Need help on using trig functions to verify identities

**MajorJohnson** · March 22nd 2011, 04:52 AM

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

**masters** · March 22nd 2011, 04:58 AM

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....

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

**MajorJohnson** · March 22nd 2011, 06:39 AM

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....

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

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 =

answer:
sinx ^ 2 / cos x ^ 2

**e^(i*pi)** · March 22nd 2011, 06:50 AM

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

**masters** · March 22nd 2011, 07:50 AM

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

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

Using your Pythagorean Identities, you can simplify this way:

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

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 =

answer:
sinx ^ 2 / cos x ^ 2

For this one...

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

**MajorJohnson** · March 22nd 2011, 12:29 PM

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

**skeeter** · March 22nd 2011, 02:00 PM

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

**MajorJohnson** · March 24th 2011, 11:52 AM

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) $sin x * tan x + cos x$

$sin x * sin x / cos x + cos x$

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

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

$1 - cos x \div 1 + cos x + 1 + cos x \div 1 - cos x = Answer: 2$

For this problem, does it matter if the right side's answer is the same value on the left side but flipped?

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

$sin ^ 2 t/ cos t$

$sin t / cos t (sin t)$

$tan t sec t$

**skeeter** · March 24th 2011, 12:50 PM

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) $sin x * tan x + cos x$

$sin x * sin x / cos x + cos x$

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

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

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

I'm confused with this one:

3) $sin t * tan t = 1 - cos ^ 2 t / cos t$
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?
$sin ^ 2 t/ cos t$ \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 $\dfrac{\sin{x}}{\cos{x}}$

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