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arcsin question doubt

Hey guys I have a doubt
Reply 1
Avatar for I'm God
I'm God
19
Original post by Stormragexox
Hey guys I have a doubt


What's your doubt?
Original post by Stormragexox
Hey guys I have a doubt


Sorry if the quality is not good enough
PHOTO-2019-04-25-20-37-20.jpg43.3KB
Reply 3
Avatar for I'm God
I'm God
19
Original post by Stormragexox
Sorry if the quality is not good enough


How far did you get with the question?
Original post by I'm God
How far did you get with the question?


So far in part a i got up to sin y =x but i am clueless about what to do next
Reply 5
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I'm God
19
Original post by Stormragexox
So far in part a i got up to sin y =x but i am clueless about what to do next


Well, I haven't done this in a while so I had a look around.
I thought that sin y = cos(y - pi/2), but apparently it's sin y = cos(pi/2 - y)
Not sure which one is correct (though I think I am), but once you get there, you can do x = cos(y - pi/2) which will give you arccos in terms of y.
Original post by I'm God
Well, I haven't done this in a while so I had a look around.
I thought that sin y = cos(y - pi/2), but apparently it's sin y = cos(pi/2 - y)
Not sure which one is correct (though I think I am), but once you get there, you can do x = cos(y - pi/2) which will give you arccos in terms of y.

thanks for replying xD how do you get pi/2?
Reply 7
Avatar for I'm God
I'm God
19
Original post by Stormragexox
thanks for replying xD how do you get pi/2?


I got it since cosy is pi/2 radians behind siny
sinA = cos(pi/2 - A)
cosA = sin(pi/2 - A)

cosA actually not quite pi/2 BEHIND.

the function transforms to get from sinA to cosA are : reflect in y-axis, translate pi/2 to the right.
Original post by I'm God
I got it since cosy is pi/2 radians behind siny
Reply 9
Avatar for I'm God
I'm God
19
Original post by begbie68
sinA = cos(pi/2 - A)
cosA = sin(pi/2 - A)

cosA actually not quite pi/2 BEHIND.

the function transforms to get from sinA to cosA are : reflect in y-axis, translate pi/2 to the right.


Fair enough, that makes sense.
I haven't done this stuff in years, so I forgot.
Why can't I just shift by pi/2 though?
cos(pi/2-x) = cos(x-pi/2)
so you're correct as well. cos is symmetric (even) about 0.

Original post by I'm God
Fair enough, that makes sense.
I haven't done this stuff in years, so I forgot.
Why can't I just shift by pi/2 though?
Original post by mqb2766
cos(pi/2-x) = cos(x-pi/2)
so you're correct as well. cos is symmetric (even) about 0.

True, it's an even function :proud:
Thanks for clearing it up :smile:
There are a few ways of arguing about which is more obvious, but @begbie68 is probably the one. Imagine you have a right angled triangle where the other two angles are
x and (pi/2-x)
Then by definition (ratio of the same two sides) you must have
cos(pi/2-x) = sin(x)
In fact, that's how it gets its name COmplement angle SIN.

Original post by I'm God
True, it's an even function :proud:
Thanks for clearing it up :smile:

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