# Thread: Need help on using trig functions to verify identities

1. ## Need help on using trig functions to verify identities

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

2. Originally Posted by MajorJohnson
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
Hi MajorJohnson,

I don't see an identity here, but if you just want to simplify....

$\displaystyle \cot z \sin z + \tan z \cos z = \dfrac{\cos z}{\sin z}\sin z+\dfrac{\sin z}{\cos z}\cos z= \boxed{\cos z + sin z}$

3. Originally Posted by masters
Originally Posted by MajorJohnson
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

Hi MajorJohnson,

I don't see an identity here, but if you just want to simplify....

$\displaystyle \cot z \sin z + \tan z \cos z = \dfrac{\cos z}{\sin z}\sin z+\dfrac{\sin z}{\cos z}\cos z= \boxed{\cos z + sin z}$
Sorry that cot z should be a cos 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.

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 =
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?

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 =

sinx ^ 2 / cos x ^ 2

4. $\displaystyle \cos z \sin z + \dfrac{\sin z}{\cos z} \cos z = \cos z \sin z + \sin z = \sin z(\cos z + 1)$

5. Originally Posted by MajorJohnson
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 =
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:

$\displaystyle \dfrac{1-\sin^2 \theta}{\csc^2 \theta -1}=\dfrac{\cos^2 \theta}{\cot^2\theta}=\dfrac{\cos^2\theta}{\frac{\ cos^2\theta}{\sin^2\theta}}=\dfrac{\cos^2\theta}{1 } \times \dfrac{\sin^2\theta}{\cos^2\theta}=\sin^2\theta$

Originally Posted by MajorJohnson
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 =

sinx ^ 2 / cos x ^ 2
For this one...

$\displaystyle \dfrac{\sec x \sin x}{\tan x}=\dfrac{\frac{1}{\cos x} \sin x}{\frac{\sin x}{\cos x}}=\dfrac{\frac{\sin x}{\cos x}}{\frac{\sin x}{\cos x}}=1$

6. okay, thank you. One more question, how do you get your equations to look like that in your post?

7. Originally Posted by MajorJohnson
okay, thank you. One more question, how do you get your equations to look like that in your post?
Latex Help Forum

8. 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:

$\displaystyle sin ^ 2 t/ cos t$

$\displaystyle sin t / cos t (sin t)$

$\displaystyle tan t sec t$

9. Originally Posted by MajorJohnson
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) $\displaystyle sin x * tan x + cos x$

$\displaystyle sin x * sin x / cos x + cos x$

$\displaystyle sin x ^ 2 / cos x + cos x = Answer: sin x ^ 2$ no ... should be sec(x)

2) $\displaystyle 1\div 1 + cos x + 1 \ div 1 - cos x =$

$\displaystyle 1 - cos x \div 1 + cos x + 1 + cos x \div 1 - cos x = Answer: 2$ no ... 2csc^2(x)

I'm confused with this one:

3) $\displaystyle sin t * tan t = 1 - cos ^ 2 t / cos t$
Doing the right side:

From this point i'm not sure about what to do next. Would i factor sin ^ 2 / cos t or convert it into tan ^ 2 t?
$\displaystyle sin ^ 2 t/ cos t$ \sin^2(t)/cos(t) = sin(t) * sin(t)/cos(t) = sin(t) * tan(t)
1. start a new thread ... this one is getting worn.

2. use parentheses or learn how to form a latex fraction ... e.g. \dfrac{\sin{x}}{\cos{x}} inside of the math tags will yield $\displaystyle \dfrac{\sin{x}}{\cos{x}}$