Question

The value of the following trigonometric identity is
${{\sin }^{3}}10{}^\circ +{{\sin }^{3}}50{}^\circ -{{\sin }^{3}}70{}^\circ =$
A. $-\dfrac{3}{2}$
B. $\dfrac{3}{4}$
C. $-\dfrac{3}{4}$
D. $-\dfrac{3}{8}$

Answer 1

Hint: We will use the identity $\sin 3A=3\sin A-4{{\sin }^{3}}A\ $and $\ \sin A-\sin B=2\cos \dfrac{A+B}{2}.\operatorname{Sin}\dfrac{A-B}{2}$ to solve this question. First we use $\sin 3A=3\sin A-4{{\sin }^{3}}A\ $to expand ${{\sin }^{3}}10{}^\circ +{{\sin }^{3}}50{}^\circ -{{\sin }^{3}}70{}^\circ $and then with the help of some simple trigonometric calculation we will solve this question.

Complete step-by-step answer:

It is given that to find the value of
${{\sin }^{3}}10{}^\circ +{{\sin }^{3}}50{}^\circ -{{\sin }^{3}}70{}^\circ ...............\left( 1 \right)$
We know that $\sin 3A=3\sin A-4{{\sin }^{3}}A\ $is an identity used in trigonometric calculation.
$\begin{align}
  & 4{{\sin }^{3}}A\ =3\sin A-\sin 3A \\
 & or \\
 & {{\sin }^{3}}A\ =\dfrac{1}{4}\left( 3\sin A-\sin 3A \right).............\left( 2 \right) \\
\end{align}$
So, with the help of this identity $\sin 3A=3\sin A-4{{\sin }^{3}}A\ $we will solve this question by putting the value of ${{\sin }^{3}}10{}^\circ +{{\sin }^{3}}50{}^\circ -{{\sin }^{3}}70{}^\circ $ in equation (2), we get,
$\begin{align}
  & {{\sin }^{3}}10{}^\circ =\dfrac{1}{4}\left( 3\sin 10{}^\circ -\sin 3\times 10{}^\circ \right) \\
 & =\dfrac{1}{4}\left( 3\sin 10{}^\circ -\sin 30{}^\circ \right) \\
 & or \\
 & {{\sin }^{3}}50{}^\circ =\dfrac{1}{4}\left( 3\sin 50{}^\circ -\sin 3\left( 50{}^\circ \right) \right) \\
 & =\dfrac{1}{4}\left( 3\sin 50{}^\circ -\sin 150{}^\circ \right) \\
 & or \\
 & {{\sin }^{3}}70{}^\circ =\dfrac{1}{4}\left( 3\sin 70{}^\circ -\sin 3\times 70{}^\circ \right) \\
 & =\dfrac{1}{4}\left( 3\sin 70{}^\circ -\sin 210{}^\circ \right) \\
\end{align}$
Now, putting the value of ${{\sin }^{3}}10{}^\circ ,{{\sin }^{3}}50{}^\circ \ and\ {{\sin }^{3}}70{}^\circ $in equation (1), we get,
$\begin{align}
  & {{\sin }^{3}}10{}^\circ +{{\sin }^{3}}50{}^\circ -{{\sin }^{3}}70{}^\circ \\
 & =\dfrac{1}{4}\left( 3\sin 10{}^\circ -\sin 30{}^\circ \right)+\dfrac{1}{4}\left( 3\sin 50{}^\circ -\sin 150{}^\circ \right)-\dfrac{1}{4}\left( 3\sin 70{}^\circ -\sin 210{}^\circ \right) \\
\end{align}$
Now, by separating the like terms, we get,
\[\begin{align}
  & =\dfrac{1}{4}\left( 3\sin 10{}^\circ +3\sin 50{}^\circ -\sin 70{}^\circ \right)+\dfrac{1}{4}\left( \sin 30{}^\circ +\sin 150{}^\circ -\sin 210{}^\circ \right) \\
 & =\dfrac{3}{4}\left( \sin 10{}^\circ +\sin 50{}^\circ -\sin 70{}^\circ \right)+\dfrac{1}{4}\left( \sin 30{}^\circ +\sin 150{}^\circ -\sin 210{}^\circ \right) \\
\end{align}\]
As from basic trigonometric calculation we may write,$\sin 150{}^\circ \ as\ \sin \left( 180{}^\circ -30{}^\circ \right)\ and\ \sin 210{}^\circ \ as\ \sin \left( 180{}^\circ +30{}^\circ \right)$ \[=\dfrac{3}{4}\left( \sin 10{}^\circ +\sin 50{}^\circ -\sin 70{}^\circ \right)+\dfrac{1}{4}\left( \sin 30{}^\circ +\sin \left( 180{}^\circ -30{}^\circ \right)-\sin \left( 180{}^\circ +30{}^\circ \right) \right).......\left( 3 \right)\]
Also,
$\begin{align}
  & \sin \left( 180{}^\circ +\theta \right)=-\sin \left( \theta \right) \\
 & and \\
 & \sin \left( 180{}^\circ -\theta \right)=\sin \left( \theta \right) \\
\end{align}$
So, from this we can write $\ \sin \left( 180{}^\circ -30{}^\circ \right)\ as\ \sin 30{}^\circ and\ \sin \left( 180{}^\circ +30{}^\circ \right)\ as\ -\sin 30{}^\circ $.
As, we know $\sin \left( \theta \right)$ nature in different quadrant as:

