Solveeit Logo
Login

Question

Question: If \(\cos \theta + {\cos ^2}\theta = 1\), prove that \[{\sin ^{12}}\theta + 3{\sin ^{10}}\theta + 3{...

If cos⁡θ+cos⁡2θ=1\cos \theta + {\cos ^2}\theta = 1, prove that sin⁡12θ+3sin⁡10θ+3sin⁡8θ+sin⁡6θ+2sin⁡4θ+2sin⁡2θ−2=1{\sin ^{12}}\theta + 3{\sin ^{10}}\theta + 3{\sin ^8}\theta + {\sin ^6}\theta + 2{\sin ^4}\theta + 2{\sin ^2}\theta - 2 = 1.

Explanation

Solution

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

Complete step-by-step answer:
According to the question:
⇒cos⁡θ+cos⁡2θ=1, ⇒cos⁡θ=1−cos⁡2θ.....(i)  \Rightarrow \cos \theta + {\cos ^2}\theta = 1, \\\ \Rightarrow \cos \theta = 1 - {\cos ^2}\theta .....(i) \\\
Now, we know that:
⇒sin⁡2θ+cos⁡2θ=1, ⇒1−cos⁡2θ=sin⁡2θ  \Rightarrow {\sin ^2}\theta + {\cos ^2}\theta = 1, \\\ \Rightarrow 1 - {\cos ^2}\theta = {\sin ^2}\theta \\\
Using this result for equation (i)(i), we’ll get:
⇒cos⁡θ=sin⁡2θ\Rightarrow \cos \theta = {\sin ^2}\theta
Squaring both sides of the equation and simplifying it further, we’ll get:
⇒cos⁡2θ=sin⁡4θ, ⇒1−sin⁡2θ=sin⁡4θ, ⇒sin⁡4θ+sin⁡2θ=1.....(ii)  \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:
⇒(sin⁡4θ+sin⁡2θ)3=13\Rightarrow {\left( {{{\sin }^4}\theta + {{\sin }^2}\theta } \right)^3} = {1^3}
We know that, (a+b)3=a3+b3+3ab(a+b){\left( {a + b} \right)^3} = {a^3} + {b^3} + 3ab\left( {a + b} \right). Applying this formula, we’ll get:
⇒sin⁡12θ+sin⁡6θ+3sin⁡4θsin⁡2θ(sin⁡4θ+sin⁡2θ)=1, ⇒sin⁡12θ+sin⁡6θ+3sin⁡10θ+3sin⁡8θ=1  \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 22 on the left hand side of the above equation:
⇒sin⁡12θ+sin⁡6θ+3sin⁡10θ+3sin⁡8θ+2(1)−2=1\Rightarrow {\sin ^{12}}\theta + {\sin ^6}\theta + 3{\sin ^{10}}\theta + 3{\sin ^8}\theta + 2(1) - 2 = 1
Putting 1=sin⁡4θ+sin⁡2θ1 = {\sin ^4}\theta + {\sin ^2}\theta from equation (ii)(ii), we’ll get:

⇒sin⁡12θ+sin⁡6θ+3sin⁡10θ+3sin⁡8θ+2(sin⁡4θ+sin⁡2θ)−2=1, ⇒sin⁡12θ+3sin⁡10θ+3sin⁡8θ+sin⁡6θ+2sin⁡4θ+2sin⁡2θ−2=1. ⇒L.H.S.=R.H.S.  \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⁡θ+cos⁡2θ=1\cos \theta + {\cos ^2}\theta = 1 which is completely in cosine form.
And the one we required to prove, sin⁡12θ+3sin⁡10θ+3sin⁡8θ+sin⁡6θ+2sin⁡4θ+2sin⁡2θ−2=1{\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⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1 to convert the given condition in purely sine form.