
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
520.2k+ views
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.
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Trending doubts
Draw a labelled sketch of the human eye class 12 physics CBSE

The final image formed by a compound microscope is class 12 physics CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

What are the major means of transport Explain each class 12 social science CBSE

Which of the following properties of a proton can change class 12 physics CBSE

What is the energy band gap of silicon and germanium class 12 physics CBSE