So, putting the value of $\ \sin \left( 180{}^\circ -30{}^\circ \right)\ as\ \sin 30{}^\circ and\ \sin \left( 180{}^\circ +30{}^\circ \right)\ as\ -\sin 30{}^\circ $ in equation (3), we get,
\[=\dfrac{3}{4}\left( \sin 10{}^\circ +\sin 50{}^\circ -\sin 70{}^\circ \right)-\dfrac{1}{4}\left( \sin 30{}^\circ +\sin 30{}^\circ +\sin 30{}^\circ \right).......\left( 4 \right)\]
Also, we know that the value of \[\sin 30{}^\circ =\dfrac{1}{2}\]. So, putting the value of \[\sin 30{}^\circ \]in equation (4), we get,
\[\begin{align}
  & =\dfrac{3}{4}\left( \sin 10{}^\circ +\sin 50{}^\circ -\sin 70{}^\circ \right)-\dfrac{1}{4}\left( \dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2} \right) \\
 & =\dfrac{3}{4}\left( \sin 10{}^\circ +\sin 50{}^\circ -\sin 70{}^\circ \right)-\dfrac{1}{4}\times \dfrac{3}{2} \\
 & =\dfrac{3}{4}\left( \sin 10{}^\circ +\sin 50{}^\circ -\sin 70{}^\circ \right)-\dfrac{3}{8}............\left( 5 \right) \\
\end{align}\]
Also, we know the formula of $\sin C+\sin D\ as\ 2\cos \dfrac{\left( C+D \right)}{2}.\cos \dfrac{\left( C-D \right)}{2}$ .
So, applying $\sin C+\sin D\ $to $\sin 50{}^\circ -\ \sin 70{}^\circ $, we get,
$\begin{align}
  & \sin 50{}^\circ -\ \sin 70{}^\circ =2\cos \dfrac{\left( 50{}^\circ +70{}^\circ \right)}{2}.\sin \dfrac{\left( 50{}^\circ -70{}^\circ \right)}{2} \\
 & \Rightarrow \sin 50{}^\circ -\ \sin 70{}^\circ =2\cos \dfrac{\left( 120{}^\circ \right)}{2}.\sin \dfrac{\left( -20{}^\circ \right)}{2} \\
 & \Rightarrow \sin 50{}^\circ -\ \sin 70{}^\circ =2\cos 60{}^\circ .\sin \left( -10{}^\circ \right) \\
 & \Rightarrow \sin 50{}^\circ -\ \sin 70{}^\circ =-2\cos 60{}^\circ .\sin 10{}^\circ \\
\end{align}$
Putting the value of $\sin 50{}^\circ -\ \sin 70{}^\circ $as $-2\cos 60{}^\circ .\sin 10{}^\circ $, in equation (5), we get,
\[=\dfrac{3}{4}\left( \sin 10{}^\circ -2\cos 60{}^\circ .\sin 10{}^\circ \right)-\dfrac{3}{8}..........\left( 6 \right)\]
We know that the value of $\cos 60{}^\circ $ is equal to $\dfrac{1}{2}$. So, putting the value of $\cos 60{}^\circ $ as $\dfrac{1}{2}$ in equation (6), we get,
\[=\dfrac{3}{4}\left( \sin 10{}^\circ -2\times \left( \dfrac{1}{2} \right).\sin 10{}^\circ \right)-\dfrac{3}{8}..........\left( 7 \right)\]
\[=\dfrac{3}{4}\left( \sin 10{}^\circ -1.\sin 10{}^\circ \right)-\dfrac{3}{8}..........\left( 8 \right)\]
Cancelling the similar terms in equation (8), we get,
$\begin{align}
  & \Rightarrow \dfrac{3}{4}\left( 0 \right)-\dfrac{3}{8} \\
 & \Rightarrow 0-\dfrac{3}{8} \\
 & \Rightarrow -\dfrac{3}{8} \\
\end{align}$
Thus, the value of ${{\sin }^{3}}10{}^\circ +{{\sin }^{3}}50{}^\circ -{{\sin }^{3}}70{}^\circ $ is $-\dfrac{3}{8}$ and option (D) is the correct answer.

Note: An alternate method to solve this question is by putting the values of $\sin \theta $ directly in question and by performing some basic mathematical operation. This method is only possible if you are provided with trigonometric tables or are permitted to use a calculator. You can solve this question in just a few lines.
$\begin{align}
  & value\ of\ \sin 10{}^\circ =0.17 \\
 & value\ of\ \sin 50{}^\circ =0.76 \\
 & value\ of\ \sin 70{}^\circ =0.93 \\
\end{align}$
Also,
$\begin{align}
  & value\ of\ {{\sin }^{3}}10{}^\circ =0.004 \\
 & value\ of\ {{\sin }^{3}}50{}^\circ =0.438 \\
 & value\ of\ {{\sin }^{3}}70{}^\circ =0.80 \\
\end{align}$
Putting value of ${{\sin }^{3}}10{}^\circ ,{{\sin }^{3}}50{}^\circ \ and\ {{\sin }^{3}}70{}^\circ $in the given equation ${{\sin }^{3}}10{}^\circ +{{\sin }^{3}}50{}^\circ -{{\sin }^{3}}70{}^\circ $, we get,
$\begin{align}
  & {{\sin }^{3}}10{}^\circ +{{\sin }^{3}}50{}^\circ -{{\sin }^{3}}70{}^\circ \\
 & =0.004+0.438-0.80 \\
 & =0.442-0.80 \\
 & =-0.358 \\
\end{align}$
We get the value as -0.358 because we have considered two decimal places only.