# Show that ${\left( {\cos \theta } \right)^3}\left( {\sin 3\theta } \right) + {\left( {\sin \theta } \right)^3}\left( {\cos 3\theta } \right) = \dfrac{3}{4}\left( {\sin 4\theta } \right)$.

Last updated date: 27th Mar 2023

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Answer

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Hint: Here, we will be proceeding with the help of the trigonometric formulas which are $\sin 3\theta = 3\sin \theta - 4{\left( {\sin \theta } \right)^3}$ and $\cos 3\theta = 4{\left( {\cos \theta } \right)^3} - 3\cos \theta $ in order to simplify the LHS of the equation which needs to be proved.

Complete step-by-step answer:

To show: ${\left( {\cos \theta } \right)^3}\left( {\sin 3\theta } \right) + {\left( {\sin \theta } \right)^3}\left( {\cos 3\theta } \right) = \dfrac{3}{4}\left( {\sin 4\theta } \right)$

As we know that $\sin 3\theta = 3\sin \theta - 4{\left( {\sin \theta } \right)^3}$ and $\cos 3\theta = 4{\left( {\cos \theta } \right)^3} - 3\cos \theta $

Taking LHS of the equation which needs to be proved, we have

\[{\text{LHS}} = {\left( {\cos \theta } \right)^3}\left( {\sin 3\theta } \right) + {\left( {\sin \theta } \right)^3}\left( {\cos 3\theta } \right)\]

By substituting the formulas for \[\left( {\sin 3\theta } \right)\] and \[\left( {\cos 3\theta } \right)\] in the above equation, we get

\[

\Rightarrow {\text{LHS}} = {\left( {\cos \theta } \right)^3}\left[ {3\sin \theta - 4{{\left( {\sin \theta } \right)}^3}} \right] + {\left( {\sin \theta } \right)^3}\left[ {4{{\left( {\cos \theta } \right)}^3} - 3\cos \theta } \right] \\

\Rightarrow {\text{LHS}} = 3\left( {\sin \theta } \right){\left( {\cos \theta } \right)^3} - 4{\left( {\sin \theta } \right)^3}{\left( {\cos \theta } \right)^3} + 4{\left( {\cos \theta } \right)^3}{\left( {\sin \theta } \right)^3} - 3\left( {\cos \theta } \right){\left( {\sin \theta } \right)^3} \\

\Rightarrow {\text{LHS}} = 3\left( {\sin \theta } \right){\left( {\cos \theta } \right)^3} - 3\left( {\cos \theta } \right){\left( {\sin \theta } \right)^3} \\

\]

Now, taking \[3\left( {\sin \theta } \right)\left( {\cos \theta } \right)\] common in the above equation, we get

\[ \Rightarrow {\text{LHS}} = 3\left( {\sin \theta } \right)\left( {\cos \theta } \right)\left[ {{{\left( {\cos \theta } \right)}^2} - {{\left( {\sin \theta } \right)}^2}} \right]\]

Also we know that ${\left( {\cos \theta } \right)^2} - {\left( {\sin \theta } \right)^2} = \cos 2\theta $

$ \Rightarrow {\text{LHS}} = 3\left( {\sin \theta } \right)\left( {\cos \theta } \right)\left( {\cos 2\theta } \right){\text{ }} \to {\text{(1)}}$

Since we know that $\sin 2\theta = 2\left( {\sin \theta } \right)\left( {\cos \theta } \right)$

Let us multiply and divide the RHS of equation (1) by 2, we get

$

\Rightarrow {\text{LHS}} = 3\left( {\sin \theta } \right)\left( {\cos \theta } \right)\left( {\cos 2\theta } \right) \\

\Rightarrow {\text{LHS}} = \dfrac{3}{2}\left[ {2\left( {\sin \theta } \right)\left( {\cos \theta } \right)} \right]\left( {\cos 2\theta } \right) \\

$

By substituting the formula for $\sin 2\theta $ in the above equation, we get

$ \Rightarrow {\text{LHS}} = \dfrac{3}{2}\left( {\sin 2\theta } \right)\left( {\cos 2\theta } \right)$

Again multiply and divide the RHS of above equation by 2 and using the formula$\sin 2\theta = 2\left( {\sin \theta } \right)\left( {\cos \theta } \right)$, we get

\[

\Rightarrow {\text{LHS}} = \dfrac{3}{2}\left( {\sin 2\theta } \right)\left( {\cos 2\theta } \right) \\

\Rightarrow {\text{LHS}} = \dfrac{3}{4}\left[ {2\left( {\sin 2\theta } \right)\left( {\cos 2\theta } \right)} \right] \\

\Rightarrow {\text{LHS}} = \dfrac{3}{4}\left( {\sin 4\theta } \right) = {\text{RHS}} \\

\]

Clearly, from the above equation it is clear that the LHS of the equation which needs to be proved is equal to its RHS.

Hence, ${\left( {\cos \theta } \right)^3}\left( {\sin 3\theta } \right) + {\left( {\sin \theta } \right)^3}\left( {\cos 3\theta } \right) = \dfrac{3}{4}\left( {\sin 4\theta } \right)$

Note: In this particular problem, we will somehow convert the LHS of the equation which needs to be proved in terms of some trigonometric function with angle $4\theta $ which is there in RHS of the equation which needs to be proved by using the formulas which are $\sin 2\theta = 2\left( {\sin \theta } \right)\left( {\cos \theta } \right)$ and ${\left( {\cos \theta } \right)^2} - {\left( {\sin \theta } \right)^2} = \cos 2\theta $.

