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Question: Show that \({\left( {\cos \theta } \right)^3}\left( {\sin 3\theta } \right) + {\left( {\sin \theta }...

Show that (cos⁡θ)3(sin⁡3θ)+(sin⁡θ)3(cos⁡3θ)=34(sin⁡4θ){\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).

Explanation

Solution

Hint: Here, we will be proceeding with the help of the trigonometric formulas which are sin⁡3θ=3sin⁡θ−4(sin⁡θ)3\sin 3\theta = 3\sin \theta - 4{\left( {\sin \theta } \right)^3} and cos⁡3θ=4(cos⁡θ)3−3cos⁡θ\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: (cos⁡θ)3(sin⁡3θ)+(sin⁡θ)3(cos⁡3θ)=34(sin⁡4θ){\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⁡3θ=3sin⁡θ−4(sin⁡θ)3\sin 3\theta = 3\sin \theta - 4{\left( {\sin \theta } \right)^3} and cos⁡3θ=4(cos⁡θ)3−3cos⁡θ\cos 3\theta = 4{\left( {\cos \theta } \right)^3} - 3\cos \theta
Taking LHS of the equation which needs to be proved, we have
LHS=(cos⁡θ)3(sin⁡3θ)+(sin⁡θ)3(cos⁡3θ){\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 (sin⁡3θ)\left( {\sin 3\theta } \right) and (cos⁡3θ)\left( {\cos 3\theta } \right) in the above equation, we get

⇒LHS=(cos⁡θ)3[3sin⁡θ−4(sin⁡θ)3]+(sin⁡θ)3[4(cos⁡θ)3−3cos⁡θ] ⇒LHS=3(sin⁡θ)(cos⁡θ)3−4(sin⁡θ)3(cos⁡θ)3+4(cos⁡θ)3(sin⁡θ)3−3(cos⁡θ)(sin⁡θ)3 ⇒LHS=3(sin⁡θ)(cos⁡θ)3−3(cos⁡θ)(sin⁡θ)3  \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(sin⁡θ)(cos⁡θ)3\left( {\sin \theta } \right)\left( {\cos \theta } \right) common in the above equation, we get
⇒LHS=3(sin⁡θ)(cos⁡θ)[(cos⁡θ)2−(sin⁡θ)2]\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 (cos⁡θ)2−(sin⁡θ)2=cos⁡2θ{\left( {\cos \theta } \right)^2} - {\left( {\sin \theta } \right)^2} = \cos 2\theta
⇒LHS=3(sin⁡θ)(cos⁡θ)(cos⁡2θ) →(1)\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⁡2θ=2(sin⁡θ)(cos⁡θ)\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
⇒LHS=3(sin⁡θ)(cos⁡θ)(cos⁡2θ) ⇒LHS=32[2(sin⁡θ)(cos⁡θ)](cos⁡2θ)  \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⁡2θ\sin 2\theta in the above equation, we get
⇒LHS=32(sin⁡2θ)(cos⁡2θ)\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 formulasin⁡2θ=2(sin⁡θ)(cos⁡θ)\sin 2\theta = 2\left( {\sin \theta } \right)\left( {\cos \theta } \right), we get

⇒LHS=32(sin⁡2θ)(cos⁡2θ) ⇒LHS=34[2(sin⁡2θ)(cos⁡2θ)] ⇒LHS=34(sin⁡4θ)=RHS  \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, (cos⁡θ)3(sin⁡3θ)+(sin⁡θ)3(cos⁡3θ)=34(sin⁡4θ){\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 4θ4\theta which is there in RHS of the equation which needs to be proved by using the formulas which are sin⁡2θ=2(sin⁡θ)(cos⁡θ)\sin 2\theta = 2\left( {\sin \theta } \right)\left( {\cos \theta } \right) and (cos⁡θ)2−(sin⁡θ)2=cos⁡2θ{\left( {\cos \theta } \right)^2} - {\left( {\sin \theta } \right)^2} = \cos 2\theta .