Question

write in factored form:
1: (cos^2x)+(2cosx+1)
2. 1-2sinx+(1-cos^2x)
3. cosx-2sin^2x+1
4. 4tan^2x-[4/(cotx)]+sinx(cscx)

Write each expression as an algebraic expression of a single trigonometric function.

1. (1-sin^2x)/(1+sinx)
2. sin^2x/(1+cosx)

Answer 1

I'll do the first two:
1.
cos^2(x) + 2cos(x) + 1 happens to be a perfect square in cos(x), so
=(cos(x)+1)^2
2.
1-2sin(x)+(1-cos^2(x))
1-2sin(x)+sin^2(x)
Again, it's a perfect square, so
(1-sin(x))^2
The third one is similar to the first two, and the fourth one, convert
sin(x) csc(x) = sin(x)/cos(x) = tan(x)
and note that 1/cot(x) = tan(x)

Answer 2

for 2 i have 2-2sinx-cos^2x

Answer 3

For 2, you do not want to introduce cos(x), use
(1-cos^2(x))=sin^2(x), then you can factorize.

Answer 4

the third one, also, my answer is that it is not factorable

Answer 5

What did you have?

Answer 6

cosx-2sin^2x+1
=cosx-2(1-cos^2(x))+1
=2cos^2(x)+cos(x)-1
=(2cos(x)-1)(cos(x)+1)

Answer 7

i am very confused :( I am just not getting this chapter

Answer 8

For this one, you need to know that
sin^2(x)+cos^2(x)=1, which allows you to convert sin^2(x) to cos^2(x) or vice versa.

Once you have them all in (either) sin(x) or cos(x), it's just like a variable called s or c.
So #3 can be looked at as:
cosx-2sin^2x+1
=cosx-2(1-cos^2(x))+1
=2cos^2(x)+cos(x)-1
=2c^2+c-1 (now we need to factorize)
=(2c-1)(c+1) now put it back to cos(x)
=(2cos(x)-1)(cos(x)+1)
Hope that's better!

[I have to be away from the keyboard for the next hour or so, sorry, but I'll be back later. If you still have problems and are in a rush, make a new post]