Question

If ${{T}_{n}}={{\sin }^{n}}\theta +{{\cos }^{n}}\theta $, then prove the following expression, $6{{T}_{10}}-15{{T}_{8}}+10{{T}_{6}}-1=0$.

Answer 1

Hint: To solve this question, we will consider the values of ${{T}_{10}}$ as ${{\sin }^{10}}\theta +{{\cos }^{10}}\theta $, ${{T}_{8}}$ as ${{\sin }^{8}}\theta +{{\cos }^{8}}\theta $, and ${{T}_{6}}$ as ${{\sin }^{6}}\theta +{{\cos }^{6}}\theta $ in the given expression, $6{{T}_{10}}-15{{T}_{8}}+10{{T}_{6}}-1=0$ to prove that the left hand side or the LHS is equal to the right hand side or the RHS or LHS = RHS = 0.

Complete step-by-step answer:

It is given in the question that, ${{T}_{n}}={{\sin }^{n}}\theta +{{\cos }^{n}}\theta $. We have to prove the expression,

 $6{{T}_{10}}-15{{T}_{8}}+10{{T}_{6}}-1=0$. We will consider the left hand side or the LHS of the expression, that is, $6{{T}_{10}}-15{{T}_{8}}+10{{T}_{6}}-1$. We can write ${{T}_{10}}$ as ${{\sin }^{10}}\theta +{{\cos }^{10}}\theta $, ${{T}_{8}}$ as ${{\sin }^{8}}\theta +{{\cos }^{8}}\theta $, and ${{T}_{6}}$ as ${{\sin }^{6}}\theta +{{\cos }^{6}}\theta $. So, by substituting the values of the terms in the expression, we get the LHS as, $=6\left( {{\sin }^{10}}\theta +{{\cos }^{10}}\theta \right)-15\left( {{\sin }^{8}}\theta +{{\cos }^{8}}\theta \right)+10\left( {{\sin }^{6}}\theta +{{\cos }^{6}}\theta \right)-1$

$=6\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)\left( {{\sin }^{8}}\theta +{{\cos }^{8}}\theta -{{\sin }^{2}}\theta {{\cos }^{2}}\theta \right)-15\left( {{\sin }^{8}}\theta +{{\cos }^{8}}\theta \right)+10\left( {{\sin }^{6}}\theta +{{\cos }^{6}}\theta \right)-1$

We know that, ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, so by substituting it in the above expression, we get the LHS as,

$\begin{align}

  & =6\left( {{\sin }^{8}}\theta +{{\cos }^{8}}\theta -{{\sin }^{2}}\theta {{\cos }^{2}}\theta \right)-15\left( {{\sin }^{8}}\theta +{{\cos }^{8}}\theta \right)+10\left( {{\sin }^{6}}\theta +{{\cos }^{6}}\theta \right)-1 \\

 & =6{{\sin }^{8}}\theta +6{{\cos }^{8}}\theta -6{{\sin }^{2}}\theta {{\cos }^{2}}\theta -15{{\sin }^{8}}\theta -15{{\cos }^{8}}\theta +10\left( {{\sin }^{6}}\theta +{{\cos }^{6}}\theta \right)-1 \\

 & =-6{{\sin }^{2}}\theta {{\cos }^{2}}\theta -9{{\sin }^{8}}\theta -9{{\cos }^{8}}\theta +10\left( {{\sin }^{6}}\theta +{{\cos }^{6}}\theta \right)-1 \\

 & =-6{{\sin }^{2}}\theta {{\cos }^{2}}\theta -9\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)\left( {{\sin }^{6}}\theta +{{\cos }^{6}}\theta -{{\sin }^{2}}\theta {{\cos }^{2}}\theta \right)+10\left( {{\sin }^{6}}\theta +{{\cos }^{6}}\theta \right)-1 \\

\end{align}$

We know that, ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, by substituting the same in the above expression, we get,

$\begin{align}

  & =-6{{\sin }^{2}}\theta {{\cos }^{2}}\theta -9\left( 1 \right)\left( {{\sin }^{6}}\theta +{{\cos }^{6}}\theta -{{\sin }^{2}}\theta {{\cos }^{2}}\theta \right)+10\left( {{\sin }^{6}}\theta +{{\cos }^{6}}\theta \right)-1 \\

 & =-6{{\sin }^{2}}\theta {{\cos }^{2}}\theta -9{{\sin }^{6}}\theta -9{{\cos }^{6}}\theta +9{{\sin }^{2}}\theta {{\cos }^{2}}\theta +10{{\sin }^{6}}\theta +9{{\cos }^{6}}\theta -1 \\

 & =3{{\sin }^{2}}\theta {{\cos }^{2}}\theta +{{\sin }^{6}}\theta +{{\cos }^{6}}\theta -1 \\

 & =3{{\sin }^{2}}\theta {{\cos }^{2}}\theta +\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)\left( {{\sin }^{4}}\theta +{{\cos }^{4}}\theta -{{\sin }^{2}}\theta {{\cos }^{2}}\theta \right)-1 \\

\end{align}$

Again, by substituting, ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, we get the LHS as,

$\begin{align}

  & =3{{\sin }^{2}}\theta {{\cos }^{2}}\theta +{{\sin }^{4}}\theta +{{\cos }^{4}}\theta -{{\sin }^{2}}\theta {{\cos }^{2}}\theta -1 \\

 & ={{\left( {{\sin }^{2}}\theta \right)}^{2}}+2{{\sin }^{2}}\theta {{\cos }^{2}}\theta +{{\left( {{\cos }^{2}}\theta \right)}^{2}}-1 \\

 & ={{\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)}^{2}}-1 \\

\end{align}$

Now, by substituting, ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ in the above expression, we get,

$\begin{align}

  & ={{\left( 1 \right)}^{2}}-1 \\

 & =1-1 \\

 & =0 \\

\end{align}$

Which is the value given in the right hand side or the RHS of the expression given in the question.

Therefore, LHS = RHS, and hence we have proved the expression given in the question.

Note: Usually most of the students skip this question during an examination, as it involves trigonometric functions with powers greater than 2, and they are afraid of solving the same. But, if we try to solve the question, step by step, then it is easy to solve any trigonometric function with power n. The mistakes that the students make while doing this question is that they get stuck as they simply multiply the given trigonometric function with the given number and solve without applying any trigonometric formula. But it is advisable to apply the derived formulas as it reduces the effort and time to solve the question.