Verify if sin^3 x - cos^3 x = (sinx - cosx)(1 + sinxcosx)

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william1941's profile pic

Posted on

To solve this problem let's multiply (sin x - cos x) and (1 + sin x* cos x).

(sin x - cos x)*(1 + sin x*cos x)

= sin x + (sin x)^2 * cos x - cos x - sin x * (cos x )^2

Now (sin x)^2 + (cos x)^2 = 1 or (sin x)^2 = 1- (cos x)^2 and (cos x)^2 = 1 - (sin x)^2.

So sin x + (sin x)^2 * cos x - cos x - sin x * (cos x )^2

= sin x + [1- (cos x)^2] * cos x - cos x - sin x *[ 1 - (sin x)^2]

= sin x + cos x - (cos x)^3 - cos x - sin x + ( sin x)^3

subtracting common terms

= ( sin x)^3 - (cos x)^3

Therefore sin^3 x - cos^3 x = (sin x - cos x)(1 + sin x*cos x)

 

 

neela's profile pic

Posted on

To prove  sin^3 x-cos^3x = (sinx-cosx)(1+sinx+cosx).

We  use the identity  a^3-b^3 = (a-b)(a^2+ab+b^2).

and  the trigonometric identity sin^2x+cos^2x = 1 to prove the given enunciation.

Therefore using the first identity,

sin^3x -cos^3x = (sinx-cosx){ sin^2x+sinx*cosx+cos^2x)

sin^3x-cos^3x  = (sinx-cosx)[(sin^2x+cos^2x)+ sinxcosx}

sin^3x-cos^3x  = (sinx -cosx)(1 +sinxcosx) , using the second identity above.

giorgiana1976's profile pic

Posted on

We'll have to verify if the expression from the left side is equal to the expression from the right side.

We'll start by re-writting the difference of the cubes from the left side. We'll use the formula:

a^3 - b^3 = (a - b)(a^2 + ab + b^2)

We'll substitute a and b by sin x and cos x and we'll get:

(sin x)^3 -  (cos x)^3 = (sin x - cos x)[(sin x)^2 + sin x*cos x + (cos x)^2]

But the sum (sin x)^2 + (cos x)^2 = 1, from the fundamental formula of trigonometry.

We'll substitute the sum of squares by the value 1.

(sin x)^3 -  (cos x)^3 = (sin x - cos x)(1 + sin x*cos x) q.e.d

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