Question

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

Answer 1

To prove:

$\sin^3 x-\cos^3 x=(1+\sin x \cos x)(\sin x- \cos x)$

Verification

It is an identity. It can be shown by two identities.

• $\sin^2 x+\cos^2 x =1$, Trigonometric Identity
• $a^3-b^3=(a-b)(a^2+ab+b^2)$, Algebraic Identity

$\text{LHS}$

$=\sin^3 x-\cos^3 x$

$=(\sin x -\cos x )(\sin^2 x +\sin x\cos x +\cos^2 x)$ [Algebraic]

$=(\sin x -\cos x )(1 +\sin x\cos x)$ [Trigonometric]

$=\text{RHS}$

Therefore it is verified.

More Information

$\sin^2 x+\cos^2 x =1$ can show different identities for other trigonometric ratios.

(I) Dividing by $\sin^2 x$

$\implies 1+\tan^2 x =\csc^2 x$

(II) Dividing by $\cos^2 x$

$\implies 1+\cot^2 x =\sec^2 x$

Answer 2

Answer:

To prove:

\sin^3 x-\cos^3 x=(1+\sin x \cos x)(\sin x- \cos x)sin

3

x−cos

3

x=(1+sinxcosx)(sinx−cosx)

Verification

It is an identity. It can be shown by two identities.

\sin^2 x+\cos^2 x =1sin

2

x+cos

2

x=1 , Trigonometric Identity

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

3

−b

3

=(a−b)(a

2

+ab+b

2

) , Algebraic Identity

\text{LHS}LHS

=\sin^3 x-\cos^3 x=sin

3

x−cos

3

x

=(\sin x -\cos x )(\sin^2 x +\sin x\cos x +\cos^2 x)=(sinx−cosx)(sin

2

x+sinxcosx+cos

2

x) [Algebraic]

=(\sin x -\cos x )(1 +\sin x\cos x)=(sinx−cosx)(1+sinxcosx) [Trigonometric]

=\text{RHS}=RHS

Therefore it is verified.

More Information

\sin^2 x+\cos^2 x =1sin

2

x+cos

2

x=1 can show different identities for other trigonometric ratios.

(I) Dividing by \sin^2 xsin

2

x

\implies 1+\tan^2 x =\csc^2 x⟹1+tan

2

x=csc

2

x

(II) Dividing by \cos^2 xcos

2

x

\implies 1+\cot^2 x =\sec^2 x⟹1+cot

2

x=sec

2

x

hope it helps you