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.
$\begin{aligned} & 4{{\sin }^{3}}A\ =3\sin A-\sin 3A \\\ & or \\\ & {{\sin }^{3}}A\ =\dfrac{1}{4}\left( 3\sin A-\sin 3A \right).............\left( 2 \right) \\\ \end{aligned}$
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,
$\begin{aligned} & {{\sin }^{3}}10{}^\circ =\dfrac{1}{4}\left( 3\sin 10{}^\circ -\sin 3\times 10{}^\circ \right) \\\ & =\dfrac{1}{4}\left( 3\sin 10{}^\circ -\sin 30{}^\circ \right) \\\ & or \\\ & {{\sin }^{3}}50{}^\circ =\dfrac{1}{4}\left( 3\sin 50{}^\circ -\sin 3\left( 50{}^\circ \right) \right) \\\ & =\dfrac{1}{4}\left( 3\sin 50{}^\circ -\sin 150{}^\circ \right) \\\ & or \\\ & {{\sin }^{3}}70{}^\circ =\dfrac{1}{4}\left( 3\sin 70{}^\circ -\sin 3\times 70{}^\circ \right) \\\ & =\dfrac{1}{4}\left( 3\sin 70{}^\circ -\sin 210{}^\circ \right) \\\ \end{aligned}$
Now, putting the value of ${{\sin }^{3}}10{}^\circ ,{{\sin }^{3}}50{}^\circ \ and\ {{\sin }^{3}}70{}^\circ$in equation (1), we get,
$\begin{aligned} & {{\sin }^{3}}10{}^\circ +{{\sin }^{3}}50{}^\circ -{{\sin }^{3}}70{}^\circ \\\ & =\dfrac{1}{4}\left( 3\sin 10{}^\circ -\sin 30{}^\circ \right)+\dfrac{1}{4}\left( 3\sin 50{}^\circ -\sin 150{}^\circ \right)-\dfrac{1}{4}\left( 3\sin 70{}^\circ -\sin 210{}^\circ \right) \\\ \end{aligned}$
Now, by separating the like terms, we get,

& =\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{aligned}$$ As from basic trigonometric calculation we may write,$\sin 150{}^\circ \ as\ \sin \left( 180{}^\circ -30{}^\circ \right)\ and\ \sin 210{}^\circ \ as\ \sin \left( 180{}^\circ +30{}^\circ \right)$ $$=\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)$$ Also, $\begin{aligned} & \sin \left( 180{}^\circ +\theta \right)=-\sin \left( \theta \right) \\\ & and \\\ & \sin \left( 180{}^\circ -\theta \right)=\sin \left( \theta \right) \\\ \end{aligned}$ So, from this we can write $\ \sin \left( 180{}^\circ -30{}^\circ \right)\ as\ \sin 30{}^\circ and\ \sin \left( 180{}^\circ +30{}^\circ \right)\ as\ -\sin 30{}^\circ $. As, we know $\sin \left( \theta \right)$ nature in different quadrant as: So, putting the value of $\ \sin \left( 180{}^\circ -30{}^\circ \right)\ as\ \sin 30{}^\circ and\ \sin \left( 180{}^\circ +30{}^\circ \right)\ as\ -\sin 30{}^\circ $ in equation (3), we get, $$=\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{aligned} & =\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{aligned}$$ Also, we know the formula of $\sin C+\sin D\ as\ 2\cos \dfrac{\left( C+D \right)}{2}.\cos \dfrac{\left( C-D \right)}{2}$ . So, applying $\sin C+\sin D\ $to $\sin 50{}^\circ -\ \sin 70{}^\circ $, we get, $\begin{aligned} & \sin 50{}^\circ -\ \sin 70{}^\circ =2\cos \dfrac{\left( 50{}^\circ +70{}^\circ \right)}{2}.\sin \dfrac{\left( 50{}^\circ -70{}^\circ \right)}{2} \\\ & \Rightarrow \sin 50{}^\circ -\ \sin 70{}^\circ =2\cos \dfrac{\left( 120{}^\circ \right)}{2}.\sin \dfrac{\left( -20{}^\circ \right)}{2} \\\ & \Rightarrow \sin 50{}^\circ -\ \sin 70{}^\circ =2\cos 60{}^\circ .\sin \left( -10{}^\circ \right) \\\ & \Rightarrow \sin 50{}^\circ -\ \sin 70{}^\circ =-2\cos 60{}^\circ .\sin 10{}^\circ \\\ \end{aligned}$ Putting the value of $\sin 50{}^\circ -\ \sin 70{}^\circ $as $-2\cos 60{}^\circ .\sin 10{}^\circ $, in equation (5), we get, $$=\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, $\begin{aligned} & \Rightarrow \dfrac{3}{4}\left( 0 \right)-\dfrac{3}{8} \\\ & \Rightarrow 0-\dfrac{3}{8} \\\ & \Rightarrow -\dfrac{3}{8} \\\ \end{aligned}$ 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. $\begin{aligned} & value\ of\ \sin 10{}^\circ =0.17 \\\ & value\ of\ \sin 50{}^\circ =0.76 \\\ & value\ of\ \sin 70{}^\circ =0.93 \\\ \end{aligned}$ Also, $\begin{aligned} & value\ of\ {{\sin }^{3}}10{}^\circ =0.004 \\\ & value\ of\ {{\sin }^{3}}50{}^\circ =0.438 \\\ & value\ of\ {{\sin }^{3}}70{}^\circ =0.80 \\\ \end{aligned}$ 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, $\begin{aligned} & {{\sin }^{3}}10{}^\circ +{{\sin }^{3}}50{}^\circ -{{\sin }^{3}}70{}^\circ \\\ & =0.004+0.438-0.80 \\\ & =0.442-0.80 \\\ & =-0.358 \\\ \end{aligned}$ We get the value as -0.358 because we have considered two decimal places only.