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NCERT Exemplar · Q9

Q.Prove that sin⁡4A=4sin⁡Acos⁡3A−4cos⁡Asin⁡3A\sin 4A = 4\sin A\cos^3 A - 4\cos A\sin^3 A.

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We prove the identity by expanding sin⁡4A\sin 4A using the double angle formula sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x \cos x, and then further expanding sin⁡2A\sin 2A and cos⁡2A\cos 2A using their respective double angle formulas, specifically choosing cos⁡2A=cos⁡2A−sin⁡2A\cos 2A = \cos^2 A - \sin^2 A to directly obtain the desired terms.

To prove a trigonometric identity, we typically start with one side (usually the more complex one) and manipulate it using known trigonometric formulas until it transforms into the other side. In this problem, the Left Hand Side (LHS) is sin⁡4A\sin 4A, which is a multiple angle. The Right Hand Side (RHS) involves sin⁡A\sin A and cos⁡A\cos A raised to powers. This tells us we need to break down sin⁡4A\sin 4A into expressions involving sin⁡A\sin A and cos⁡A\cos A.

The most natural way to approach sin⁡4A\sin 4A is to see it as sin⁡(2×2A)\sin(2 \times 2A). This allows us to use the double angle formula for sine. After the first application, we will have terms involving sin⁡2A\sin 2A and cos⁡2A\cos 2A, which can then be further expanded using their respective double angle formulas. The key will be selecting the appropriate form for cos⁡2A\cos 2A to simplify directly to the RHS.

Here are the standard double angle formulas we will use:

sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x \cos x

cos⁡2x=cos⁡2x−sin⁡2x\cos 2x = \cos^2 x - \sin^2 x

cos⁡2x=2cos⁡2x−1\cos 2x = 2\cos^2 x - 1

cos⁡2x=1−2sin⁡2x\cos 2x = 1 - 2\sin^2 x

Let's start with the LHS and transform it:

  1. Express sin⁡4A\sin 4A as a double angle: We can write 4A4A as 2×2A2 \times 2A. Applying the double angle formula for sine, sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x \cos x, where x=2Ax = 2A:

sin⁡4A=sin⁡(2⋅2A)=2sin⁡2Acos⁡2A\sin 4A = \sin(2 \cdot 2A) = 2\sin 2A \cos 2A

  1. Expand sin⁡2A\sin 2A: The term sin⁡2A\sin 2A can be directly expanded using the double angle formula for sine:

sin⁡2A=2sin⁡Acos⁡A\sin 2A = 2\sin A \cos A

  1. Expand cos⁡2A\cos 2A: For cos⁡2A\cos 2A, we have three common forms. We need to choose the one that will most directly lead to the RHS, which contains terms like sin⁡Acos⁡3A\sin A \cos^3 A and cos⁡Asin⁡3A\cos A \sin^3 A. The form cos⁡2A=cos⁡2A−sin⁡2A\cos 2A = \cos^2 A - \sin^2 A is ideal because it introduces both cos⁡2A\cos^2 A and sin⁡2A\sin^2 A terms, which, when multiplied by the sin⁡Acos⁡A\sin A \cos A from sin⁡2A\sin 2A, will naturally generate the cubic terms we need. cos⁡2A=cos⁡2A−sin⁡2A\cos 2A = \cos^2 A - \sin^2 A …

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