1. ## trigonometric identities

I think I have this problem down but then again I'm not sure. If you could check my work, please:

cot^2x = 1-2cscx+csc^2x
csc^2x-1 = 1-2cscx+csc^2x (from the pythag. identity)
0 = 2-2cscx + csc^2x-csc^2x
0 = 2-2cscx
0 = -1+cscx (divided all by -2)
cscx = 1

Also, I am so stuck on the following two. I have to use the double (or half, but I'm sure it's double) angle formulas but I don't know how to properly incorporate them:

1. 2cos^2x-1 = cosx

and

2. cos2x + 6sin^2x = 2

Help would be very much appreciated. Thank you!

2. Originally Posted by augustfai
Also, I am so stuck on the following two. I have to use the double angle formulas but I don't know how to properly incorporate them:

1. 2cos^2x-1 = cosx

and

2. cos2x + 6sin^2x = 2

Help would be very much appreciated. Thank you!
For the first one this may help. let $\displaystyle u=\cos(x)$

$\displaystyle \underbrace{2\cos^2(x)}_{2u^2}-1=\underbrace{\cos(x)}_{u}$

$\displaystyle 2u^2-1=u \iff 2u^2-u-1=0 \iff (2u+1)(u-1)$

so $\displaystyle u=-\frac{1}{2}$ and $\displaystyle u=1$

but $\displaystyle u=cos(x)$ so we get

$\displaystyle \cos(x)=-\frac{1}{2}$ and $\displaystyle \cos(x)=1$

All that is left is to solve these.

for the 2nd one use this identity
using the identity $\displaystyle \cos(2x)=1-2\sin^{2}(x)$

and try the sub $\displaystyle u=\sin(x)$

Good luck.

3. Originally Posted by augustfai
I think I have this problem down but then again I'm not sure. If you could check my work, please:

cot^2x = 1-2cscx+csc^2x
csc^2x-1 = 1-2cscx+csc^2x (from the pythag. identity)
0 = 2-2cscx + csc^2x-csc^2x
0 = 2-2cscx
0 = -1+cscx (divided all by -2)
cscx = 1

Also, I am so stuck on the following two. I have to use the double angle formulas but I don't know how to properly incorporate them:

1. 2cos^2x-1 = cosx

and

2. cos2x + 6sin^2x = 2

Help would be very much appreciated. Thank you!
Your work on the problem above is correct.

For 1. $\displaystyle 2 \cos{^2}{x} - 1 = \cos{(2x)}$. I would show this using $\displaystyle \cos{(2x)} = \cos{(x+x)} = \cos{x}\cdot{\cos{x}} - \sin{x}\cdot{\sin{x}} = \cos{^2}{x} - \sin{^2}{x}$.

4. Originally Posted by icemanfan
Your work on the problem above is correct.

For 1. $\displaystyle 2 \cos{^2}{x} - 1 = \cos{(2x)}$. I would show this using $\displaystyle \cos{(2x)} = \cos{(x+x)} = \cos{x}\cdot{\cos{x}} - \sin{x}\cdot{\sin{x}} = \cos{^2}{x} - \sin{^2}{x}$.
Thank you, but I'm a little confused on this bit: how did $\displaystyle 2 \cos{^2}{x} - 1 = \cos{(x)}$ become $\displaystyle 2 \cos{^2}{x} - 1 = \cos{(2x)}$?

5. If the equation in 1 is written correctly, you may disregard my first post.

For 2) use the double-angle identity that I worked out in the first post and substitute $\displaystyle 2 \cos{^2}{x} + 6 \sin{^2}{x} = 3$. Then:

$\displaystyle 2 \cos{^2}{x} + 2 \sin{^2}{x} + 4 \sin{^2}{x} = 3$

$\displaystyle 2 + 4 \sin{^2}{x} = 3$

$\displaystyle 4 \sin{^2}{x} = 1$

$\displaystyle \sin{^2}{x} = \frac{1}{4}$ and so

$\displaystyle \sin{x} = \frac{1}{2}$ or $\displaystyle \sin{x} = -\frac{1}{2}$

6. Originally Posted by augustfai
Thank you, but I'm a little confused on this bit: how did $\displaystyle 2 \cos{^2}{x} - 1 = \cos{(x)}$ become $\displaystyle 2 \cos{^2}{x} - 1 = \cos{(2x)}$?
The equation that I worked out is an identity. It wasn't actually solving an equation. Since it looked so close to the identity I just assumed you had copied it incorrectly, but that was my mistake.

7. Originally Posted by icemanfan
The equation that I worked out is an identity. It wasn't actually solving an equation. Since it looked so close to the identity I just assumed you had copied it incorrectly, but that was my mistake.
Ah, alright. Thank you. I actually don't know whether I'm supposed to use the double or the half-angle formulas, I assumed double since half doesn't seem to work out.

EDIT:
I think I sort of figured out the first one--

$\displaystyle 2 \cos{^2}{x} - 1 = \cos{(x)}$

turns into cos2x = cos x

What, then, from there?

8. Hello,

Originally Posted by augustfai
Ah, alright. Thank you. I actually don't know whether I'm supposed to use the double or the half-angle formulas, I assumed double since half doesn't seem to work out.

EDIT:
I think I sort of figured out the first one--

$\displaystyle 2 \cos{^2}{x} - 1 = \cos{(x)}$

turns into cos2x = cos x

What, then, from there?
cos(a)=cos(b) <=> a=b+2k pi, OR a=-b+2k pi

So... ?

9. I figured it out! Thanks, everyone.