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C3 help needed edexcel

Show that arcsin x + arccos x = pi/2

Don't know where to start
Reply 1
Has it got anything to do with x being = 1/root2
Reply 2
Original post by Vorsah
Has it got anything to do with x being = 1/root2


How do you know what x is? Aren't you supposed to be proving it for general x?
Reply 3
Original post by Vorsah
Has it got anything to do with x being = 1/root2


Wait is x=12x=\dfrac{1}{\sqrt{2}} a given in the question?

so arcsin(x)=π4arc\sin(x)=\dfrac{\pi}{4}

and arccos(x)=π4arc\cos(x)=\dfrac{\pi}{4}

Add them, the way I got those results is by recalling that cos(π4)=12\cos\left(\dfrac{\pi}{4} \right)=\dfrac{1}{ \sqrt{2}} and sin(π4)=12\sin\left(\dfrac{\pi}{4} \right)=\dfrac{1}{ \sqrt{2}} as sin(x)=cos(π2x)\sin\left(x \right)=\cos\left(\dfrac{\pi}{2}-x \right)
(edited 10 years ago)
Reply 4
Original post by Vorsah
Show that arcsin x + arccos x = pi/2

Don't know where to start


One way of doing it is if we let a=arcsinx a = \arcsin x and b=arccosx b = \arccos x , then x=... x = ...
I won't spoil it, but it's to do with the fact that sin(x)cos(π2x)sin(x) \equiv cos(\frac{\pi}{2} - x).
(edited 10 years ago)
Reply 6
Original post by Robbie242
Wait is x=12x=\dfrac{1}{\sqrt{2}} a given in the question?

so arcsin(x)=π4arc\sin(x)=\dfrac{\pi}{4}

and arccos(x)=π4arc\cos(x)=\dfrac{\pi}{4}

Add them, the way I got those results is by recalling that cos(π4)=12\cos\left(\dfrac{\pi}{4} \right)=\dfrac{1}{ \sqrt{2}} and sin(π4)=12\sin\left(\dfrac{\pi}{4} \right)=\dfrac{1}{ \sqrt{2}} as sin(x)=cos(π2x)\sin\left(x \right)=\cos\left(\dfrac{\pi}{2}-x \right)


It's not given in the question
Reply 7
Original post by Vorsah
It's not given in the question

hmm
(edited 10 years ago)

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