{\sin }^3\theta (1-{\sin }^2\theta )+{\cos }^3\theta (1-{\cos }^2\theta )}{(\sin \theta +\cos \theta )}}$

Further Using the property ${\displaystyle {\sin }^2\theta +{\cos }^2\theta =1}$ we get

${\displaystyle {\cos }^2\theta =1-{\sin }^2\theta }$

${\displaystyle {\sin }^2\theta =1-{\cos }^2\theta }$

So,

${\displaystyle \frac{{\sin }^3\theta (1-{\sin }^2\theta )+{\cos }^3\theta (1-{\cos }^2\theta )}{\sin \theta +\cos \theta } =\frac{{\sin }^3\theta {\cos }^2\theta +{\cos }^3\theta {\sin }^2\theta }{\sin \theta +\cos \theta }}$

${\displaystyle =\frac{{\sin }^2\theta {\cos }^2\theta (\sin \theta +\cos \theta )}{\sin \theta +\cos \theta }}$

${\displaystyle ={\sin }^2\theta {\cos }^2\theta }$

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

${\displaystyle \frac{T_5-T_7}{T_3}=\frac{({\sin }^5\theta +coi{s}^5)({\sin }^7\theta +{\cos }^2\theta )}{{\sin }^3\theta +{\cos }^3\theta }}$

So,

${\displaystyle \frac{({\sin }^5\theta +{\cos }^5\theta )-({\sin }^7\theta +{\cos }^7\theta )}{{\sin }^3\theta +{\cos }^3\theta }=\frac{{\sin }^5\theta -{\sin }^7\theta +{\cos }^5\theta -{\cos }^7\theta }{{\sin }^3\theta +{\cos }^3\theta }}$

${\displaystyle =\frac{{\sin }^5\theta (1 -{\sin }^2\theta )+{\cos }^5\theta (1+{\cos }^2\theta )}{{\sin }^3\theta +{\cos }^3\theta }}$

${\displaystyle ={\sin }^2\theta {\cos }^2\theta }$

Hence proved