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    if sina = 4/5
    cosa = 3/5
    tana = 4/c

    how do i solve cos((90-a)/2)?
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    (Original post by alesha98)
    if sina = 4/5
    cosa = 3/5
    tana = 4/c

    how do i solve cos((90-a)/2)?
    Could you post a picture of the question?
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    (Original post by Zacken)
    Could you post a picture of the question?
    Here is the question, the one i am having difficulty with is from question 8(b):
    Attached Images
     
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    (Original post by alesha98)
    Here is the question, the one i am having difficulty with is from question 8(b):
    That's a lot different to your original question! Please always post the full question.

    I recommend you watch this video : it should help you with force by a string on a pulley questions:

    https://www.youtube.com/watch?v=rC2IQ3R5im4


    If you want us to help you with your attempt, please post your working in full.
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    (Original post by alesha98)
    Here is the question, the one i am having difficulty with is from question 8(b):
    It's a mechanics question (this is why you always need to post the full question!!) and the intended method (I think) is instead of evaluating it exactly is to simply find \alpha by doing \alpha = \tan^{-1} \frac{4}{3} and then plugging this in into your cosine using the 'ANS' button on your calculator. (I'm assuming you already know how to find the force on a pulley since you seem to have gotten 2T\cos \left(\frac{90 - \alpha}{2}\right).
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    (Original post by Zacken)
    It's a mechanics question (this is why you always need to post the full question!!) and the intended method (I think) is instead of evaluating it exactly is to simply find \alpha by doing \alpha = \tan^{-1} \frac{4}{3} and then plugging this in into your cosine using the 'ANS' button on your calculator. (I'm assuming you already know how to find the force on a pulley since you seem to have gotten 2T\cos \left(\frac{90 - \alpha}{2}\right).
    I really should have read the question first - I thought this was a standard question with a given angle.

    But finding the angle using arctan is a simple way to do this, as you say.
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    (Original post by notnek)
    That's a lot different to your original question! Please always post the full question.

    I recommend you watch this video : it should help you with force by a string on a pulley questions:

    https://www.youtube.com/watch?v=rC2IQ3R5im4


    If you want us to help you with your attempt, please post your working in full.
    i got the right equation, but somehow i solved the equation wrong , the only part i got wrong is what i posted.
    i have set an equation :2(T)(cos((90-a)/2))
    =2(2g/3)(cos((90-a)/2))
    then here is where i got wrong: =2(2g/3)(cos(45-a/2))
    =2(2g/3)((cos45)-((3/5)/2))
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    (Original post by alesha98)
    i got the right equation, but somehow i solved the equation wrong , the only part i got wrong is what i posted.
    i have set an equation :2(T)(cos((90-a)/2))
    =2(2g/3)(cos((90-a)/2))
    then here is where i got wrong: =2(2g/3)(cos(45-a/2))
    =2(2g/3)((cos45)-((3/5)/2))
    \tan \alpha = \frac{4}{3} \Rightarrow \alpha = \tan^{-1}\frac{4}{3}

    You can use your calculator and then plug \alpha into

    \displaystyle \cos \left(\frac{90-\alpha}{2}\right)
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    (Original post by Zacken)
    It's a mechanics question (this is why you always need to post the full question!!) and the intended method (I think) is instead of evaluating it exactly is to simply find \alpha by doing \alpha = \tan^{-1} \frac{4}{3} and then plugging this in into your cosine using the 'ANS' button on your calculator. (I'm assuming you already know how to find the force on a pulley since you seem to have gotten 2T\cos \left(\frac{90 - \alpha}{2}\right).
    Thankyou so much, now i got the right answer . I was trying to solve this for an hour and got nothing ...
    Re picture:sorry about that because i have a really bad laptop...
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    (Original post by notnek)
    \tan \alpha = \frac{4}{3} \Rightarrow \alpha = \tan^{-1}\frac{4}{3}

    You can use your calculator and then plug \alpha into

    \displaystyle \cos \left(\frac{90-\alpha}{2}\right)
    i didnt find the value of a first, so i was struggling
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    (Original post by alesha98)
    i didnt find the value of a first, so i was struggling
    Yes it's common to use exact trig values for these kinds of questions so I can see why this question caused you problems.
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    (Original post by notnek)
    I really should have read the question first - I thought this was a standard question with a given angle.

    But finding the angle using arctan is a simple way to do this, as you say.
    The markscheme (June 2015) uses the exact value method, presumably from double-angle formulae and ugly manipulation, thought that was quite weird. :erm:


    (Original post by alesha98)
    ...
    No worries. Even if a picture wasn't possible, a simple link to the June 2015 paper would have sufficed.
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    (Original post by notnek)
    Yes it's common to use exact trig values for these kinds of questions so I can see why this question caused you problems.
    Thanks, i will remember to find the angle first for this kind of questions
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    (Original post by Zacken)
    The markscheme (June 2015) uses the exact value method, presumably from double-angle formulae and ugly manipulation, thought that was quite weird. :erm:




    No worries. Even if a picture wasn't possible, a simple link to the June 2015 paper would have sufficed.
    yeah that is why i was trying to solve the equation instead of finding the value of a first.
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    (Original post by Zacken)
    The markscheme (June 2015) uses the exact value method, presumably from double-angle formulae and ugly manipulation, thought that was quite weird. :erm:
    Wouldn't using \cos\left ( 90-\alpha \right ) \equiv \sin \alpha and then using a right angled triangle help us find the exact ratio?
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    (Original post by aymanzayedmannan)
    Wouldn't using \cos\left ( 90-\alpha \right ) \equiv \sin \alpha and then using a right angled triangle help us find the exact ratio?
    Yeah, but here we have \cos \left(45^{\circ} - \frac{\alpha}{2}\right) which is rather trickier to work with. Not that it can't be done, just that it wouldn't be worth the two marks required for it.
 
 
 
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