Maths C3 - Trigonometry... Help??

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    (Original post by ValerieKR)
    sec(theta)=-2.5
    cos(theta)=1/sec(theta)=1/(-2.5)=-0.4
    Thank you so much!!
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    Ok so I think my Maths knowledge is lacking... :/

    Rewrite the following as a power of sec, cosec, or cot
    \frac{1-\sin^2\theta}{\sin^2 \theta}

    I realise now that I'm suppose to factorise but what from this equation indicates that I am meant to start off by doing that? I feel like I don't always know what I'm suppose to start off doing when solving equations.
    I know the whole BODMAS thingy but I don't know how to always apply it.
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    (Original post by RDKGames)
    Eating lunch while doing maths has been scientifically been proven to lead to confusion. Don't do it.
    source?
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    (Original post by Philip-flop)
    Ok so I think my Maths knowledge is lacking... :/

    Rewrite the following as a power of sec, cosec, or cot
    \frac{1-\sin^2\theta}{\sin^2 \theta}

    I realise now that I'm suppose to factorise but what from this equation indicates that I am meant to start off by doing that? I feel like I don't always know what I'm suppose to start off doing when solving equations.
    I know the whole BODMAS thingy but I don't know how to always apply it.
    1-sin^2(x)=cos^2(x)


    You should know those identity and you should be able to carry on with that
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    (Original post by metrize)
    1-sin^2(x)=cos^2(x)


    You should know those identity and you should be able to carry on with that
    Thank you!! I forgot about that identity!! Managed to work out that the answer is \cot^2 \theta
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    (Original post by Philip-flop)
    Thank you!! I forgot about that identity!! Managed to work out that the answer is \cot^2 \theta
    No offense but that is not an identity you 'forget'. Many others are derived straight from \sin^2x+\cos^2x \equiv 1. Try not to forget it next time.
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    (Original post by RDKGames)
    No offense but that is not an identity you 'forget'. Many others are derived straight from \sin^2x+\cos^2x \equiv 1. Try not to forget it next time.
    You're right! It's not an identity I forgot, it's an identity that I didn't spot to use for this example. Obviously my AS-level Maths knowledge is a bit rusty after the summer (although my knowledge was never amazing :P) I will do better to try and think more strategically instead of panicking and thinking that I can't solve something
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    Ok, so yet again I find myself struggling with Trig Identities I've just come across this question which is...

    Q) Write down the value of cotx ....
    5sinx=4cosx

    I don't seem to know what to start with first
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    (Original post by Philip-flop)
    Ok, so yet again I find myself struggling with Trig Identities I've just come across this question which is...

    Q) Write down the value of cotx ....
    5sinx=4cosx

    I don't seem to know what to start with first
    Find  \sin x / \cos x and then take reciprocal.
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    (Original post by B_9710)
    Find  \sin x / \cos x and then take reciprocal.
    I don't quite understand

    So far I've done...

    5sinx=4cosx

     \frac{5sinx}{4cosx}=0

    Which then becomes this...?

    \frac{5}{4}\times\frac{1}{tanx}

    \frac{5}{4} \cot x

    Have I done this right so far? If so, what do I do next?
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    (Original post by Philip-flop)
    I don't quite understand

    So far I've done...

    5sinx=4cosx

     \frac{5sinx}{4cosx}=0

    Which then becomes this...?

    \frac{5}{4}\times\frac{1}{tanx}

    \frac{5}{4} \cot x

    Have I done this right so far? If so, what do I do next?
    That's not true! \frac{4\cos x}{4\cos x} = 1 not 0. Like \frac{1}{1} = 0.

    So with 5 \sin x = 4\cos x divide both sides by \cos x: this gives you \frac{5\sin x}{\cos x} = \frac{4\cos x}{\cos x}.

    Now remember that \frac{\cos x}{\cos x} = 1. So you get 5 \frac{\sin x}{\cos x} = 4.

    But you know that \tan x = \frac{\sin x}{\cos x}, so you have 5 \tan x = 4.

    Now divide both sides by 5 to get \frac{5\tan x}{5} = \frac{4}{5} but you know that \frac{5}{5} = 1 so you have

    \frac{5\tan x}{5} = \tan x, which means \tan x = \frac{4}{5}. Now take the reciprocal of both sides to get \displaystyle \frac{1}{\tan x} = \frac{1}{\frac{4}{5}} = \frac{5}{4}
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    (Original post by Zacken)
    That's not true! \frac{4\cos x}{4\cos x} = 1 not 0. Like \frac{1}{1} = 0.

    So with 5 \sin x = 4\cos x divide both sides by \cos x: this gives you \frac{5\sin x}{\cos x} = \frac{4\cos x}{\cos x}.

    Now remember that \frac{\cos x}{\cos x} = 1. So you get 5 \frac{\sin x}{\cos x} = 4.

    But you know that \tan x = \frac{\sin x}{\cos x}, so you have 5 \tan x = 4.

    Now divide both sides by 5 to get \frac{5\tan x}{5} = \frac{4}{5} but you know that \frac{5}{5} = 1 so you have

    \frac{5\tan x}{5} = \tan x, which means \tan x = \frac{4}{5}. Now take the reciprocal of both sides to get \displaystyle \frac{1}{\tan x} = \frac{1}{\frac{4}{5}} = \frac{5}{4}
    OMG I am going to kick myself. When I can't work something out I go into panic mode and start making Maths up!! I originally had...

