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#1 2009-12-02 09:05:08
- raindropsonroses
- Member
- Registered: 2009-12-02
- Posts: 3
Trig Identities
Hello everyone, I was wondering if someone could help me prove a couple trig identities for am assignment?
Here they are:
(cosx/1+sinx) + (cosx/1-sinx)= 2/cosx
(cosx/1+sinx)=(1-sinx/cosx)
(sinx+tanx/cosx+1)=tanx
1+(sinx+cosx)(cosx-sinx)=2cos*squared*x
(cosx+sinxtanx)/tanx=1/sinx
1/sin*squared*x + 1/ cos*squared*x = 1/sin*squared*xcos*squared*x
I know it's a lot and I hope it's not too confusing the the squares and whatnot.
Help would be greatly appreciated!
Thanks :)
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#2 2009-12-02 09:06:29
- raindropsonroses
- Member
- Registered: 2009-12-02
- Posts: 3
Re: Trig Identities
Oops! For the second equation, the sad face should be = then bracket...
Fixed
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#3 2009-12-02 09:25:08
- bobbym
- bumpkin
- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606
Re: Trig Identities
Hi raindropsonroses;
For the last one #6:
Just add the two trig rational forms:
Since cos^2(x)+sin^2(x)=1 you are done.
For #5)
Use the simple identity tan(x) = sin(x) / cos(x)
Add the 2 fractions in the numerator of the LHS.
Simplify the LHS a little:
Multiply top and bottom of the LHS by cos(x)
The rest of them aren't any harder than these 2, so please try them for yourself. If you get stuck then repost.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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#4 2009-12-02 09:25:14
- mathsyperson
- Moderator
- Registered: 2005-06-22
- Posts: 4,900
Re: Trig Identities
For the first one, try making the two fractions on the left into one, and work from there. That's always a good thing to do, but since the denominators here are A+B and A-B, they're really asking to be joined up.
The second one is easy enough if you can get it into a more recognisable form.
Try multiplying both sides by both denominators and see what you're left with.
For the the third, write every term on the left as something that involves cos x, and see if you can simplify from there.
Why did the vector cross the road?
It wanted to be normal.
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#5 2009-12-02 12:56:52
- raindropsonroses
- Member
- Registered: 2009-12-02
- Posts: 3
Re: Trig Identities
Alrighty, thank you both so much! I managed to figure out most of them and found that they aren't all very hard once you understand the basic concepts. Thanks 
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#6 2009-12-02 19:03:38
- bobbym
- bumpkin
- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606
Re: Trig Identities
Hi raindropsonroses;
Glad to help. Remeber book problems and homework always have an easy solution, look for it.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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#7 2017-09-13 16:29:37
- Monox D. I-Fly
- Member
- From: Indonesia
- Registered: 2015-12-02
- Posts: 2,000
Re: Trig Identities
For the fourth question:
1+(sinx+cosx)(cosx-sinx)=1+(cosx+sinx)(cosx-sinx)=1+cos*squared*x-sin*squared*x=cos*squared*+sin*squared*x+cos*squared*x-sin*squared*x=2cos*squared*x
For the fifth question:
(cosx+sinxtanx)/tanx=(cosx/tanx)+sinx=(cos*squared*x/sinx)+(sin*squared*x/sinx)=cos*squared*x+sin*squaredx/sinx=1/sinx
Actually I never watch Star Wars and not interested in it anyway, but I choose a Yoda card as my avatar in honor of our great friend bobbym who has passed away.
May his adventurous soul rest in peace at heaven.
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