Trig help, please!

squeefish

All right, so I have a test on trig identities tomorrow and I'm a bit confused by some equations on the review (lost a page of my notes.) Here's an example of the kind of thing that's confusing me...

Solve for x over the Reals: 2sinxcosx=sinx.

I'd really appreciate it if someone could set me on the right track. Thanks guys.

Edit: Phwuh, I can't manage to prove this either: sinx+cosxcotx/cosxcscx=secx I've gotten up to this point: (sinx+cosx(cosx/sinx))/(cosx/sinx)=1/cosx and now I'm stuck.

Deswa

The easiest thing is to try to simplify the problem first.

1st

Divide by sin(x) => 2cos(x) = 1
Divide by 2 => cos(x) = 1/2
Now Im assuming you are using radians, so:
x = pi/3 +/- 2pi * N and -pi/3 +/- 2pi * N

Also sin(x) = 0 will make it equal

2nd

(sin(x) + cos*cot) = sec 
       cos*csc

Okay, first thing is to rewrite it all in cos's and sin's.

(sin(x) + cos*(cos/sin)) = sec 
       (cos/sin)

From there multiply by (sin/cos)/(sin/cos). This equals 1 so you can do this.

(sin^2/cos) + cos

Then we multiply by cos/cos to get a common denominator

sin^2 + cos^2
       cos

The top part = 1, as you know
1/cos = sec 

Thanatos

squeefish wrote: »

All right, so I have a test on trig identities tomorrow and I'm a bit confused by some equations on the review (lost a page of my notes.) Here's an example of the kind of thing that's confusing me...

Solve for x over the Reals: 2sinxcosx=sinx.

I'd really appreciate it if someone could set me on the right track. Thanks guys.

Edit: Phwuh, I can't manage to prove this either: sinx+cosxcotx/cosxcscx=secx I've gotten up to this point: (sinx+cosx(cosx/sinx))/(cosx/sinx)=1/cosx and now I'm stuck.

It's been awhile, but isn't there a trig identity for cos^2x? Like, 1-sinx, or something?

Aero

For the first one, remember that sin(x) can equal to zero. So you can't only consider the cos(x) part of the solution.

Deswa

cos^2 + sin^2 = 1

cos^2 = 1 - sin^2

squeefish

Thanatos wrote: »

It's been awhile, but isn't there a trig identity for cos^2x? Like, 1-sinx, or something?

There is (I've had to use that one in almost every question he's given us :P) but it doesn't seem to help in this case... I don't have a cos^2 or a sin^2 in either of them. And thanks Deswa, I got the first question now.

Thanatos

squeefish wrote: »

Thanatos wrote: »

It's been awhile, but isn't there a trig identity for cos^2x? Like, 1-sinx, or something?

There is (I've had to use that one in almost every question he's given us :P) but it doesn't seem to help in this case... I don't have a cos^2 or a sin^2 in either of them. And thanks Deswa, I got the first question now.

Dude, sin(x) * sin(x) = sin^2(x). cos(x) * cos(x) = cos^2(x).

I mean, from that, you should be able to solve it.

Edit: Damn, actually, looking at that, the way I was going to do it, you end up with tan(x) = tan(x)

Deswa

Okay I see what you were saying on the 2nd one, also Aero is totally right, I forgot about the sin thing.

I'll fix it in my post.

Edit:
Than, you cant change the right side. Also I fixed my first post.

Thanatos

Deswa wrote: »

Okay I see what you were saying on the 2nd one, also Aero is totally right, I forgot about the sin thing.

I'll fix it in my post.

Edit:
Than, you cant change the right side. Also I fixed my first post.

Heh, really? That sucks.

If you multiply both sides by sin(x), it all comes out beautifully. However, using the identity for cos^2(x) works almost as well. Here it is with the multiplying, if you really can't get it without the multiplying, I can show you, but you'll be better off figuring it out on your own:

(sinx+cosx(cosx/sinx))/(cosx/sinx)=1/cosx
((sinx+cos^2x/sinx))/(cosx/sinx))*sinx=1/cosx*sinx
(sin^2x+(cos^2x*sinx)/sinx)/(cosx/sinx)=sinx/cosx
((sin^2x+cos^2x)/(cosx/sinx)=sinx/cosx
sin^2x+cos^x=1
1/(cosx/sinx)=sinx/cosx
sinx/cosx=sinx/cosx

Ninja Bot

Also, if you're not allowed to have notes on the test, putting text in your calculator (as a program) can be very valuable. I've put in trig identities, formulas and the entire unit circle into my calculator and I wouldn't have been able to half ass my way to a solid 65% without them.

rockmonkey

or if you don't want to cheat just try to memorize them. even if you can't be looking over them right before the test and soon as you get it write them all down in the corner. I'm talking before you even jot your name down.

Chief1138

Ninja Bot wrote: »

Also, if you're not allowed to have notes on the test, putting text in your calculator (as a program) can be very valuable. I've put in trig identities, formulas and the entire unit circle into my calculator and I wouldn't have been able to half ass my way to a solid 65% without them.

This is considered cheating in most circles

Deswa

Thanatos wrote: »

Deswa wrote: »

Okay I see what you were saying on the 2nd one, also Aero is totally right, I forgot about the sin thing.

I'll fix it in my post.

Edit:
Than, you cant change the right side. Also I fixed my first post.

Heh, really? That sucks.

If you multiply both sides by sin(x), it all comes out beautifully. However, using the identity for cos^2(x) works almost as well. Here it is with the multiplying, if you really can't get it without the multiplying, I can show you, but you'll be better off figuring it out on your own:

You can multiply, its just that you cannot alter the right side of the equation. In situations like this you typically multiply by one, like cos/cos. It can change the left hand, but you didnt actually do anything but times it by 1, which is meaningless.

cooljammer00

there should be a sheet for this stuff. with all the identities on it.


that may be what you lost, so uh...i dunno.

squeefish

Heh, nah, they're merciful in my class, we're getting a little formula sheet. I just find proving identities to get overwhelming pretty quickly, I'm not exactly the most organized person in the world and it all kinda turns into a mess on my page. I just need to study more; already did a few worksheets and I have lunch+spare before my test tomorrow, thank god. Thanks for all your help, guys.

lordswing

Isn't the identify sin2x = 2sinxcosx? From there, you know that x is just equal to 2 right?

cooljammer00

the trick is to stick to one side, usually.