Where is wrong?!!

Geometry Level 3

Where is the start of the wrong in these step/s ?

Prove: (sin x)[(cos^3)x] - (cos x)[(sin^3)x] = (1/4)(sin 4x)

Step 1: (sin x)[(cos^3)x] - (cos x)[(sin^3)x] = (1/4)(sin 4x)

Step 2: (sin x)(cos x)[(cos^2)x - (sin^2)x] = (1/4)(sin 4x)

Step 3: (sin x)(cos x)[1-2(sin^2)x] = (1/4)(sin 4x)

Step 4: (sin x)(cos x)(cos 2x) = (1/4)(2sin 2x)(cos 2x)

Step 5: (sin x)(cos x)(cos 2x) = (1/4)(2sin x)(cos x)(cos 2x)

Step 6: 1 = 2 ?

Where is the start of the wrong? :D

Take note: If you get the answer, show your solution/Proof .

Step 5 Step 6 Step 4 Step 1

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Christian Daang by Christian Daang , 20, Philippines

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2 solutions

Noel Lo
Jul 23, 2017

I thought 1 4 2 sin ⁡ 2 x cos ⁡ 2 x = 1 4 2 ( 2 sin ⁡ x cos ⁡ x ) cos ⁡ 2 x = sin ⁡ x cos ⁡ x cos ⁡ 2 x \frac{1}{4} 2 \sin {2x} \cos{2x}=\frac{1}{4} 2(2\sin{x}\cos{x})\cos{2x}=\sin{x}\cos{x}\cos{2x} ?

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Christian Daang
Nov 2, 2014

The start is in Step 5 because of, From Step 4, on the Right Side of the equation, the (2sin 2x) thing should go to 2[(2 sin x)(cos x)] or (4 sin x)(cos x) not (2sin x)(cos x).

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