Question

If \[{{T}_{n}}={{\sin }^{n}}\theta +{{\cos }^{n}}\theta \], prove that:-
\[2{{T}_{6}}-3{{T}_{4}}+1=0\]

Answer 1

Hint: In such questions, we prove them by either making the left hand side that is L.H.S. or by making the right hand side that is R.H.S. equal to the other in order to prove the proof that has been asked.

Complete step-by-step answer:
The most important formula that will be used in the solution is as follows
\[{{\sin }^{2}}x+{{\cos }^{2}}x=1\]
Some other important formulae that might be used in solving this question is as follows
\[\begin{align}
  & {{(a+b)}^{3}}-3a\cdot b\cdot (a+b)={{a}^{3}}+{{b}^{3}} \\
 & {{(a+b)}^{2}}-2a\cdot b={{a}^{2}}+{{b}^{2}} \\
\end{align}\]

Now, these are the results that would be used to prove the proof mentioned in this question as using these identities, we would convert the left hand side that is L.H.S. or the right hand side that is R.H.S. to make either of them equal to the other.
In this particular question, we will first write the value of \[{{T}_{n}}\] according to the value of n and then we will simplify the obtained expression by using the formulae that are mentioned above and hence, we will try to prove the given statement.
As mentioned in the question, we have to prove the given expression.
Now, we will start with the left hand side that is L.H.S. and try to make the necessary changes that are given in the hint, first, as follows
\[\begin{align}
  & =2{{T}_{6}}-3{{T}_{4}}+1 \\
 & =2\left( {{\sin }^{6}}\theta +{{\cos }^{6}}\theta \right)-3\left( {{\sin }^{4}}\theta +{{\cos }^{4}}\theta \right)+1 \\
\end{align}\]
Now, on using the expressions
\[\begin{align}
  & {{(a+b)}^{3}}-3a\cdot b\cdot (a+b)={{a}^{3}}+{{b}^{3}} \\
 & {{(a+b)}^{2}}-2a\cdot b={{a}^{2}}+{{b}^{2}} \\
\end{align}\]
We get the following result
\[\begin{align}
  & =2\left( {{\sin }^{6}}\theta +{{\cos }^{6}}\theta \right)-3\left( {{\sin }^{4}}\theta +{{\cos }^{4}}\theta \right)+1 \\
 & {{(a+b)}^{3}}-3a\cdot b\cdot (a+b)={{a}^{3}}+{{b}^{3}} \\
 & {{(a+b)}^{2}}-2a\cdot b={{a}^{2}}+{{b}^{2}} \\
 & =2\left( {{\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)}^{3}}-3\cdot {{\sin }^{2}}\theta \cdot {{\cos }^{2}}\theta \left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right) \right)-3\left( {{\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)}^{2}}-2\cdot {{\sin }^{2}}\theta \cdot {{\cos }^{2}}\theta \right)+1 \\
 & ({{\sin }^{2}}x+{{\cos }^{2}}x=1) \\
 & =2\left( {{\left( 1 \right)}^{3}}-3\cdot {{\sin }^{2}}\theta \cdot {{\cos }^{2}}\theta \left( 1 \right) \right)-3\left( {{\left( 1 \right)}^{2}}-2\cdot {{\sin }^{2}}\theta \cdot {{\cos }^{2}}\theta \right)+1 \\
 & =2\left( 1-3\cdot {{\sin }^{2}}\theta \cdot {{\cos }^{2}}\theta \right)-3\left( 1-2\cdot {{\sin }^{2}}\theta \cdot {{\cos }^{2}}\theta \right)+1 \\
 & =2-6\cdot {{\sin }^{2}}\theta \cdot {{\cos }^{2}}\theta -3+6\cdot {{\sin }^{2}}\theta \cdot {{\cos }^{2}}\theta +1 \\
 & =2+1-3=0 \\
\end{align}\]
(Using the identities that are mentioned in the hint)
Now, as the right hand side that is R.H.S. is equal to the left hand side that is L.H.S., hence, the expression has been proved.

Note: -Another method of attempting this question is by converting the right hand side that is R.H.S. to the left hand side that is L.H.S. by using the relations that are given in the hint. Through this method also, we could get to the correct answer and hence, we would be able to prove the required proof.
Another method of doing this question is by directly putting any known angle in place of \[\theta \] and hence, we can easily get the solution.