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    Need to simplify
    Sine inverse (x) +Cos inverse (x)
    This is what i did,

    x= Sin inverse x
    Therefore Sin x = x
    Same for cos i.e Cos x = x
    Squaring both equations and adding them, sin^2(x) + Cos ^2(x)= x^2 + x ^2
    2x^2 = 1
    x= +or - √0.5
    The textbook answer is (1/2)pi
    I have no idea where the pi come from
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    Actually i don't know if what i did is allowed
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    (Original post by Bilbao)
    Need to simplify
    Sine inverse (x) +Cos inverse (x)
    This is what i did,

    x= Sin inverse x
    Therefore Sin x = x
    Same for cos i.e Cos x = x
    Squaring both equations and adding them, sin^2(x) + Cos ^2(x)= x^2 + x ^2
    2x^2 = 1
    x= +or - √0.5
    The textbook answer is (1/2)pi
    I have no idea where the pi come from
    No that's not allowed. Firstly, if you're going to assign variables, you want to choose one that isn't being used! So say y=\arcsin x instead of x= \arcsin x.

    Anyway, the correct way to approach this would be to start with the above and say that x=\sin y Then note that \sin y = \cos (\frac{\pi}{2}-y) which means that x=\cos(\frac{\pi}{2}-y) which hence yields us the value of \arccos x. Then it is clear where the answer comes from.
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    (Original post by Bilbao)
    Need to simplify
    Sine inverse (x) +Cos inverse (x)
    This is what i did,

    x= Sin inverse x
    Therefore Sin x = x
    Same for cos i.e Cos x = x
    Squaring both equations and adding them, sin^2(x) + Cos ^2(x)= x^2 + x ^2
    2x^2 = 1
    x= +or - √0.5
    The textbook answer is (1/2)pi
    I have no idea where the pi come from


    (Original post by RDKGames)
    No that's not allowed. Firstly, if you're going to assign variables, you want to choose one that isn't being used! So say y=\arcsin x instead of x= \arcsin x.

    Anyway, the correct way to approach this would be to start with the above and say that x=\sin y Then note that \sin y = \cos (\frac{\pi}{2}-y) which means that x=\cos(\frac{\pi}{2}-y) which hence yields us the value of \arccos x. Then it is clear where the answer comes from.
    As RDKGames said, you cannot chose a variable that is already used before. Another approach to it would be to derivate
    \arccos x+\arcsin x. Then you'll notice that that function is actually constant, so you can pick one particular value to evaluate it.
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    (Original post by RDKGames)
    No that's not allowed. Firstly, if you're going to assign variables, you want to choose one that isn't being used! So say y=\arcsin x instead of x= \arcsin x.

    Anyway, the correct way to approach this would be to start with the above and say that x=\sin y Then note that \sin y = \cos (\frac{\pi}{2}-y) which means that x=\cos(\frac{\pi}{2}-y) which hence yields us the value of \arccos x. Then it is clear where the answer comes from.
    Sin A = x
    A= arcsin x
    Sin A = Cos ( 90° - A)
    Therefore Cos ( 90° - A) = x
    Cos inverse x = 90° - A
    Replacing all values in the original eqn,
    A + ( 90° - A) = 1/2 pi
    Great all good but i tried the other way around where Cos A = Sin ( 90° + A) but this does not give the same answer as above

    Cos A = x
    A= cos inverse x
    Arcsin x = 90° + A
    90° + A + A = 1/2 pi +2 A
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    (Original post by Bilbao)
    Sin A = x
    A= arcsin x
    Sin A = Cos ( 90° - A)
    Therefore Cos ( 90° - A) = x
    Cos inverse x = 90° - A
    Replacing all values in the original eqn,
    A + ( 90° - A) = 1/2 pi
    Great all good but i tried the other way around where Cos A = Sin ( 90° + A) but this does not give the same answer as above

    Cos A = x
    A= cos inverse x
    Arcsin x = 90° + A
    90° + A + A = 1/2 pi +2 A
    Note that you're working with inverse functions here, which are not periodic unlike \cos A and \sin B. So, we need to restrict the domains of \sin B and \cos A to -\frac{\pi}{2} \leq B \leq \frac{\pi}{2} and 0 \leq A \leq \pi in order to take inverses.

    Applying this to your idea, we have \cos A =x, and you then say that \cos A = \sin (\frac{\pi}{2}+A) where B=\frac{\pi}{2}+A which fine as an identity when A,B are unrestricted, but note here that this is not an identity when 0 \leq A \leq \pi as then B is not bounded between -\frac{\pi}{2} and \frac{\pi}{2} like we agreed. (i.e. when A=\frac{\pi}{2} then B=\pi which is outside our agreed interval)

    Alternatively, if you say that \cos A = x then \cos A = \sin (\frac{\pi}{2}-A), we get that B=\frac{\pi}{2}-A and indeed -\frac{\pi}{2} \leq B \leq \frac{\pi}{2} for all allowed A.
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    (Original post by Bilbao)
    Sin A = x
    A= arcsin x
    Sin A = Cos ( 90° - A)
    Therefore Cos ( 90° - A) = x
    Cos inverse x = 90° - A
    Replacing all values in the original eqn,
    A + ( 90° - A) = 1/2 pi
    Great all good but i tried the other way around where Cos A = Sin ( 90° + A) but this does not give the same answer as above

    Cos A = x
    A= cos inverse x
    Arcsin x = 90° + A
    90° + A + A = 1/2 pi +2 A
    To add onto RDK, ditch degrees. It's very hard to follow when you flip between writing 90^\circ and \pi/2.
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    (Original post by RDKGames)
    Note that you're working with inverse functions here, which are not periodic unlike \cos A and \sin B. So, we need to restrict the domains of \sin B and \cos A to -\frac{\pi}{2} \leq B \leq \frac{\pi}{2} and 0 \leq A \leq \pi in order to take inverses.

    Applying this to your idea, we have \cos A =x, and you then say that \cos A = \sin (\frac{\pi}{2}+A) where B=\frac{\pi}{2}+A which fine as an identity when A,B are unrestricted, but note here that this is not an identity when 0 \leq A \leq \pi as then B is not bounded between -\frac{\pi}{2} and \frac{\pi}{2} like we agreed. (i.e. when A=\frac{\pi}{2} then B=\pi which is outside our agreed interval)

    Alternatively, if you say that \cos A = x then \cos A = \sin (A-\frac{\pi}{2}), we get that B=A-\frac{\pi}{2} and indeed -\frac{\pi}{2} \leq B \leq \frac{\pi}{2} for all allowed A.
    Hmm i tried to draw Sin ( A - 90°) and it does not match Cos x where 0 < x < 180 ° for cos inverse (x) to exist.
    However Sin(90° - A) does match Cos X
    where x is within the inverse's range.
    Well im aware that if the sign changes it does equal one another
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    (Original post by Bilbao)
    Hmm i tried to draw Sin ( A - 90°) and it does not match Cos x where 0 < x < 180 ° for cos inverse (x) to exist.
    However Sin(90° - A) does match Cos X
    where x is within the inverse's range.
    Well im aware that if the sign changes it does equal one another
    Yes, I've meant to write 90-A instead.
 
 
 
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