Question

# If $\sin \theta +{{\sin }^{2}}\theta =1$, find the value of ${{\cos }^{12}}\theta +3{{\cos }^{10}}\theta +3{{\cos }^{8}}\theta +{{\cos }^{6}}\theta +2{{\cos }^{4}}\theta +2{{\cos }^{2}}\theta -2$

Hint: First of all, use the formula ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ in the given equation to get $\sin \theta ={{\cos }^{2}}\theta$. Now use ${{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right)$ in the given expression to express it in terms of $\left( \sin \theta +{{\sin }^{2}}\theta \right)$ whose value is 1.

Here, we are given that $\sin \theta +{{\sin }^{2}}\theta =1$ and we have to find the value of ${{\cos }^{12}}\theta +3{{\cos }^{10}}\theta +3{{\cos }^{8}}\theta +{{\cos }^{6}}\theta +2{{\cos }^{4}}\theta +2{{\cos }^{2}}\theta -2$
First of all let us take the given equation, that is, $\sin \theta +{{\sin }^{2}}\theta =1$
We know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$
Or, ${{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta$
By putting the value of ${{\sin }^{2}}\theta$ in the given equation, we get,
$\sin \theta +1-{{\cos }^{2}}\theta =1$
By cancelling 1 from both sides, we get,
$\sin \theta -{{\cos }^{2}}\theta =0$
Or, $\sin \theta ={{\cos }^{2}}\theta ....\left( i \right)$
Now, let us consider the expression whose value is to be found as,
$A={{\cos }^{12}}\theta +3{{\cos }^{10}}\theta +3{{\cos }^{8}}\theta +{{\cos }^{6}}\theta +2{{\cos }^{4}}\theta +2{{\cos }^{2}}\theta -2$
We can also write the above expression as,
$A={{\left( {{\cos }^{4}}\theta \right)}^{3}}+3{{\cos }^{6}}\theta \left( {{\cos }^{4}}\theta +{{\cos }^{2}}\theta \right)+{{\left( {{\cos }^{2}}\theta \right)}^{3}}+2\left( {{\cos }^{4}}\theta -{{\cos }^{2}}\theta -1 \right)$
Now by writing, ${{\cos }^{6}}\theta ={{\cos }^{2}}\theta .{{\cos }^{4}}\theta$, we get
$A={{\left( {{\cos }^{4}}\theta \right)}^{3}}+3{{\cos }^{2}}\theta .{{\cos }^{4}}\theta \left( {{\cos }^{4}}\theta +{{\cos }^{2}}\theta \right)+{{\left( {{\cos }^{2}}\theta \right)}^{3}}+2\left( {{\cos }^{4}}\theta -{{\cos }^{2}}\theta -1 \right)$
Now, we know that ${{a}^{3}}+3ab\left( a+b \right)+{{b}^{3}}={{\left( a+b \right)}^{3}}$
By taking $a={{\cos }^{4}}\theta$ and $b={{\cos }^{2}}\theta$ in the above expression, we get,
$A={{\left( {{\cos }^{4}}\theta +{{\cos }^{2}}\theta \right)}^{3}}+2\left( {{\cos }^{4}}\theta +{{\cos }^{2}}\theta -1 \right)$
By writing ${{\cos }^{4}}\theta ={{\left( {{\cos }^{2}}\theta \right)}^{2}}$ in the above expression, we get,
$A={{\left[ {{\left( {{\cos }^{2}}\theta \right)}^{2}}+\left( {{\cos }^{2}}\theta \right) \right]}^{3}}+2\left[ {{\left( {{\cos }^{2}}\theta \right)}^{2}}+{{\cos }^{2}}\theta -1 \right]$
From equation (i), we know that ${{\cos }^{2}}\theta =\sin \theta$. By applying it in the above expression, we get
$A={{\left[ {{\left( \sin \theta \right)}^{2}}+\sin \theta \right]}^{3}}+2\left( {{\left( \sin \theta \right)}^{2}}+\sin \theta -1 \right)$
As we are given that $\sin \theta +{{\sin }^{2}}\theta =1$, therefore by applying it in the above equation, we get,
\begin{align} & A={{\left[ 1 \right]}^{3}}+2\left[ 1-1 \right] \\ & A=1+2\left( 0 \right) \\ & A=1 \\ \end{align}
Therefore, we get the value of ${{\cos }^{12}}\theta +3{{\cos }^{10}}\theta +3{{\cos }^{8}}\theta +{{\cos }^{6}}\theta +2{{\cos }^{4}}\theta +2{{\cos }^{2}}\theta -2=1$.

Note: Take special care of powers of each term while writing them and always cross-check after writing each equation. Whenever you get the higher powers, always try using the formulas like ${{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right)$or ${{\left( a-b \right)}^{3}}={{a}^{3}}-{{b}^{3}}-3ab\left( a-b \right)$ according to the question. Here, while solving the given equation, some students finally write it as $\sin \theta =\cos \theta$ or ${{\sin }^{2}}\theta ={{\cos }^{2}}\theta$ instead of $\sin \theta ={{\cos }^{2}}\theta$. So this mistake must be avoided.