Factoring Trigonometric Expressions...4

[nycmathguy]

Section 5.1

Question 32

sin^2 x + 3 cos x + 3

Let sin^2 x = 1 - cos^2 x.

1 - cos^2 x + 3 cos x + 3

-cos^2 x + 3 cos x + 4

Let u = cos x

-u^2 + 3u + 3

Factor by grouping.

-1(4) = -4

4 + (-1) = 3

-u^2 + 4u - u + 3

-u^2 + 4u becomes u(-u + 4)

-u + 3 stays the same.

u(-u + 4) - u + 3

Back-substitute for u.

cos x[-cos x + 4 - cos x + 3]

Not too sure about this one.

You say?

 

[MathLover1]

this was enough

sin^2 (x )+ 3cos (x) + 3
1 - cos^2( x) + 3 cos (x) + 3
-cos^2 (x) + 3 cos (x) + 4
Could not simplify further.

 

[nycmathguy]

this was enough

sin^2 (x )+ 3cos (x) + 3
1 - cos^2( x) + 3 cos (x) + 3
-cos^2 (x) + 3 cos (x) + 4
Could not simplify further.

In other words, I went too far. There is more than one correct form for each answer. Doesn't this mean that it can be simplified further?

 

[MathLover1]

yes, that is simplest form

you got cos x[-cos x + 4 - cos x + 3] which is incorrect because it is equal to 7 cos(x) - 2 cos^2(x), and that is not given expression

 

[nycmathguy]

yes, that is simplest form

you got cos x[-cos x + 4 - cos x + 3] which is incorrect because it is equal to 7 cos(x) - 2 cos^2(x), and that is not given expression

Thanks. I like to be corrected. It's the only way to learn.