# Solve the equation \[{\left( {\sin \theta } \right)^3}\left( {\cos \theta } \right) - {\left( {\cos \theta } \right)^3}\left( {\sin \theta } \right) = \dfrac{1}{4}\] for the value of $\theta $.

Answer

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Hint- Here, we will be simplifying the LHS of the given equation by using the formulas which are \[\sin 2\theta = 2\left( {\sin \theta } \right)\left( {\cos \theta } \right)\] and \[\cos 2\theta = \left[ {{{\left( {\cos \theta } \right)}^2} - {{\left( {\sin \theta } \right)}^2}} \right]\]. Then, with the help of the condition i.e., when \[\sin x = - 1\] then \[x = 2n\pi - \dfrac{\pi }{2}\] we will find the required values of $\theta $.

“Complete step-by-step answer:”

The given equation is \[{\left( {\sin \theta } \right)^3}\left( {\cos \theta } \right) - {\left( {\cos \theta } \right)^3}\left( {\sin \theta } \right) = \dfrac{1}{4}\].

This equation can be simplified as under

\[

\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 \[\sin 2\theta = 2\left( {\sin \theta } \right)\left( {\cos \theta } \right)\] and \[\cos 2\theta = \left[ {{{\left( {\cos \theta } \right)}^2} - {{\left( {\sin \theta } \right)}^2}} \right]\]

Using the above formulas, equation (1) becomes,

\[

\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 \[\sin 2\theta = 2\left( {\sin \theta } \right)\left( {\cos \theta } \right)\] replace \[\theta \] by \[2\theta \], we get

\[

\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

\[

\Rightarrow - \left[ {\sin 4\theta } \right] = 1 \\

\Rightarrow \sin 4\theta = - 1 \\

\]

Also we know that when \[\sin x = - 1\] then \[x = 2n\pi - \dfrac{\pi }{2}\] where $n \in Z$ (Z is set of integers)

Replacing x in the above condition by \[4\theta \], we get

\[

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., \[\theta \] in this case) in the given equation to the trigonometric functions of larger angle (i.e. \[4\theta \] in this case). Also, in this problem \[\sin 4\theta = - 1\] is coming after simplification which means the values of \[4\theta \] which are possible are \[2n\pi - \dfrac{\pi }{2}\] where n belongs to the set of integers.

“Complete step-by-step answer:”

The given equation is \[{\left( {\sin \theta } \right)^3}\left( {\cos \theta } \right) - {\left( {\cos \theta } \right)^3}\left( {\sin \theta } \right) = \dfrac{1}{4}\].

This equation can be simplified as under

\[

\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 \[\sin 2\theta = 2\left( {\sin \theta } \right)\left( {\cos \theta } \right)\] and \[\cos 2\theta = \left[ {{{\left( {\cos \theta } \right)}^2} - {{\left( {\sin \theta } \right)}^2}} \right]\]

Using the above formulas, equation (1) becomes,

\[

\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 \[\sin 2\theta = 2\left( {\sin \theta } \right)\left( {\cos \theta } \right)\] replace \[\theta \] by \[2\theta \], we get

\[

\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

\[

\Rightarrow - \left[ {\sin 4\theta } \right] = 1 \\

\Rightarrow \sin 4\theta = - 1 \\

\]

Also we know that when \[\sin x = - 1\] then \[x = 2n\pi - \dfrac{\pi }{2}\] where $n \in Z$ (Z is set of integers)

Replacing x in the above condition by \[4\theta \], we get

\[

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., \[\theta \] in this case) in the given equation to the trigonometric functions of larger angle (i.e. \[4\theta \] in this case). Also, in this problem \[\sin 4\theta = - 1\] is coming after simplification which means the values of \[4\theta \] which are possible are \[2n\pi - \dfrac{\pi }{2}\] where n belongs to the set of integers.

Last updated date: 20th Sep 2023

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