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Question: Prove that: \({\sin ^3}x + {\cos ^3}x = (\sin x + \cos x)(1 - \sin x\cos x)\)....

Prove that: sin⁡3x+cos⁡3x=(sin⁡x+cos⁡x)(1−sin⁡xcos⁡x){\sin ^3}x + {\cos ^3}x = (\sin x + \cos x)(1 - \sin x\cos x).

Explanation

Solution

In this question, we are given an equation in which we are asked to prove that LHS = RHS. Start with expanding the LHS of the question. Use the cubic formula and simplify the equation. After a step, you will notice sin⁡2x+cos⁡2x{\sin ^2}x + {\cos ^2}x in the second part of the formulae. Use trigonometric identity to simplify it and it will prove LHS = RHS.

Formula used: 1) a3+b3=(a+b)(a2−ab+b2){a^3} + {b^3} = (a + b)({a^2} - ab + {b^2})
2) sin⁡2x+cos⁡2x=1{\sin ^2}x + {\cos ^2}x = 1

Complete step-by-step answer:
We are given an equation in which we have to prove that LHS = RHS.
⇒sin⁡3x+cos⁡3x=(sin⁡x+cos⁡x)(1−sin⁡xcos⁡x)\Rightarrow {\sin ^3}x + {\cos ^3}x = (\sin x + \cos x)(1 - \sin x\cos x)
The LHS of the given equation resembles (a3+b3{a^3} + {b^3}). So, we will use the formula a3+b3=(a+b)(a2−ab+b2){a^3} + {b^3} = (a + b)({a^2} - ab + {b^2})
Expanding the Left Hand Side (LHS),
⇒sin⁡3x+cos⁡3x=(sin⁡x+cos⁡x)(sin⁡2x−sin⁡xcos⁡x+cos⁡2x)\Rightarrow {\sin ^3}x + {\cos ^3}x = (\sin x + \cos x)({\sin ^2}x - \sin x\cos x + {\cos ^2}x)
Rearranging the terms,
⇒sin⁡3x+cos⁡3x=(sin⁡x+cos⁡x)(sin⁡2x+cos⁡2x−sin⁡xcos⁡x)\Rightarrow {\sin ^3}x + {\cos ^3}x = (\sin x + \cos x)({\sin ^2}x + {\cos ^2}x - \sin x\cos x)
Now, we know that sin⁡2x+cos⁡2x=1{\sin ^2}x + {\cos ^2}x = 1. Substituting this is the above equation,
⇒sin⁡3x+cos⁡3x=(sin⁡x+cos⁡x)(1−sin⁡xcos⁡x)\Rightarrow {\sin ^3}x + {\cos ^3}x = (\sin x + \cos x)(1 - \sin x\cos x)= RHS
Therefore, LHS=RHS.
Hence, proved.

Note: We can solve this problem in another way. Here below is an alternate method of solution for this problem.
If you do not remember the cubic formula, then you can use the following method-
⇒sin⁡3x+cos⁡3x=(sin⁡x+cos⁡x)(sin⁡2x−sin⁡xcos⁡x+cos⁡2x)\Rightarrow {\sin ^3}x + {\cos ^3}x = (\sin x + \cos x)({\sin ^2}x - \sin x\cos x + {\cos ^2}x) (given)
We can write x3{x^3} as xx2x{x^2}. Similarly, we will expand the LHS of the given equation.
⇒sin⁡3x+cos⁡3x=sin⁡xsin⁡2x+cos⁡xcos⁡2x\Rightarrow {\sin ^3}x + {\cos ^3}x = \sin x{\sin ^2}x + \cos x{\cos ^2}x
We know that sin⁡2x+cos⁡2x=1{\sin ^2}x + {\cos ^2}x = 1 ⇒\Rightarrow sin⁡2x=1−cos⁡2x{\sin ^2}x = 1 - {\cos ^2}x and cos⁡2x=1−sin⁡2x{\cos ^2}x = 1 - {\sin ^2}x
Using these two,
⇒sin⁡3x+cos⁡3x=sin⁡x(1−cos⁡2x)+cos⁡x(1−sin⁡2x)\Rightarrow {\sin ^3}x + {\cos ^3}x = \sin x(1 - {\cos ^2}x) + \cos x(1 - {\sin ^2}x)
Opening the brackets,
⇒sin⁡3x+cos⁡3x=sin⁡x−sin⁡xcos⁡2x+cos⁡x−cos⁡xsin⁡2x\Rightarrow {\sin ^3}x + {\cos ^3}x = \sin x - \sin x{\cos ^2}x + \cos x - \cos x{\sin ^2}x
Rearranging,
⇒sin⁡3x+cos⁡3x=(sin⁡x+cos⁡x)−sin⁡xcos⁡x(sin⁡x+cos⁡x)\Rightarrow {\sin ^3}x + {\cos ^3}x = (\sin x + \cos x) - \sin x\cos x(\sin x + \cos x)
Taking (sin⁡x+cos⁡x)\left( {\sin x + \cos x} \right) common,
⇒sin⁡3x+cos⁡3x=(sin⁡x+cos⁡x)(1−sin⁡xcos⁡x)\Rightarrow {\sin ^3}x + {\cos ^3}x = (\sin x + \cos x)(1 - \sin x\cos x)= RHS
Therefore, LHS = RHS.
sin⁡3x+cos⁡3x=(sin⁡x+cos⁡x)(1−sin⁡xcos⁡x){\sin ^3}x + {\cos ^3}x = (\sin x + \cos x)(1 - \sin x\cos x)
Hence, proved.