Question

If $\sin \theta +{{\sin }^{2}}\theta +{{\sin }^{3}}\theta =1$ , then show that ${{\cos }^{6}}\theta -4{{\cos }^{4}}\theta +8{{\cos }^{2}}\theta =4$ .

Answer 1

Hint: Use the property that ${{\cos }^{2}}A+{{\sin }^{2}}A=1$ to simplify the equation $\sin \theta +{{\sin }^{2}}\theta +{{\sin }^{3}}\theta =1$ . Start by simplification of the equation followed by squaring both sides of the equation. Finally, convert all the terms to the cosine form to get the required result.

Complete step-by-step answer:
Before moving to the solution, let us discuss the periodicity of sine and cosine function, which we would be using in the solution. All the trigonometric ratios, including sine and cosine, are periodic functions. We can better understand this using the graph of sine and cosine.
First, let us start with the graph of sinx.

Next, let us see the graph of cosx.

Looking at both the graphs, we can say that the graphs are repeating after a fixed period i.e. $2{{\pi }^{c}}$ . So, we can say that the fundamental period of the cosine function and the sine function is $2{{\pi }^{c}}=360{}^\circ $
Now to start with the solution to the above question, we will try to simplify the equation given in the question.
$\sin \theta +{{\sin }^{2}}\theta +{{\sin }^{3}}\theta =1$
$\Rightarrow \sin \theta +{{\sin }^{3}}\theta =1-{{\sin }^{2}}\theta $
Now we know that $1-{{\sin }^{2}}\theta ={{\cos }^{2}}\theta $ . So, we get
$\sin \theta \left( 1+{{\sin }^{2}}\theta \right)={{\cos }^{2}}\theta $
Again using the formula $1-{{\cos }^{2}}\theta ={{\sin }^{2}}\theta $ , we get
$\sin \theta \left( 1+1-{{\cos }^{2}}\theta \right)={{\cos }^{2}}\theta $
$\Rightarrow \sin \theta \left( 2-{{\cos }^{2}}\theta \right)={{\cos }^{2}}\theta $
Now, if we square both sides of the equation, we get
${{\sin }^{2}}\theta {{\left( 2-{{\cos }^{2}}\theta \right)}^{2}}={{\cos }^{4}}\theta $
Again using the formula $1-{{\cos }^{2}}\theta ={{\sin }^{2}}\theta $ and the formula ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ , we get
$\left( 1-{{\cos }^{2}}\theta \right)\left( {{2}^{2}}+{{\cos }^{4}}\theta -4{{\cos }^{2}}\theta \right)={{\cos }^{4}}\theta $
$\Rightarrow \left( 1-{{\cos }^{2}}\theta \right)\left( 4+{{\cos }^{4}}\theta -4{{\cos }^{2}}\theta \right)={{\cos }^{4}}\theta $
$\Rightarrow 4+{{\cos }^{4}}\theta -4{{\cos }^{2}}\theta -4{{\cos }^{2}}\theta -{{\cos }^{6}}\theta +4{{\cos }^{4}}\theta ={{\cos }^{4}}\theta $
 Now, if we rearrange the terms of the equation according to our need, we get
${{\cos }^{6}}\theta -4{{\cos }^{4}}\theta +8{{\cos }^{2}}\theta =4$
So, we can conclude that the equation ${{\cos }^{6}}\theta -4{{\cos }^{4}}\theta +8{{\cos }^{2}}\theta =4$ is proved.

Note: Be careful about the calculation and the signs of the formulas you use as the signs in the formulas are very confusing and are very important for solving the problems. Also, it would help if you remember the properties related to complementary angles and trigonometric ratios.