Question

If sin x + cos x = a, evaluate ${{\sin }^{6}}x+{{\cos }^{6}}x$.

Answer 1

Hint: For solving this problem, first we convert ${{\sin }^{6}}x+{{\cos }^{6}}x$ in terms of exponent 3 as ${{\left( {{\sin }^{2}}x \right)}^{3}}+{{\left( {{\cos }^{2}}x \right)}^{3}}$ to apply the identity of ${{a}^{3}}+{{b}^{3}}$. Now, we simplify the expression by using the identity ${{a}^{3}}+{{b}^{3}}={{\left( a+b \right)}^{3}}-3ab\left( a+b \right)$ . Again, using the algebraic identity ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$, we further simplify the expression to obtain value in terms of a.
Complete step-by-step answer:
According to the problem statement, we are given sin x + cos x = a and we are required to evaluate ${{\sin }^{6}}x+{{\cos }^{6}}x$. Considering the evaluation part ${{\sin }^{6}}x+{{\cos }^{6}}x$ and converting it to exponent 3 as ${{\left( {{\sin }^{2}}x \right)}^{3}}+{{\left( {{\cos }^{2}}x \right)}^{3}}$, we get
$\Rightarrow {{\sin }^{6}}x+{{\cos }^{6}}x={{\left( {{\sin }^{2}}x \right)}^{3}}+{{\left( {{\cos }^{2}}x \right)}^{3}}$
By using the algebraic identity, we know that ${{a}^{3}}+{{b}^{3}}={{\left( a+b \right)}^{3}}-3ab\left( a+b \right)$.
Simplifying the evaluation part by using the above formula and putting $a={{\sin }^{2}}x\text{ and }b\text{=}{{\cos }^{2}}x$, we get
$\Rightarrow {{\left( {{\sin }^{2}}x \right)}^{3}}+{{\left( {{\cos }^{2}}x \right)}^{3}}={{\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)}^{3}}-3\left( {{\sin }^{2}}x \right)\left( {{\cos }^{2}}x \right)\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)$
One of the important trigonometric identities used is ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$.
$\begin{align}
  & \Rightarrow {{\left( 1 \right)}^{3}}-3\left( {{\sin }^{2}}x \right)\left( {{\cos }^{2}}x \right)\left( 1 \right) \\
 & \Rightarrow 1-3\left( {{\sin }^{2}}x \right)\left( {{\cos }^{2}}x \right) \\
 & \Rightarrow 1-3{{\left( \sin x\cdot \cos x \right)}^{2}}\ldots \left( 1 \right) \\
\end{align}$
To further simplify the expression, use the algebraic identity with some manipulation as:
\[\begin{align}
  & {{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab \\
 & 2ab={{\left( a+b \right)}^{2}}-\left( {{a}^{2}}+{{b}^{2}} \right) \\
 & ab=\dfrac{{{\left( a+b \right)}^{2}}-\left( {{a}^{2}}+{{b}^{2}} \right)}{2} \\
\end{align}\]
Now, operating the above formulation in equation (1) and replacing $a=\sin x\text{ and }b=\cos x$, we get
\[\begin{align}
  & \Rightarrow 1-3{{\left[ \dfrac{{{\left( \sin x+\cos x \right)}^{2}}-\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)}{2} \right]}^{2}} \\
 & \Rightarrow 1-\dfrac{3}{{{2}^{2}}}{{\left[ {{\left( \sin x+\cos x \right)}^{2}}-{{\left( 1 \right)}^{2}} \right]}^{2}} \\
 & \Rightarrow 1-\dfrac{3}{4}{{\left[ {{\left( \sin x+\cos x \right)}^{2}}-1 \right]}^{2}} \\
\end{align}\]
At last, substitute sin x + cos x = a, we get
\[\begin{align}
  & \Rightarrow 1-\dfrac{3}{4}{{\left[ {{\left( a \right)}^{2}}-1 \right]}^{2}} \\
 & \Rightarrow 1-\dfrac{3}{4}{{\left[ {{a}^{2}}-1 \right]}^{2}} \\
\end{align}\]
Now, by using the identity ${{\left( x-y \right)}^{2}}={{x}^{2}}+{{y}^{2}}-2xy$, we get
\[\begin{align}
  & \Rightarrow 1-\dfrac{3}{4}{{\left[ {{a}^{2}}-1 \right]}^{2}} \\
  & \Rightarrow 1-\dfrac{3}{4}\left[ {{a}^{4}}+1-2{{a}^{2}} \right] \\
 & \Rightarrow 1-\dfrac{3}{4}-\dfrac{3}{4}\left( {{a}^{4}}-2{{a}^{2}} \right) \\
 & \Rightarrow \dfrac{1}{4}-\dfrac{3{{a}^{2}}}{4}\left( {{a}^{2}}-2 \right) \\
\end{align}\]
So, the obtained answer is \[\dfrac{1}{4}-\dfrac{3{{a}^{2}}}{4}\left( {{a}^{2}}-2 \right)\].
Note: Students must have the knowledge of algebraic expansions for solving this problem. Substitution of different values must be done in a way to simplify the expression. Students must memorise these steps to minimise errors and save time.