Simplify: (sin^3 x + cos^3 x) / (sin x + cos x)?

Question

a. sin^2 x + 2 sin x cos x + cos^2 x
b. 1 + 2 sin x cos x
c. 1 - sin x cos x
d. Both sin^2 x + 2 sin x cos x + cos^2 x and 1 + 2 sin x cos x
e. none of these

Answer 1

This is basically just factoring cubics.
To factor cubics, you pretty much just have to know that
x^3 + y^3 = (x + y)(x^2 - xy + y^2)
So in this problem youre dividing (x^3 + y^3) by (x + y).
So youre gonna get (x^2 - xy + y^2).
Translating that back into sin and cos, you'll get
(sin^2 x - sinxcosx + cos^2 x)
There is a trigonometric identity that states that sin^2 x + cos^2 x = 1. Therefore by switching things around and substituting, you're gonna come out with...
1- sinxcosx.
Read and understand the process! Don't just copy the answer ;)

Answer 2

sin^2 x -sin x * cos x- cos^2 x +cos^2 x
sin x +cos x\sin^3 x +0sin^2 x*cos x + 0sin x*cos^2 x +cos^3 x
-(sin^3 x + sin^2 x *cos x)
-sin^2 x *cos x +0sin x *cos^2 x
-(-sin^2 x *cos x - sin x cos x^2)
sin x *cos^2 x +cos^3 x
-(sin x cos^2 x +cos^3 x)
0


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

sin^2 x -sin x * cos x- cos^2 x +cos^2 x
sin^2 x +cos^2 x -sin x * cos x
1-sin x * cos x

c. 1 - sin x cos x
is correct.

Answer 3

I think it will be c. 1-sinx cosx

Here it goes,

simplify sin^3x + cos^3x = (sinx+cosx)(sin^2x-sinxcosx+cos^2x)
Check the formual a^3+b^3=(a+b)(a^2-ab-b^2)

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

Answer 4

sin ? [ sin(3?) + sin (5?) ] = sin ? [ sin (4? - ?) + sin(4? + ?) ] use the id sin(A + B) + sin(A - B) = 2 sin A cos B = sin ? [ 2sin (4?) cos ? ] = cos ? [ 2 sin (4?) sin ? ] use the id 2 sin A sin B = cos (A - B) - cos (A + B) = cos ? [ cos(4? - ?)- cos(4? + ?) ] = cos ? [ cos (3?) - cos (5?) ]

Answer 5

(sin³x + cos³x) / (sin x + cos x)
= (sin x + cos x)[sin²x - (sin x)(cos x) + cos²x] / (sin x + cos x)
= sin²x - (sin x)(cos x) + cos²x
= (sin²x + cos²x) - (sin x)(cos x)
= 1 - (sin x)(cos x)

The answer is
c. 1 - sin x cos x

Answer 6

C