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Question: Prove the following: \(2\sin A{{\cos }^{3}}A-2{{\sin }^{3}}A\cos A=\dfrac{1}{2}\sin 4A\)....

Prove the following: 2sin⁡Acos⁡3A−2sin⁡3Acos⁡A=12sin⁡4A2\sin A{{\cos }^{3}}A-2{{\sin }^{3}}A\cos A=\dfrac{1}{2}\sin 4A.

Explanation

Solution

Hint: To solve this question, we have to use the trigonometric formulas that we have studied in the chapter trigonometry. In trigonometry, we have some formulas that can be used to convert a trigonometric function from single to double angle. One of these formulas is 2 sinA cosA = sin2A. Another formula is cos⁡2A−sin⁡2A=cos⁡2A{{\cos }^{2}}A-{{\sin }^{2}}A=\cos 2A. Using these formulas, we can solve this question.

Complete step by step answer:
Before proceeding with the question, we must know all the formulas that will be required to solve this question.
In trigonometry, we have some formulas that relate trigonometric functions having an angle as an argument to the trigonometric functions having arguments as twice of these angles. One of these formulas is,
2 sinA cosA = sin2A . . . . . . . . . . . . (1)
Another formula is,
cos⁡2A−sin⁡2A=cos⁡2A{{\cos }^{2}}A-{{\sin }^{2}}A=\cos 2A . . . . . . . . . . . . . . . . (2)
In the question, we are required to prove 2sin⁡Acos⁡3A−2sin⁡3Acos⁡A=12sin⁡4A2\sin A{{\cos }^{3}}A-2{{\sin }^{3}}A\cos A=\dfrac{1}{2}\sin 4A. Let us consider the left side of the equation.
2sin⁡Acos⁡3A−2sin⁡3Acos⁡A2\sin A{{\cos }^{3}}A-2{{\sin }^{3}}A\cos A
This can be also written as,
2sin⁡Acos⁡A(cos⁡2A−sin⁡2A)2\sin A\cos A\left( {{\cos }^{2}}A-{{\sin }^{2}}A \right)
Using formula (1), substituting 2 sinA cosA = sin2A in the above equation, we get,
sin⁡2A(cos⁡2A−sin⁡2A)\sin 2A\left( {{\cos }^{2}}A-{{\sin }^{2}}A \right)
Using formula (2), substituting cos⁡2A−sin⁡2A=cos⁡2A{{\cos }^{2}}A-{{\sin }^{2}}A=\cos 2A in the above equation, we get,
sin⁡2A.cos⁡2A\sin 2A.\cos 2A
Let us multiply and divide the above equation by 2.
12(2sin⁡2Acos⁡2A)\dfrac{1}{2}\left( 2\sin 2A\cos 2A \right)
If we substitute A = 2A in formula (1), we can write 2 sin2A cos2A = sin4A. Substituting this in the above equation, we get,
12(2sin⁡2Acos⁡2A)=12sin⁡4A\dfrac{1}{2}\left( 2\sin 2A\cos 2A \right)=\dfrac{1}{2}\sin 4A

Hence, we have proved the left side of the equation given in the question to its right side.

Note: This is an easy question which can be solved using the basic knowledge of trigonometry. The only possible mistake that one can commit in this question is applying wrong formulas. There is a possibility that one my write cos⁡2A−sin⁡2A=1{{\cos }^{2}}A-{{\sin }^{2}}A=1 since in trigonometry, we have an identity cos⁡2A+sin⁡2A=1{{\cos }^{2}}A+{{\sin }^{2}}A=1 which may create a confusion. This mistake will lead us to an incorrect answer.