Question

How do you express $\sin \left( {3\theta } \right)$ in terms of trigonometric functions of $\theta $.

Answer 1

Hint: The given problem requires us to express the sine of angle $\left( {3\theta } \right)$ in terms of trigonometric functions of angle $\theta $. So, we will first use the compound angle formulae of sine to expand the expression of $\sin \left( {3\theta } \right)$. Then, we will use the double angle formulae of sine and cosine to get to the required answer. Double angle formulae for sine and cosine are: $\sin \left( {2x} \right) = 2\sin (x)\cos (x)$ and \[\cos \left( {2x} \right) = \left( {1 - 2{{\sin }^2}x} \right) = \left( {2{{\cos }^2}x - 1} \right)\] respectively.

Complete step-by-step answer:
For simplifying $\sin \left( {3\theta } \right)$ to trigonometric functions of unit $\theta $, we first use the compound angle formulae of sine to convert $\sin \left( {3\theta } \right)$ to trigonometric functions of unit $\left( {2\theta } \right)$ and $\theta $. So, we know the compound angle formula for the sum of two angles is $\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B$. So, first splitting up the angle in two parts, we get,
$\sin \left( {3\theta } \right) = \sin \left( {2\theta + \theta } \right)$
Now, using the compound angle formula, we get,
$ \Rightarrow \sin \left( {3\theta } \right) = \sin 2\theta \cos \theta + \cos 2\theta \sin \theta $
Now, we have the expression of $\sin \left( {3\theta } \right)$ in terms of trigonometric functions of $\theta $ and $\left( {2\theta } \right)$. So, we will convert the functions involving the angle $\left( {2\theta } \right)$ into trigonometric functions involving angle $\theta $ using the double angle formulae of sine and cosine. We know that the double angle formulae of sine and cosine are: $\sin \left( {2x} \right) = 2\sin (x)\cos (x)$ and \[\cos \left( {2x} \right) = \left( {1 - 2{{\sin }^2}x} \right) = \left( {2{{\cos }^2}x - 1} \right)\] respectively. So, we get,
$ \Rightarrow \sin \left( {3\theta } \right) = \left( {2\sin \theta \cos \theta } \right)\cos \theta + \left( {1 - 2{{\sin }^2}\theta } \right)\sin \theta $
Now, simplifying the expression, we get,
\[ \Rightarrow \sin \left( {3\theta } \right) = 2\sin \theta {\cos ^2}\theta + \sin \theta - 2{\sin ^3}\theta \]
Now, we use the trigonometric identity \[{\cos ^2}\theta + {\sin ^2}\theta = 1\]. So, we get,
\[ \Rightarrow \sin \left( {3\theta } \right) = 2\sin \theta \left( {1 - {{\sin }^2}\theta } \right) + \sin \theta - 2{\sin ^3}\theta \]
Opening the brackets, we get,
\[ \Rightarrow \sin \left( {3\theta } \right) = 2\sin \theta - 2{\sin ^3}\theta + \sin \theta - 2{\sin ^3}\theta \]
Adding up the like terms, we get,
\[ \Rightarrow \sin \left( {3\theta } \right) = 3\sin \theta - 4{\sin ^3}\theta \]
Hence, $\sin \left( {3\theta } \right)$ in terms of trigonometric functions of unit $\theta $ is$\left( {3\sin \theta - 4{{\sin }^3}\theta } \right)$.

Note: The above question can also be solved by using compound angle formulae instead of double angle formulae such as $\sin (A + B) = \left( {\sin A\cos B + \cos A\sin B} \right)$ and $\cos (A + B) = \left( {\cos A\cos B - \sin A\sin B} \right)$. This method can also be used to get to the answer of the given problem but involves more calculations. We must take care of simplifying rules and calculations while solving such problems.