Question

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

Answer 1

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 $${(a+b)}^3=a^3+b^3+3ab(a+b)$$ in the given expression to express it in terms of $$(\sin \theta +{\sin }^2\theta )$$ whose value is 1.

Complete step-by-step answer:
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,
$$A={\left({\cos }^4\theta \right)}^3+3{\cos }^6\theta ({\cos }^4\theta +{\cos }^2\theta )+{\left({\cos }^2\theta \right)}^3+2({\cos }^4\theta -{\cos }^2\theta -1)$$
Now by writing, $${\cos }^6\theta ={\cos }^2\theta .{\cos }^4\theta$$, we get
$$A={\left({\cos }^4\theta \right)}^3+3{\cos }^2\theta .{\cos }^4\theta ({\cos }^4\theta +{\cos }^2\theta )+{\left({\cos }^2\theta \right)}^3+2({\cos }^4\theta -{\cos }^2\theta -1)$$
Now, we know that $$a^3+3ab(a+b)+b^3={(a+b)}^3$$
By taking $$a={\cos }^4\theta$$ and $$b={\cos }^2\theta$$ in the above expression, we get,
$$A={({\cos }^4\theta +{\cos }^2\theta )}^3+2({\cos }^4\theta +{\cos }^2\theta -1)$$
By writing $${\cos }^4\theta ={\left({\cos }^2\theta \right)}^2$$ in the above expression, we get,
$$A={[{\left({\cos }^2\theta \right)}^2+\left({\cos }^2\theta \right)]}^3+2[{\left({\cos }^2\theta \right)}^2+{\cos }^2\theta -1]$$
From equation (i), we know that $${\cos }^2\theta =\sin \theta$$. By applying it in the above expression, we get
$$A={[{(\sin \theta )}^2+\sin \theta ]}^3+2({(\sin \theta )}^2+\sin \theta -1)$$
As we are given that $$\sin \theta +{\sin }^2\theta =1$$, therefore by applying it in the above equation, we get,
$$\begin{array}{rl}& A={\left[1\right]}^3+2[1-1]\\ & A=1+2\left(0\right)\\ & A=1\end{array}$$
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 $${(a+b)}^3=a^3+b^3+3ab(a+b)$$or $${(a-b)}^3=a^3-b^3-3ab(a-b)$$ 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.