1. ## Help with proofs

ᴓ=pheta

1. (sinᴓ + cosᴓ)² = 1+sin2ᴓ

2.cot2ᴓ=cos²ᴓ-sin²ᴓ over 2sinᴓ cosᴓ

3.tan2ᴓ=2 over cotᴓ-tanᴓ

3. Originally Posted by TJ123456
ᴓ=pheta

1. (sinᴓ + cosᴓ)² = 1+sin²ᴓ

2.cotᴓ=cos²ᴓ-sin²ᴓ over 2sinᴓ cosᴓ

3.tan2ᴓ=2 over cotᴓ-tanᴓ
1. Expand like you would any other function:

$sin^2(\theta) + 2sin(\theta)cos(\theta) + cos^2(\theta) = 1+ 2sin(\theta)$

It would appear number 1 is not an identity. (try 90deg)

2. $cot(\theta) = \frac{cos^2(\theta)-sin^2(\theta)}{2sin(\theta)cos(\theta)} = \frac{cos(2\theta)}{sin(2\theta)} = cot(2(\theta))$

2 is not an identity either (try 45 degrees)

3. $\frac{2}{cot(\theta)-tan(\theta)} = \frac{2}{\frac{cos^2(\theta)-sin^2(\theta)}{cos(\theta)sin(\theta)}}$

Looking at the denominator: $cos(\theta)sin(\theta) = \frac{1}{2}sin(2\theta)$

$\frac{cos^2(\theta)-sin^2(\theta)}{cos(\theta)sin(\theta)} = \frac{2cos(2\theta)}{sin(2\theta} = 2cot(2\theta)$

The original expression becomes: $\frac{2}{2cot(2\theta)} = tan(2\theta)$

4. thanks,my teacher said they would all work,so i went back and check my typing(i was in hurry this morning) and i did made a typo,so thats why they didnt work,i edit them just now if u wanna help again but you dont have too,again thanks man