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Question: Show that \[4\sin {\rm{\theta co}}{{\rm{s}}^3}{\rm{\theta }} - 4\cos {\rm{\theta }}{\sin ^3}{\rm{\th...

Show that 4sin⁡θcos3θ−4cos⁡θsin⁡3θ=4\sin {\rm{\theta co}}{{\rm{s}}^3}{\rm{\theta }} - 4\cos {\rm{\theta }}{\sin ^3}{\rm{\theta }} =

Explanation

Solution

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 4sin⁡θcos3θ−4cos⁡θsin⁡3θ4\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 take2sin⁡θcosθ2\sin {\rm{\theta cos\theta }} common from both the terms of the equation. Therefore, we get
⇒4sin⁡θcos3θ−4cos⁡θsin⁡3θ=2×2sin⁡θcosθ(cos2θ−sin⁡2θ)\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 that2sin⁡θcosθ=sin2θ2\sin {\rm{\theta cos\theta = sin 2\theta }}. So we have to put this value in the above equation, we get
⇒4sin⁡θcos3θ−4cos⁡θsin⁡3θ=2×sin2θ(cos2θ−sin⁡2θ)\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 thatcos2θ−sin⁡2θ=cos2θ{\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
⇒4sin⁡θcos3θ−4cos⁡θsin⁡3θ=2×sin2θ×cos2θ\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. 2sin⁡θcosθ=sin2θ2\sin {\rm{\theta cos\theta = sin 2\theta }}. Then the equation becomes
⇒4sin⁡θcos3θ−4cos⁡θsin⁡3θ=sin4θ\Rightarrow 4\sin {\rm{\theta co}}{{\rm{s}}^3}{\rm{\theta }} - 4\cos {\rm{\theta }}{\sin ^3}{\rm{\theta }} = {\rm{sin 4\theta }}

Hence, sin4θ{\rm{sin 4\theta }} is the value of the given equation.
So, 4sin⁡θcos3θ−4cos⁡θsin⁡3θ=sin4θ4\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: 2sin⁡θcosθ=sin2θ2\sin {\rm{\theta cos\theta = sin 2\theta }}andcos2θ−sin⁡2θ=cos2θ{\rm{co}}{{\rm{s}}^2}{\rm{\theta }} - {\sin ^2}{\rm{\theta = cos 2\theta }}