I'm not sure if i'm doing this problem correctly. :
cot z sin z + tan z cos z
Heres one way i did it:
cotz = (cos / sin) sin z
= cos z
= cot z + (sin/cos) cos z
This is what i'm left with:
= cot z + cos z
I'm not sure if i'm doing this problem correctly. :
cot z sin z + tan z cos z
Heres one way i did it:
cotz = (cos / sin) sin z
= cos z
= cot z + (sin/cos) cos z
This is what i'm left with:
= cot z + cos z
Sorry that cot z should be a cos zHi MajorJohnson,
I don't see an identity here, but if you just want to simplify....
$\displaystyle \cot z \sin z + \tan z \cos z = \dfrac{\cos z}{\sin z}\sin z+\dfrac{\sin z}{\cos z}\cos z= \boxed{\cos z + sin z}$
And the method you did, was the other way i had tried to solve it and got the same answer. I just wasn't sure if it was right or not.
Thanks.
I've i try converting them using fundamental I.D. they won't cancel out, it'll just be subtracted to equal 0. Would that be the answer then?Code:Here's another one i'm stuck on: 1 - sin^2 / csc^2 - 1 = Heres what i've gottten so far: 1/csc ^ 2 - sin ^ 2 / 1 =
Here's another one
sec x * sin x/tan x =
1 / cos x * sin x / sin x /cos x
(sin x / cos x ) (sin x / cos x) =
sin x /cos x ^ 2 * sin x =
answer:
sinx ^ 2 / cos x ^ 2
Hello again MajorJohnson,
If I'm reading your expressions correctly, you have
$\displaystyle \dfrac{1-\sin^2 \theta}{\csc^2 \theta -1}$
Using your Pythagorean Identities, you can simplify this way:
$\displaystyle \dfrac{1-\sin^2 \theta}{\csc^2 \theta -1}=\dfrac{\cos^2 \theta}{\cot^2\theta}=\dfrac{\cos^2\theta}{\frac{\ cos^2\theta}{\sin^2\theta}}=\dfrac{\cos^2\theta}{1 } \times \dfrac{\sin^2\theta}{\cos^2\theta}=\sin^2\theta$
For this one...
$\displaystyle \dfrac{\sec x \sin x}{\tan x}=\dfrac{\frac{1}{\cos x} \sin x}{\frac{\sin x}{\cos x}}=\dfrac{\frac{\sin x}{\cos x}}{\frac{\sin x}{\cos x}}=1$
Thanks,
I might just keep using this thread to post more problems, up just to make sure i understand this. After working through these for some time, i feel like i'm almost comfortable with proving trig identities.
1) $\displaystyle sin x * tan x + cos x$
$\displaystyle sin x * sin x / cos x + cos x$
$\displaystyle sin x ^ 2 / cos x + cos x = Answer: sin x ^ 2 $
2) $\displaystyle 1\div 1 + cos x + 1 \ div 1 - cos x = $
$\displaystyle 1 - cos x \div 1 + cos x + 1 + cos x \div 1 - cos x = Answer: 2 $
For this problem, does it matter if the right side's answer is the same value on the left side but flipped?
3) $\displaystyle sin t * tan t = 1 - cos ^ 2 t / cos t $
Doing the right side:
$\displaystyle sin ^ 2 t/ cos t $
$\displaystyle sin t / cos t (sin t) $
$\displaystyle tan t sec t $