Further Pure Mathematics 2 (FP2) Help Thread

Hi akhl,

I need some help with the Maths problem below. I tried using sin(x) = cos[(pi/2) - x] and ended with arccos(x) and - arccos(x), which cancels down?



Thanks.

Originally posted by: akhl

Let sin(theta) = x --------------------(1)

Then arcsin(x) = theta -------------------(2)
sin(theta) = cos[(pi/2) - theta] --------------(3)

From equations (1) and (2),
cos[(pi/2) - theta] = x
Therefore, arccos(x) = (pi/2) - theta ------------(4)

Adding equations (2) and (4), we get
arcsin(x) + arccos(x) = theta + (pi/2) - theta
arcsin(x) + arccos(x) = pi/2
(Proved)

Thanks, akhl.😊 Another question on inverse trignometric functions.

sin pi = 0
arcsin(sin pi) = arcsin(0) = 0
Therefore arcsin(sin pi) not= pi
(Proved)

Originally posted by: akhl

arcsin(sinx) = x if x = 0

Thanks, Akhl.😃

Zero is not the only solution to arcsin(sinx) = x, right?

Originally posted by: akhl

Right. I thought only an example was needed.

Let me give complete answer.

arcsin(sinx) = x if

-pi/2 <= x < pi/2

How do you know this?

As you know, unless otherwise specified, by arcsin, we mean principal value. And principal value of arcsin is between -pi/2 and +pi/2