Question

The expression $3\left[ {{\sin }^{4}}\left\{ \dfrac{3\pi }{2}-\alpha \right\}+{{\sin }^{4}}\left( 3\pi +\alpha \right) \right]-2\left[ {{\sin }^{6}}\left\{ \dfrac{\pi }{2}+\alpha \right\}+{{\sin }^{6}}\left( 5\pi -\alpha \right) \right]$ is equal to:
A. 0
B. 1
C. 3
D. None

Answer 1

Hint: For the above question we will use the quadrant properties of the trigonometric function. We know that in $I{{I}^{nd}}$ and ${{I}^{st}}$ quadrant $\sin \theta $ is always positive otherwise it is negative. Also, we will use the identity as follows:
$\begin{align}
  & \sin \left( \dfrac{\pi }{2}+\theta \right)=\cos \theta \\
 & \sin \left( \dfrac{3\pi }{2}-\theta \right)=-\cos \theta \\
\end{align}$
So, by using the above properties and identity we will get the value of the expression.

Complete step-by-step answer:

We have been asked to find the value of the expression,
$3\left[ {{\sin }^{4}}\left\{ \dfrac{3\pi }{2}-\alpha \right\}+{{\sin }^{4}}\left( 3\pi +\alpha \right) \right]-2\left[ {{\sin }^{6}}\left\{ \dfrac{\pi }{2}+\alpha \right\}+{{\sin }^{6}}\left( 5\pi -\alpha \right) \right]$
We know that $\sin \left( \dfrac{3\pi }{2}-\theta \right)$ lies in the $II{{I}^{rd}}$quadrant and equals to $\left( -\cos \theta \right)$.
$\Rightarrow {{\sin }^{4}}\left( \dfrac{3\pi }{2}-\alpha \right)={{\left( -\cos \alpha \right)}^{4}}={{\cos }^{4}}\alpha $
Again, we know that \[\sin \left( 3\pi +\theta \right)\] lies in the $II{{I}^{rd}}$quadrant and equals $-\sin \theta $.
$\Rightarrow {{\sin }^{4}}\left( 3\pi +\alpha \right)={{\left( -\sin \alpha \right)}^{4}}={{\sin }^{4}}\alpha $
We know that $\sin \left( \dfrac{\pi }{2}+\theta \right)$ lies in the $I{{I}^{nd}}$ quadrant and equals to $\cos \theta $.
$\Rightarrow {{\sin }^{6}}\left( \dfrac{\pi }{2}+d \right)={{\left( \cos \alpha \right)}^{6}}={{\cos }^{6}}\alpha $
Also, we know that $\sin \left( \pi -\theta \right)$ lies in the $I{{I}^{nd}}$ quadrant and equals $\sin \theta $.
$\Rightarrow {{\sin }^{6}}\left( 5\pi -\alpha \right)={{\left( \sin \alpha \right)}^{6}}={{\sin }^{6}}\alpha $
On substituting these values in the given expression, we get,
$=3\left[ {{\cos }^{4}}\alpha +{{\sin }^{4}}\alpha \right]-2\left[ {{\cos }^{6}}\alpha +{{\sin }^{6}}\alpha \right]$
Now, using the identity as follows;
$\begin{align}
  & {{a}^{4}}+{{b}^{4}}={{\left( {{a}^{2}}+{{b}^{2}} \right)}^{2}}-2{{a}^{2}}{{b}^{2}}\ and \\
 & {{a}^{6}}+{{b}^{6}}={{\left( {{a}^{2}}+{{b}^{2}} \right)}^{3}}-3{{a}^{2}}{{b}^{2}}\left( {{a}^{2}}+{{b}^{2}} \right) \\
 & =3\left[ {{\left( {{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha \right)}^{2}}-2{{\cos }^{2}}\alpha {{\sin }^{2}}\alpha \right]-2\left[ {{\left( {{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha \right)}^{3}}-3{{\cos }^{2}}\alpha {{\sin }^{2}}\alpha \times \left( {{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha \right) \right] \\
\end{align}$Since, we know the trigonometric identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$.
$\begin{align}
  & \Rightarrow 3\left[ {{\left( 1 \right)}^{2}}-2{{\cos }^{2}}\alpha {{\sin }^{2}}\alpha \right]-2\left[ {{\left( 1 \right)}^{3}}-3{{\cos }^{2}}\alpha {{\sin }^{2}}\alpha \left( 1 \right) \right] \\
 & =3-6{{\cos }^{2}}\alpha {{\sin }^{2}}\alpha -2+6{{\cos }^{2}}\alpha {{\sin }^{2}}\alpha \\
 & =\left( 3-2 \right)-6{{\cos }^{2}}\alpha {{\sin }^{2}}\alpha +6{{\cos }^{2}}\alpha {{\sin }^{2}}\alpha \\
 & =1 \\
\end{align}$
Hence, the given expression is equal to 1.
Therefore, the correct option is option B.

Note: Be careful while using the trigonometric properties and their identities also take care of the sign mistake during calculation in each step.
Also, remember the property that, $\sin \left( \dfrac{\left( 2n+1 \right)\pi }{2}+\theta \right)={{\left( -1 \right)}^{n}}\cos \theta $ where ‘n’ is any whole number and $\sin \left( n\pi +\theta \right)={{\left( -1 \right)}^{n}}\sin \theta $. These properties will help you in these types of questions.