Question

If $\sin \theta +\cos \theta =x$ , Prove that ${{\sin }^{6}}\theta +{{\cos }^{6}}\theta =\dfrac{4-3{{\left( {{x}^{2}}-1 \right)}^{2}}}{4}$

Answer 1

Hint: In this question,we will use some algebraic formulas such that ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$, ${{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab(a+b)$ and trigonometric identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$.


Complete step-by-step answer:

It is given that, $\sin \theta +\cos \theta =x$
Squaring the given equation, we get
${{\left( \sin \theta +\cos \theta \right)}^{2}}={{x}^{2}}$
Appling the formula ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$, we get
${{\sin }^{2}}\theta +2\sin \theta \cos \theta +{{\cos }^{2}}\theta ={{x}^{2}}$
Rearranging the terms, we get
$\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)+2\sin \theta \cos \theta ={{x}^{2}}$
We know that, the trigonometric identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$
$1+2\sin \theta \cos \theta ={{x}^{2}}$
$2\sin \theta \cos \theta ={{x}^{2}}-1$
Dividing both sides by 2, we get
$\sin \theta \cos \theta =\dfrac{{{x}^{2}}-1}{2}$
Again, squaring on both sides, we get
${{\left( \sin \theta \cos \theta \right)}^{2}}={{\left( \dfrac{{{x}^{2}}-1}{2} \right)}^{2}}$
${{\sin }^{2}}\theta {{\cos }^{2}}\theta =\dfrac{{{({{x}^{2}}-1)}^{2}}}{4}................(1)$
Let us consider the L. H. S.
${{\sin }^{6}}\theta +{{\cos }^{6}}\theta ={{\left( {{\sin }^{2}}\theta \right)}^{3}}+{{\left( {{\cos }^{2}}\theta \right)}^{3}}$
Appling the formula ${{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab(a+b)$ and it is rearranging as follow${{a}^{3}}+{{b}^{3}}={{\left( a+b \right)}^{3}}-3ab(a+b)$, we get
${{\sin }^{6}}\theta +{{\cos }^{6}}\theta ={{\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)}^{3}}-3{{\sin }^{2}}\theta {{\cos }^{2}}\theta \left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right).........(2)$
Now put the value from equation (1) ${{\sin }^{2}}\theta {{\cos }^{2}}\theta =\dfrac{{{({{x}^{2}}-1)}^{2}}}{4}$ in the equation (2) and trigonometric identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, we get
\[{{\sin }^{6}}\theta +{{\cos }^{6}}\theta ={{\left( 1 \right)}^{3}}-3\dfrac{{{({{x}^{2}}-1)}^{2}}}{4}\left( 1 \right)\]
Rearranging the terms, we get
\[{{\sin }^{6}}\theta +{{\cos }^{6}}\theta =1-3\dfrac{{{({{x}^{2}}-1)}^{2}}}{4}\]
Taking the LCM on the right side, we get
\[{{\sin }^{6}}\theta +{{\cos }^{6}}\theta =\dfrac{4-3{{({{x}^{2}}-1)}^{2}}}{4}\]
This is the desired result.

Note: We might get confused the algebraic expansion of ${{(a+b)}^{3}}$ and ${{a}^{3}}-{{b}^{3}}$. The algebraic expansion of ${{(a+b)}^{3}}$ is ${{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}$ and algebraic expansion of ${{a}^{3}}-{{b}^{3}}$ is ${{\left( a+b \right)}^{3}}-3ab(a+b)$. Both the algebraic expansions are not equal.