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Question: Solve the equation \[{\left( {\sin \theta } \right)^3}\left( {\cos \theta } \right) - {\left( {\cos ...

Solve the equation (sin⁡θ)3(cos⁡θ)−(cos⁡θ)3(sin⁡θ)=14{\left( {\sin \theta } \right)^3}\left( {\cos \theta } \right) - {\left( {\cos \theta } \right)^3}\left( {\sin \theta } \right) = \dfrac{1}{4} for the value of θ\theta .

Explanation

Solution

Hint- Here, we will be simplifying the LHS of the given equation 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⁡2θ=[(cos⁡θ)2−(sin⁡θ)2]\cos 2\theta = \left[ {{{\left( {\cos \theta } \right)}^2} - {{\left( {\sin \theta } \right)}^2}} \right]. Then, with the help of the condition i.e., when sin⁡x=−1\sin x = - 1 then x=2nπ−π2x = 2n\pi - \dfrac{\pi }{2} we will find the required values of θ\theta .

“Complete step-by-step answer:”
The given equation is (sin⁡θ)3(cos⁡θ)−(cos⁡θ)3(sin⁡θ)=14{\left( {\sin \theta } \right)^3}\left( {\cos \theta } \right) - {\left( {\cos \theta } \right)^3}\left( {\sin \theta } \right) = \dfrac{1}{4}.
This equation can be simplified as under

⇒(sin⁡θ)(cos⁡θ)[(sin⁡θ)2−(cos⁡θ)2]=14 ⇒−4(sin⁡θ)(cos⁡θ)[(cos⁡θ)2−(sin⁡θ)2]=1 ⇒−2[2(sin⁡θ)(cos⁡θ)][(cos⁡θ)2−(sin⁡θ)2]=1 →(1)  \Rightarrow \left( {\sin \theta } \right)\left( {\cos \theta } \right)\left[ {{{\left( {\sin \theta } \right)}^2} - {{\left( {\cos \theta } \right)}^2}} \right] = \dfrac{1}{4} \\\ \Rightarrow - 4\left( {\sin \theta } \right)\left( {\cos \theta } \right)\left[ {{{\left( {\cos \theta } \right)}^2} - {{\left( {\sin \theta } \right)}^2}} \right] = 1 \\\ \Rightarrow - 2\left[ {2\left( {\sin \theta } \right)\left( {\cos \theta } \right)} \right]\left[ {{{\left( {\cos \theta } \right)}^2} - {{\left( {\sin \theta } \right)}^2}} \right] = 1{\text{ }} \to {\text{(1)}} \\\

As we know that sin⁡2θ=2(sin⁡θ)(cos⁡θ)\sin 2\theta = 2\left( {\sin \theta } \right)\left( {\cos \theta } \right) and cos⁡2θ=[(cos⁡θ)2−(sin⁡θ)2]\cos 2\theta = \left[ {{{\left( {\cos \theta } \right)}^2} - {{\left( {\sin \theta } \right)}^2}} \right]
Using the above formulas, equation (1) becomes,

⇒−2[sin⁡2θ][cos⁡2θ]=1 ⇒−[2(sin⁡2θ)(cos⁡2θ)]=1 →(2)  \Rightarrow - 2\left[ {\sin 2\theta } \right]\left[ {\cos 2\theta } \right] = 1 \\\ \Rightarrow - \left[ {2\left( {\sin 2\theta } \right)\left( {\cos 2\theta } \right)} \right] = 1{\text{ }} \to {\text{(2)}} \\\

In the formula sin⁡2θ=2(sin⁡θ)(cos⁡θ)\sin 2\theta = 2\left( {\sin \theta } \right)\left( {\cos \theta } \right) replace θ\theta by 2θ2\theta , we get

⇒sin⁡2(2θ)=2(sin⁡2θ)(cos⁡2θ) ⇒sin⁡4θ=2(sin⁡2θ)(cos⁡2θ) →(3)  \Rightarrow \sin 2\left( {2\theta } \right) = 2\left( {\sin 2\theta } \right)\left( {\cos 2\theta } \right) \\\ \Rightarrow \sin 4\theta = 2\left( {\sin 2\theta } \right)\left( {\cos 2\theta } \right){\text{ }} \to {\text{(3)}} \\\

Using equation (3) in equation (2), we get

⇒−[sin⁡4θ]=1 ⇒sin⁡4θ=−1  \Rightarrow - \left[ {\sin 4\theta } \right] = 1 \\\ \Rightarrow \sin 4\theta = - 1 \\\

Also we know that when sin⁡x=−1\sin x = - 1 then x=2nπ−π2x = 2n\pi - \dfrac{\pi }{2} where n∈Zn \in Z (Z is set of integers)
Replacing x in the above condition by 4θ4\theta , we get

4θ=2nπ−π2 ⇒θ=2nπ4−π2×4 ⇒θ=nπ2−π8 where n∈Z  4\theta = 2n\pi - \dfrac{\pi }{2} \\\ \Rightarrow \theta = \dfrac{{2n\pi }}{4} - \dfrac{\pi }{{2 \times 4}} \\\ \Rightarrow \theta = \dfrac{{n\pi }}{2} - \dfrac{\pi }{8}{\text{ where }}n \in Z \\\

Note- In these types of problems, we convert the trigonometric functions of smaller angle (i.e., θ\theta in this case) in the given equation to the trigonometric functions of larger angle (i.e. 4θ4\theta in this case). Also, in this problem sin⁡4θ=−1\sin 4\theta = - 1 is coming after simplification which means the values of 4θ4\theta which are possible are 2nπ−π22n\pi - \dfrac{\pi }{2} where n belongs to the set of integers.