Trigonometric Identities Help!!

Er, just turn tan(x) into sinx over cosx. Don't try to go from sin or cos into tan right away, deal with the basic functions first. The first term should then read sin squared over cos which is okay as far as the RHS is concerned, then deal with the second term and get it under the same denominator.

Ok so now I've got sin^2x/cosx + cos^2x/cosx = 1/cosx. How do I finish the question without just crossing out the cosx's?

Why would you want to do that? Think about it! You're starting from the left hand side which is in terms of tans, sines and cosines and you wnat to show that's the same thing as the right hand side which is made up only of cosines. So the logical thing to do in the RHS is to turn the tan into a sine and cosine, and then hopefully get rid of sines altogether.

So try: $\sin x \frac{\sin x}{\cos x} + \cos x = \frac{\sin^2 x}{\cos x} + \frac{\cos x}{1}$ - now do common denominators, then add the numerator and you get something nice in the numerator that simplies very nicely and gets you what you want!

Ok so I'm a bit confused. Where has the 1/cosx on the RHS gone? Sorry to be so annoying but I really don't understand this topic

Ok so I'm a bit confused. Where has the 1/cosx on the RHS gone? Sorry to be so annoying but I really don't understand this topic

Why would you want to do that? Think about it! You're starting from the left hand side which is in terms of tans, sines and cosines and you wnat to show that's the same thing as the right hand side which is made up only of cosines. So the logical thing to do in the RHS is to turn the tan into a sine and cosine, and then hopefully get rid of sines altogether.

So try: $\sin x \frac{\sin x}{\cos x} + \cos x = \frac{\sin^2 x}{\cos x} + \frac{\cos x}{1}$ - now do common denominators, then add the numerator and you get something nice in the numerator that simplies very nicely and gets you what you want!

No, no. He is not showing the actual RHS on his RHS. He is simply showing what the actual LHS turns into, maybe it would make sense if you replace the equals sign in his post with an arrow? To show what it goes to.

She does have feelings "Zac"!!!! Alex, just have faith in the lord and you will find the correct identity for you!! God bless

Guys I'm so confused. I've got sin^2x+cos^2x/cos^2x = 1/cosx I think I'm going to burn my house down if I don't answer this question soon

Guys I'm so confused. I've got sin^2x+cos^2x/cos^2x = 1/cosx I think I'm going to burn my house down if I don't answer this question soon

Okay. So the way your answer should be is this:



since $\sin^2 x+ \cos^2 x= 1$. If this is what you have, then you're all set. You're done. Finito, completed. That's all the question wants.

Just to make sure, you should not have this:

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

$\displaystyle \sin x \frac{\sin x}{\cos x} + \cos x= \frac{1}{\cos x}$

$\displaystyle \frac{\sin^2 x}{\cos x} + \frac{\cos^2 x}{\cos x} = \frac{1}{\cos x}$

$\displaystyle \frac{\sin^2 x + \cos^2 x}{\cos x} = \frac{1}{\cos x}$

I've hit another brick wall as predicted. For question 5, I've turned the 1-cosx into sinx, so that the RHS equals 1. Now I'm stuck... again. Please can you come to the rescue zac and RDKgames?!! <3

What??? Where did that come from? Firstly, leave the RHS ALONE. Secondly, that's not even true because the cosine is not squared.

Start with LHS:

$\displaystyle \frac{1}{\sin{x}}-\frac{1}{\tan{x}} \Rightarrow ....$

RHS has denominator of sinx, so we need to get this LHS under sinx. We see the first term is already under sinx, so move onto the second term. How can you get 1/tan in terms of sin and cos?

Would 1/tanx equal cosx/sinx?

To me it does, but okay.

Yes

Wow I think I've done it! Is the answer 1-cosx/sinx = 1-cosx/sinx?

You seem to have trouble understanding logical implications. To prove an identity (or more generally, a statement) you need to deduce the identity from something true, not the other way round.

Note that, by your logic, this argument is valid:
Start with $2=-2$. Squaring both sides, we get $4=4$. This is a true statement, so it follows that $2=-2$ is also a true statement.

I don't mean any offence, but you should avoid helping someone with a question unless you're completely sure you know what you're talking about.

lol wat

Well, looks like it yeah.



Yes Zacken pointed it out. Dunno, felt completely sure about it until I was pointed out something I overlooked.