# Proving Identities and Double angles problems

1. Oct 30, 2005

### sens_freak_18

I've been working at proving these identites and I just cant seem to figure them out, some of them just come to me others I work on them for 10 mintues or more and just get more and more bogged down.

1) csc2x + cot2x = cotx so far i have
1/sin2x + cos2x/sin2x = cosx/sinx
1+cos2x/sinx = cosx/sinx

2) cos^2x + 4cosx + 3 = cosx + 3 so far i have
4cos^2x + 3 = cosx + 3

3)1+cosx/1-cosx - 1-cosx/1+cosx = 4cotxcscx so far i have
1+cosx/sin^2x - sin^2x/1+cosx = 4cosx/4sinx*1/sinx

4)cos^2x-csc^2x/cot^2x = sin^2x - sec^2x so far i have
cos^2x - 1/sin^2x/ cos^2X/sin^2x
cos^2x - 1/sin^2x * sin^2X/cos^2x
cos^2x - 1/cos^2x
cos^2x - sec^2x

5) 1 + sec^2x + cot^2x/ csc^2x = sec^2x

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

6) sin(x+y) + sin(x-y)/cos(x+y) + cos(x-y) = tanx
I have absolutely no clue what to do with this one, i've tried changing the left side to cosxsiny+sinxcosy and stuff like that but it just doesnt come to me

Thanks for any help given to me, I really need some and greatly appreciate it!

2. Oct 30, 2005

### Hurkyl

Staff Emeritus
First off, you need to learn how to use parentheses correctly:

$$a + b / c + d = a + \frac{b}{c} + d$$

but

$$(a + b) / (c + d) = \frac{a + b}{c + d}$$

Secondly, I see you're converting everything into sines and cosines: that's usually a good thing.

But you're missing two other easy steps:

(1) Clearing denominators.

For example, in order to prove

$$\frac{a}{b} = \frac{c}{d}$$

whenever it is defined all you have to do is to prove

$$ad = bc$$

whenever both b and d are nonzero. (it's okay to prove this holds even when b or d are zero)

(2) Working with only one angle.

In a lot of these, I see that you have trig functions with different arguments. E.G. in the first one, cos x and cos 2x both appear. You can often fix that.