Question

Show that \[4\sin {\rm{\theta co}}{{\rm{s}}^3}{\rm{\theta }} - 4\cos {\rm{\theta }}{\sin ^3}{\rm{\theta }} = \]

Answer 1

Hint:
Here, we have to use the basic identities of the trigonometric functions to find out the value of the given equation. So we have to apply the properties of the trigonometric function for the simplification of the equation and by solving the simplified equation we will get the value of the equation.

Complete step by step solution:
Given equation is \[4\sin {\rm{\theta co}}{{\rm{s}}^3}{\rm{\theta }} - 4\cos {\rm{\theta }}{\sin ^3}{\rm{\theta }}\]
Now we have to simplify the given equation by using the properties of trigonometric functions.
So, to simplify the equation here in the equation we can take\[2\sin {\rm{\theta cos\theta }}\] common from both the terms of the equation. Therefore, we get
\[ \Rightarrow 4\sin {\rm{\theta co}}{{\rm{s}}^3}{\rm{\theta }} - 4\cos {\rm{\theta }}{\sin ^3}{\rm{\theta }} = 2 \times 2\sin {\rm{\theta cos\theta }}({\rm{co}}{{\rm{s}}^2}{\rm{\theta }} - {\sin ^2}{\rm{\theta }})\]
Now we know that\[2\sin {\rm{\theta cos\theta = sin 2\theta }}\]. So we have to put this value in the above equation, we get
\[ \Rightarrow 4\sin {\rm{\theta co}}{{\rm{s}}^3}{\rm{\theta }} - 4\cos {\rm{\theta }}{\sin ^3}{\rm{\theta }} = 2 \times {\rm{sin 2\theta }}({\rm{co}}{{\rm{s}}^2}{\rm{\theta }} - {\sin ^2}{\rm{\theta }})\]
Also, we know that\[{\rm{co}}{{\rm{s}}^2}{\rm{\theta }} - {\sin ^2}{\rm{\theta = cos 2\theta }}\]. So by putting this value in the above equation, we get
\[ \Rightarrow 4\sin {\rm{\theta co}}{{\rm{s}}^3}{\rm{\theta }} - 4\cos {\rm{\theta }}{\sin ^3}{\rm{\theta }} = 2 \times {\rm{sin 2\theta }} \times {\rm{cos 2\theta }}\]
Now again using the same property of the trigonometric function i.e. \[2\sin {\rm{\theta cos\theta = sin 2\theta }}\]. Then the equation becomes
\[ \Rightarrow 4\sin {\rm{\theta co}}{{\rm{s}}^3}{\rm{\theta }} - 4\cos {\rm{\theta }}{\sin ^3}{\rm{\theta }} = {\rm{sin 4\theta }}\]

Hence, \[{\rm{sin 4\theta }}\] is the value of the given equation.
So, \[4\sin {\rm{\theta co}}{{\rm{s}}^3}{\rm{\theta }} - 4\cos {\rm{\theta }}{\sin ^3}{\rm{\theta }} = {\rm{sin 4\theta }}\]

Note:
We should know the different properties of the trigonometric function and also in which quadrant which function is positive or negative as in the first quadrant all the functions i.e. sin, cos, tan, cot, sec, cosec are positive. In the second quadrant, only the sin and cosec function are positive and all the other functions are negative. In the third quadrant, only tan and cot function is positive and in the fourth quadrant, only cos and sec function is positive. Also, we should know the basic properties of the trigonometric functions and with the help of this concept, this question can be easily solved.
Properties used in the question: \[2\sin {\rm{\theta cos\theta = sin 2\theta }}\]and\[{\rm{co}}{{\rm{s}}^2}{\rm{\theta }} - {\sin ^2}{\rm{\theta = cos 2\theta }}\]