Complete step-by-step answer:

To show: ${\left( {\cos \theta } \right)^3}\left( {\sin 3\theta } \right) + {\left( {\sin \theta } \right)^3}\left( {\cos 3\theta } \right) = \dfrac{3}{4}\left( {\sin 4\theta } \right)$

As we know that $\sin 3\theta = 3\sin \theta - 4{\left( {\sin \theta } \right)^3}$ and $\cos 3\theta = 4{\left( {\cos \theta } \right)^3} - 3\cos \theta $

Taking LHS of the equation which needs to be proved, we have

\[{\text{LHS}} = {\left( {\cos \theta } \right)^3}\left( {\sin 3\theta } \right) + {\left( {\sin \theta } \right)^3}\left( {\cos 3\theta } \right)\]

By substituting the formulas for \[\left( {\sin 3\theta } \right)\] and \[\left( {\cos 3\theta } \right)\] in the above equation, we get

\[

\Rightarrow {\text{LHS}} = {\left( {\cos \theta } \right)^3}\left[ {3\sin \theta - 4{{\left( {\sin \theta } \right)}^3}} \right] + {\left( {\sin \theta } \right)^3}\left[ {4{{\left( {\cos \theta } \right)}^3} - 3\cos \theta } \right] \\

\Rightarrow {\text{LHS}} = 3\left( {\sin \theta } \right){\left( {\cos \theta } \right)^3} - 4{\left( {\sin \theta } \right)^3}{\left( {\cos \theta } \right)^3} + 4{\left( {\cos \theta } \right)^3}{\left( {\sin \theta } \right)^3} - 3\left( {\cos \theta } \right){\left( {\sin \theta } \right)^3} \\

\Rightarrow {\text{LHS}} = 3\left( {\sin \theta } \right){\left( {\cos \theta } \right)^3} - 3\left( {\cos \theta } \right){\left( {\sin \theta } \right)^3} \\

\]

Now, taking \[3\left( {\sin \theta } \right)\left( {\cos \theta } \right)\] common in the above equation, we get

\[ \Rightarrow {\text{LHS}} = 3\left( {\sin \theta } \right)\left( {\cos \theta } \right)\left[ {{{\left( {\cos \theta } \right)}^2} - {{\left( {\sin \theta } \right)}^2}} \right]\]

Also we know that ${\left( {\cos \theta } \right)^2} - {\left( {\sin \theta } \right)^2} = \cos 2\theta $

$ \Rightarrow {\text{LHS}} = 3\left( {\sin \theta } \right)\left( {\cos \theta } \right)\left( {\cos 2\theta } \right){\text{ }} \to {\text{(1)}}$

Since we know that $\sin 2\theta = 2\left( {\sin \theta } \right)\left( {\cos \theta } \right)$

Let us multiply and divide the RHS of equation (1) by 2, we get

$

\Rightarrow {\text{LHS}} = 3\left( {\sin \theta } \right)\left( {\cos \theta } \right)\left( {\cos 2\theta } \right) \\

\Rightarrow {\text{LHS}} = \dfrac{3}{2}\left[ {2\left( {\sin \theta } \right)\left( {\cos \theta } \right)} \right]\left( {\cos 2\theta } \right) \\

$

By substituting the formula for $\sin 2\theta $ in the above equation, we get

$ \Rightarrow {\text{LHS}} = \dfrac{3}{2}\left( {\sin 2\theta } \right)\left( {\cos 2\theta } \right)$

Again multiply and divide the RHS of above equation by 2 and using the formula$\sin 2\theta = 2\left( {\sin \theta } \right)\left( {\cos \theta } \right)$, we get

\[

\Rightarrow {\text{LHS}} = \dfrac{3}{2}\left( {\sin 2\theta } \right)\left( {\cos 2\theta } \right) \\

\Rightarrow {\text{LHS}} = \dfrac{3}{4}\left[ {2\left( {\sin 2\theta } \right)\left( {\cos 2\theta } \right)} \right] \\

\Rightarrow {\text{LHS}} = \dfrac{3}{4}\left( {\sin 4\theta } \right) = {\text{RHS}} \\

\]

Clearly, from the above equation it is clear that the LHS of the equation which needs to be proved is equal to its RHS.

Hence, ${\left( {\cos \theta } \right)^3}\left( {\sin 3\theta } \right) + {\left( {\sin \theta } \right)^3}\left( {\cos 3\theta } \right) = \dfrac{3}{4}\left( {\sin 4\theta } \right)$

Note: In this particular problem, we will somehow convert the LHS of the equation which needs to be proved in terms of some trigonometric function with angle $4\theta $ which is there in RHS of the equation which needs to be proved by using the formulas which are $\sin 2\theta = 2\left( {\sin \theta } \right)\left( {\cos \theta } \right)$ and ${\left( {\cos \theta } \right)^2} - {\left( {\sin \theta } \right)^2} = \cos 2\theta $.

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