    5sinx=4cosx

    \frac{5sinx}{cosx}=4

    5tanx=4

    tanx=\frac{4}{5}

    But then I got stuck because I thought I went wrong somewhere as I wasn't sure how to get the equation to become cot(x)
    As I knew that \cot x \Leftrightarrow \frac{1}{tan x} I assumed that I wouldn't be able to get the answer from what I had originally had done. So I started all over again, and from there on my brain decided to make up Maths rules

    But seriously, Thanks again Zacken!! I'm sure you must be fed up with my noobie mistakes by now!! :P
    How did you identify whether to take the reciprocal of both sides?
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    (Original post by Philip-flop)
    [...] How did you identify whether to take the reciprocal of both sides?

    Like you said, you know that \cot x = \frac{1}{\tan x}, which means that whenever you know what tan x is, you automatically know what cot x it, it's just 1/(whatever tan x is). And vice-versa. Whenever you know cot x, you know tan x. Whenever you know sin x you know cosec x, whenever you know cos x, you know sec x, etc...

    So here we know that tan x = 4/5, which means we automatically know that cot x = 1/(4/5) = 5/4.
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    (Original post by Zacken)
    Like you said, you know that \cot x = \frac{1}{\tan x}, which means that whenever you know what tan x is, you automatically know what cot x it, it's just 1/(whatever tan x is). And vice-versa. Whenever you know cot x, you know tan x. Whenever you know sin x you know cosec x, whenever you know cos x, you know sec x, etc...

    So here we know that tan x = 4/5, which means we automatically know that cot x = 1/(4/5) = 5/4.
    Yeah that's it! I know these things in the back of my mind but have trouble adopting these techniques
    Maths will not defeat me!! :P

    Thanks again Zacken!
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    Ok, I'm having another dumb moment with this question...

    Q) \sec ^2(x) cos^5(x)+\cot (x) \mathrm{cosec}(x) sin^4(x)

    I think I've made a mistake somewhere, so far I've got...

    cos^5x(\frac{1}{cos^2x})+sin^4x(  \frac{cosx}{sinx})(\frac{1}{sinx  })

    cos^5x(\frac{1}{cos^2x})+sin^4x(  \frac{cosx}{sin^2x})

    cos^3x+sin^2xcosx

    Edit: Never mind I think i've got it by factorising now...
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    (Original post by Philip-flop)
    Ok, I'm having another dumb moment with this question...

    Q) \sec ^2(x) cos^5(x)+\cot (x) \mathrm{cosec}(x) sin^4(x)

    I think I've made a mistake somewhere, so far I've got...

    cos^5x(\frac{1}{cos^2x})+sin^4x(  \frac{cosx}{sinx})(\frac{1}{sinx  })

    cos^5x(\frac{1}{cos^2x})+sin^4x(  \frac{cosx}{sin^2x})

    cos^3x+sin^2xcosx
    Whwat are you meant to do? What's the question?
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    (Original post by Philip-flop)
    Ok, I'm having another dumb moment with this question...

    Q) \sec ^2(x) cos^5(x)+\cot (x) \mathrm{cosec}(x) sin^4(x)

    I think I've made a mistake somewhere, so far I've got...

    cos^5x(\frac{1}{cos^2x})+sin^4x(  \frac{cosx}{sinx})(\frac{1}{sinx  })

    cos^5x(\frac{1}{cos^2x})+sin^4x(  \frac{cosx}{sin^2x})

    cos^3x+sin^2xcosx
    factorise out a cos(x)
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    (Original post by Zacken)
    Whwat are you meant to do? What's the question?
    Never mind I think I've managed to get it by factorising now
    Q) Simplify the following expression...
    \sec ^2(x) cos^5(x)+\cot (x) \mathrm{cosec}(x) sin^4(x)

    My answer below...
    cos^5x(\frac{1}{cos^2x})+sin^4x(  \frac{cosx}{sinx})(\frac{1}{sinx  })

    cos^5x(\frac{1}{cos^2x})+sin^4x(  \frac{cosx}{sin^2x})

    cos^3x+sin^2xcosx

    Now that I have factorised I've got...
    cosx(cos^2x+sin^2x)

    using the Trig Identity cos^2 x + sin^2 x = 1 .....
    cosx(1)

    cosx

    (Original post by ValerieKR)
    factorise out a cos(x)
    Thank you so much. I stupidly didn't think of factorising at first
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    I'm stuck yet again...

    How can I get the L.H.S (Left Hand Side) to equal the R.H.S (Right Hand Side)?? ...

    (1-cosx)(1+ \sec x)=sinxtanx

    So far I've expanded the bracket to give me...

    1+ \sec x -cosx -cosx\sec x
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    (Original post by Philip-flop)
    I'm stuck yet again...

    How can I get the L.H.S (Left Hand Side) to equal the R.H.S (Right Hand Side)?? ...

    (1-cosx)(1+ \sec x)=sinxtanx

    So far I've expanded the bracket to give me...

    1+ \sec x -cosx -cosx\sec x
    simplify cos(x)sec(x) and put the remaining terms over a cos(x) denominator
 
 
 
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