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    (Original post by RDKGames)
    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?
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    (Original post by Alex.trin)
    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?
    Get them under one fraction and look for any identities you can use.
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    (Original post by Zacken)
    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
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    (Original post by Alex.trin)
    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
    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.
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    (Original post by Alex.trin)
    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
    See, that's the entire point of trig identites. The question is \sin x \tan x + \cos x = \frac{1}{\cos x}. YOU WANT TO SHOW THIS. YOU DO NOT KNOW IT IS TRUE. THE ENTIRE POINT OF THE QUESTION IS TO SHOW THIS.

    So if you ever write down \sin x \tan x + \cos x = \frac{1}{\cos x} anywhere in the middle of your proof, you are dong it wrong.

    The entire point is to start with one side, let's say \sin x \tan x + \cos x. You do not write the =\frac{1}{\cos x}. You are trying to show it, so do not assume it.

    The point is to start with one side \sin x \tan x + \cos x say and use various manipulations like:

    \sin x \tan x + \cos x = \frac{\sin^2 x}{\cos x} + \frac{\cos^2 x}{\cos x} = \frac{\sin^2 x + \cos^2 x}{\cos x} = \frac{1}{\cos x}.

    See? You started with one side, used a bunch of various manipulation to show that that side is the same thing as the other side, at no point do you assume they're the same.

    I'm not sure I've explained this very coherently, so I'd appreciate it if you told me precisely which part of my comment here is confusing you so I can clear it up in detail.

    (also, you're not annoying in the least)
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    (Original post by Zacken)
    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!
    She does have feelings "Zac"!!!! Alex, just have faith in the lord and you will find the correct identity for you!! God bless
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    (Original post by RDKGames)
    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.
    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
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    (Original post by ellahick)
    She does have feelings "Zac"!!!! Alex, just have faith in the lord and you will find the correct identity for you!! God bless
    lol wat
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    (Original post by Alex.trin)
    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
    Yes you are right! (though the denominator on the LHS is not squared, typo maybe?)

    Look at the numerator on the LHS. What identity do you know for sin squared + cos squared??
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    (Original post by Alex.trin)
    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:

    \displaystyle 

\begin{align*}\sin x \tan x + \cos  x &= \sin x\frac{\sin x}{\cos x} + \frac{\cos^2 x}{\cos x} \\ & = \frac{\sin^2 x + \cos^2 x}{\cos x} \\&= \frac{1}{\cos x} \end{align*}

    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}
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    (Original post by Zacken)
    Okay. So the way your answer should be is this:

    \displaystyle 

\begin{align*}\sin x \tan x + \cos  x &= \sin x\frac{\sin x}{\cos x} + \frac{\cos^2 x}{\cos x} \\ & = \frac{\sin^2 x + \cos^2 x}{\cos x} \\&= \frac{1}{\cos x} \end{align*}

    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}
    Wow I love you guys, thanks zac and RDKgames <3. Onto the next question...
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    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
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    (Original post by Alex.trin)
    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?
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    (Original post by RDKGames)
    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?
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    (Original post by Alex.trin)
    Would 1/tanx equal cosx/sinx?
    Yes
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    (Original post by RDKGames)
    To me it does, but okay.
    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.
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    (Original post by RDKGames)
    Yes
    Wow I think I've done it! Is the answer 1-cosx/sinx = 1-cosx/sinx?
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    (Original post by Alex.trin)
    Wow I think I've done it! Is the answer 1-cosx/sinx = 1-cosx/sinx?
    Well, looks like it yeah.

    (Original post by IrrationalRoot)
    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.
    Yes Zacken pointed it out. Dunno, felt completely sure about it until I was pointed out something I overlooked.
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    (Original post by Zacken)
    lol wat
    Join the church and only then will you be granted the ability to understand gods wonderful ways
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    (Original post by RDKGames)
    Well, looks like it yeah.



    Yes Zacken pointed it out. Dunno, felt completely sure about it until I was pointed out something I overlooked.
    god doesn't like beef, alex mind your tone please, god bless
 
 
 
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