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    how to change
    1/sin^22x to 4cosec^22x

    I would have thought it would be cosec^22x
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    (Original post by joyoustele)
    how to change
    1/sin^22x to 4cosec^22x

    I would have thought it would be cosec^22x
    It would indeed be. Did you read the question right...?
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    (Original post by RDKGames)
    It would indeed be. Did you read the question right...?
    Yep,the question was: to express sec^2x+cosec^2x in terms of sinx and cosx

    sec^2x+cosec^2x which i got \dfrac{1}{cos^2xsin^2x}

    then in part (ii) it said to show that sec^2x+cosec^2x=4cosec^22x
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    (Original post by joyoustele)
    Yep,the question was: to express sec^2x+cosec^2x in terms of sinx and cosx

    sec^2x+cosec^2x which i got \dfrac{1}{cos^2xsin^2x}

    then in part (ii) it said to show that sec^2x+cosec^2x=4cosec^22x
    Then you should have \frac{1}{\frac{1}{4}\sin^2(2x)} instead.
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    (Original post by joyoustele)
    Yep,the question was: to express sec^2x+cosec^2x in terms of sinx and cosx

    sec^2x+cosec^2x which i got \dfrac{1}{cos^2xsin^2x}

    then in part (ii) it said to show that sec^2x+cosec^2x=4cosec^22x
    \displaystyle \dfrac{1}{cos^2xsin^2x} = \dfrac{1}{\left(\cos x \sin x\right)^2}

    \displaystyle \sin 2x = 2\sin x \cos x \Rightarrow \sin x \cos x = \frac{1}{2}\sin 2x

    ...
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    (Original post by RDKGames)
    Then you should have \frac{1}{\frac{1}{4}\sin^2(2x)} instead.
    how did you get \dfrac{1}{\frac{1}{4}\sin^2(2x)}
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    (Original post by Notnek)
    \displaystyle \dfrac{1}{cos^2xsin^2x} = \dfrac{1}{\left(\cos x \sin x\right)^2}

    \displaystyle \sin 2x = 2\sin x \cos x \Rightarrow \sin x \cos x = \frac{1}{2}\sin 2x

    ...
    oh
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    BTW thank you guys very much. RDKGames Notnek thanks to you guys, and exam solutions i went from an E(predicted) to an A in maths. i even think an A* overall grade could be on the cards for me now. Thanks alot.
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    (Original post by joyoustele)
    BTW thank you guys very much. RDKGames Notnek thanks to you guys, and exam solutions i went from an E(predicted) to an A. i even think an A* overall grade could be on the cards for me now. Thanks alot.
    That's really great to hear! Keep asking for help whenever you're stuck and aim for the A*

    By the way, for trig especially you need to practice as many different questions as possible because there's so much recognition of identites and little tricks that you can only pick up with practice. For example in this question you needed to spot that you have the form \sin x \cos x which is part of the identity for \sin 2x.
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    (Original post by Notnek)
    That's really great to hear! Keep asking for help whenever you're stuck and aim for the A*

    By the way, for trig especially you need to practice as many different questions as possible because there's so much recognition of identites and little tricks that you can only pick up with practice. For example in this question you needed to spot that you have the form \sin x \cos x which is part of the identity for \sin 2x.
    Will do. Thank you very much.
    I will be getting that A*
 
 
 
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