Question

If $\cos \theta + {\cos ^2}\theta = 1$, prove that \[{\sin ^{12}}\theta + 3{\sin ^{10}}\theta + 3{\sin ^8}\theta + {\sin ^6}\theta + 2{\sin ^4}\theta + 2{\sin ^2}\theta - 2 = 1\].

Answer 1

Hint: Start with $\cos \theta + {\cos ^2}\theta = 1$, convert this equation in terms of $\sin \theta $ using ${\sin ^2}\theta + {\cos ^2}\theta = 1$. Then simplify it further to get the proof.

Complete step-by-step answer:
According to the question:
$
   \Rightarrow \cos \theta + {\cos ^2}\theta = 1, \\
   \Rightarrow \cos \theta = 1 - {\cos ^2}\theta .....(i) \\
$
Now, we know that:
$
   \Rightarrow {\sin ^2}\theta + {\cos ^2}\theta = 1, \\
   \Rightarrow 1 - {\cos ^2}\theta = {\sin ^2}\theta \\
$
Using this result for equation $(i)$, we’ll get:
$ \Rightarrow \cos \theta = {\sin ^2}\theta $
Squaring both sides of the equation and simplifying it further, we’ll get:
$
   \Rightarrow {\cos ^2}\theta = {\sin ^4}\theta , \\
   \Rightarrow 1 - {\sin ^2}\theta = {\sin ^4}\theta , \\
   \Rightarrow {\sin ^4}\theta + {\sin ^2}\theta = 1 .....(ii) \\
$
Cubing both sides of the above equation, we’ll get:
$ \Rightarrow {\left( {{{\sin }^4}\theta + {{\sin }^2}\theta } \right)^3} = {1^3}$
We know that, ${\left( {a + b} \right)^3} = {a^3} + {b^3} + 3ab\left( {a + b} \right)$. Applying this formula, we’ll get:
 $
   \Rightarrow {\sin ^{12}}\theta + {\sin ^6}\theta + 3{\sin ^4}\theta {\sin ^2}\theta \left( {{{\sin }^4}\theta + {{\sin }^2}\theta } \right) = 1, \\
   \Rightarrow {\sin ^{12}}\theta + {\sin ^6}\theta + 3{\sin ^{10}}\theta + 3{\sin ^8}\theta = 1 \\
$
To obtain the above result, we add and subtract $2$ on the left hand side of the above equation:
$ \Rightarrow {\sin ^{12}}\theta + {\sin ^6}\theta + 3{\sin ^{10}}\theta + 3{\sin ^8}\theta + 2(1) - 2 = 1$
Putting $1 = {\sin ^4}\theta + {\sin ^2}\theta $ from equation $(ii)$, we’ll get:
\[
   \Rightarrow {\sin ^{12}}\theta + {\sin ^6}\theta + 3{\sin ^{10}}\theta + 3{\sin ^8}\theta + 2({\sin ^4}\theta + {\sin ^2}\theta ) - 2 = 1, \\
   \Rightarrow {\sin ^{12}}\theta + 3{\sin ^{10}}\theta + 3{\sin ^8}\theta + {\sin ^6}\theta + 2{\sin ^4}\theta + 2{\sin ^2}\theta - 2 = 1. \\
   \Rightarrow L.H.S. = R.H.S. \\
\]
This is the required proof.

Note: The condition given in the question is $\cos \theta + {\cos ^2}\theta = 1$ which is completely in cosine form.
And the one we required to prove, \[{\sin ^{12}}\theta + 3{\sin ^{10}}\theta + 3{\sin ^8}\theta + {\sin ^6}\theta + 2{\sin ^4}\theta + 2{\sin ^2}\theta - 2 = 1\], is completely in sine form.
Thus, it’s the ideal condition for using the formula ${\sin ^2}\theta + {\cos ^2}\theta = 1$ to convert the given condition in purely sine form.