# Need help on using trig functions to verify identities

• Mar 22nd 2011, 04:52 AM
MajorJohnson
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
• Mar 22nd 2011, 04:58 AM
masters
Quote:

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....

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

• Mar 22nd 2011, 06:39 AM
MajorJohnson
Quote:

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

Quote:

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
• Mar 22nd 2011, 06:50 AM
e^(i*pi)
$\displaystyle \cos z \sin z + \dfrac{\sin z}{\cos z} \cos z = \cos z \sin z + \sin z = \sin z(\cos z + 1)$
• Mar 22nd 2011, 07:50 AM
masters
Quote:

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,

$\displaystyle \dfrac{1-\sin^2 \theta}{\csc^2 \theta -1}$

Using your Pythagorean Identities, you can simplify this way:

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

Quote:

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...

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

• Mar 22nd 2011, 12:29 PM
MajorJohnson
okay, thank you. One more question, how do you get your equations to look like that in your post?
• Mar 22nd 2011, 02:00 PM
skeeter
Quote:

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
• Mar 24th 2011, 11:52 AM
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) $\displaystyle sin x * tan x + cos x$

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

$\displaystyle sin x ^ 2 / cos x + cos x = Answer: sin x ^ 2$

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:

$\displaystyle sin ^ 2 t/ cos t$

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

$\displaystyle tan t sec t$
• Mar 24th 2011, 12:50 PM
skeeter
Quote:

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 $\displaystyle \dfrac{\sin{x}}{\cos{x}}$