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    I'm doing part a, and I dont know where to start. can someone explain this question to me please? thanks

    should I start with drawing the arccosx graph?
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    y=arccosx.
    cosy=x
    siny=??
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    (Original post by Radioactivedecay)
    y=arccosx.
    cosy=x
    siny=??
    can you show your working in getting to the answer?
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    (Original post by Maths&physics)
    I'm doing part i, and I dont know where to start. can someone explain this question to me please? thanks

    should I start with drawing the arccosx graph?
    This was one of the worst answered Edexcel pure questions of all time by the way. Nasty question.
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    What r the answers
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    (Original post by Notnek)
    This was one of the worst answered Edexcel pure questions of all time by the way. Nasty question.
    yeah, I dont get it.

    I see that cos y = x

    then I dont know where to go from there?
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    (Original post by Maths&physics)
    yeah, I dont get it.

    I see that cos y = x

    then I dont know where to go from there?
    Do you know an identity that links cos and sin (not cos^2 and sin^2)? It’s a less used identity but can be seen by observing that sin/cos are translations of each other.
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    (Original post by Notnek)
    Do you know an identity that links cos and sin (not cos^2 and sin^2)? It’s a less used identity but can be seen by observing that sin/cos are translations of each other.
    no, I dont know, what is it?
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    (Original post by Maths&physics)
    yeah, I dont get it.

    I see that cos y = x

    then I dont know where to go from there?
    Well tbh just arcsin both sides and say done
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    For part b, should be Pi/2 (90 degrees)
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    try out some numbers...

    π/3 = arccos {1/2} .... π/3 is the angle whose cos is 1/2

    π/6 = arcsin {1/2}.......π/6 is the angle whose sin is 1/2

    if the angles are acute you can see that they add up to π/2
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    (Original post by Notnek)
    Do you know an identity that links cos and sin (not cos^2 and sin^2)? It’s a less used identity but can be seen by observing that sin/cos are translations of each other.
    cos(x) = sin(x+pi/2)

    I dont know if thats an identity but looking at both graphs simultaneously, thats what I see.
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    (Original post by RDKGames)
    Well tbh just arcsin both sides and say done
    thats not the right answer though
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    (Original post by Maths&physics)
    thats not the right answer though
    It's in terms of y so technically you answer their question.

    Though if you really want to play by the rules then you need to use the identity you stated following on from \arcsin(\cos y) = \arcsin x, except you want to use \cos x \equiv \sin (\frac{\pi}{2}-x) instead.
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    Could you just do this using graphical method? If they are saying "express arcsinx in terms of y" this implies we need to do something to y=arccosx such that its the same as arcsinx. So if you do -arccosx +pi/2 you see that after this transformation the arccosx graph is exactly the same as arcsinx graph, therefore arcsinx=-y+pi/2, so for the next part we just combine all of this and do y+-y+pi/2 which is just pi/2
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    (Original post by RDKGames)
    Well tbh just arcsin both sides and say done
    Doesn’t help much with part b though
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    (Original post by Notnek)
    Do you know an identity that links cos and sin (not cos^2 and sin^2)? It’s a less used identity but can be seen by observing that sin/cos are translations of each other.
    so, I'm looking at the arcs and arcsin graphs.

    for arcsin to = arcos (y), it must reflect in the y axis (-y) and go up pi/2?

    which means arcsinx = pi/2 - y ????

    is that right way of thinking and doing it?
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    (Original post by Maths&physics)
    so, I'm looking at the arcs and arcsin graphs.

    for arcsin to = arcos (y), it must reflect in the y axis (-y) and go up pi/2?

    which means arcsinx = pi/2 - y ????

    is that right way of thinking and doing it?
    That's correct. The more standard approach would be to consider sin/cos. The identities you need are

    \cos x = \sin\left(\frac{\pi}{2}-x\right)

    \sin x = \cos\left(\frac{\pi}{2}-x\right)

    This is why e.g. sin(60) = cos(30) and cos(80) = sin(10) in other words the sin/cos angles add up to 90 or \frac{\pi}{2} radians. You need to know these identities for the exam (although they're used rarely).

    Back to the question:

    y=arccos(x)
    \Rightarrow \cos(y) = x
    \Rightarrow \sin\left(\frac{\pi}{2}-y\right) = x

    Can you finish it off to find \arcsin(x) in terms of y?
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    (Original post by Notnek)
    That's correct. The more standard approach would be to consider sin/cos. The identities you need are

    \cos x = \sin\left(\frac{\pi}{2}-x\right)

    \sin x = \cos\left(\frac{\pi}{2}-x\right)

    This is why e.g. sin(60) = cos(30) and cos(80) = sin(10) in other words the sin/cos angles add up to 90 or \frac{\pi}{2} radians. You need to know these identities for the exam (although they're used rarely).

    Back to the question:

    y=arccos(x)
    \Rightarrow \cos(y) = x
    \Rightarrow \sin\left(\frac{\pi}{2}-y\right) = x

    Can you finish it off to find \arcsin(x) in terms of y?
    thanks!! no i can't, i don't understand how you got the answer. what did you do?
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    (Original post by Maths&physics)
    thanks!! no i can't, i don't understand how you got the answer. what did you do?
    E.g.

    \sin y = x

    becomes

    y = \arcsin x

    You can do a similar thing here.
 
 
 
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