Trigonometric Identities...

seminolecc45 · New Member Offline Joined: 06 Feb 2005 Posts: 11

The problem is:
1 + sin^2x and match it with one of the 5 following answers
The answer was csc^2x - cot^2x + sin^2x

What steps is there to get to that answer?
Thanks!

**Posted:** Sun Feb 06, 2005 9:15 am

probability1.01

cos^2x + sin^2x = 1
cot^2x + 1 = csc^2x
1 = csc^2x - cot^2x
csc^2x - cot^2x = 1

**Posted:** Sun Feb 06, 2005 9:26 am

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seminolecc45 · New Member Offline Joined: 06 Feb 2005 Posts: 11

(cos^2 :theta: - sin^2 :theta: ) / (sin :theta: cos :theta: )

This idenity chapter is very hard. Can anyone help me out with this problem?
Thanks!

**Posted:** Sun Feb 06, 2005 1:20 pm

kueh · Riemann Hypothesis Offline Joined: 26 May 2004 Posts: 417

Use $\cos{2x} = \cos^2 x - \sin^2 x, \sin{2x} = 2\cos x\sin x$

**Posted:** Sun Feb 06, 2005 1:24 pm

seminolecc45 · New Member Offline Joined: 06 Feb 2005 Posts: 11

Thanks kueh, but I figured it out with using the trig identities we used so far. Can you help me with a couple others?

(cosx/sinx) + sinx/(1+cosX)

And

cotx + tanx = secx X cscx

I figured out most of the other problems, now I am into verifying Sad

**Posted:** Sun Feb 06, 2005 6:01 pm

probability1.01

Basic trick for this stuff if to convert everything into sines and cosines and then multiply/divide/add/subtract/factor to simplify. After enough of that, the identity should be obvious. Be sure to watch out for sin^2 + cos^2 = 1. It is a favorite in deriving identities with Smile

.

**Posted:** Sun Feb 06, 2005 6:06 pm

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seminolecc45 · New Member Offline Joined: 06 Feb 2005 Posts: 11

Can someone help me out with
(cosx/sinx) + (sin)/(1+cos)

I really need someones help, literally 6 people I talked to can not figure it out. Pleeease help!
Thank you so so much Very Happy

**Posted:** Sun Feb 06, 2005 7:22 pm

joml88

**Posted:** Sun Feb 06, 2005 7:38 pm

seminolecc45 · New Member Offline Joined: 06 Feb 2005 Posts: 11

Oh my god, thank you soooo much. My friend and I were working on that problem for about 90 minutes, literrally. I will post any problems I have with tommorrow's homework. Thank you everyone for helping me, I am going to do so good on this test, as opposed to everyone else.
Thanks! Very Happy

**Posted:** Sun Feb 06, 2005 8:48 pm

probability1.01

I await JBL's discovery of this topic. Confused

**Posted:** Sun Feb 06, 2005 9:06 pm

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tetrahedr0n

This one is somewhat more OK than the other few topics. I had half a mind of deleting the other one with the quadratic functions. I'm tempted to delete some of this one too.

**Posted:** Mon Feb 07, 2005 12:35 pm

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shobber

**Posted:** Thu Feb 10, 2005 2:50 am

seminolecc45 · New Member Offline Joined: 06 Feb 2005 Posts: 11

Hey, can you help me with a couple trig equations?
The first one is: tanX+1= :sqrt: 3 + :sqrt: 3cotX

And

tanX - cotX

Please help, thanks!

**Posted:** Tue Feb 15, 2005 7:54 pm

shobber

**Posted:** Wed Feb 16, 2005 3:20 am

RC-7th

$\tan(x)-\cot(x)=$

$\frac{\sin(x)}{\cos(x)}-\frac{\cos(x)}{\sin(x)}=$

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

$-\frac{\cos(2x)}{\sin(x)\cos(x)}=$

$-\frac{2\cos(2x)}{2\sin(x)\cos(x)}=$

$-2\frac{\cos(2x)}{\sin(2x)}=$

$-2\cot(2x)$

**Posted:** Wed Feb 16, 2005 4:46 pm

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mandy_pal

for the first one, factor out the sqrt(3) and isolate all the trig functions on one side. YOu should get something like...

(tanX+1)/(cotX+1)=sqrt(3)
now use the fact that tanX=sinX/cosX and cotX=cosX/sinX and sub that into the equation. You should get:

(sinX+cosX)/cosX * sinX/(sinX+cosX) = sqrt (3)
simplifying:

sinX/cosX=sqrt(3)
tanX=sqrt(3)
therefore X=60 or 240 where 0<x<2pi

Remember that when solving trig equations, it's always useful to convert everything to sines and cosines and go from there. Hope this helps.

Cheers!

**Posted:** Fri Feb 18, 2005 3:01 pm

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jli

**Posted:** Tue Feb 22, 2005 1:53 pm