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Core 3 inverse trig

In core 2 when u had an equation e.g. Sinx=2/3 u could find x and then do 180-x, without plotting graph or using cast diagram, is there a way to do a similar calculation method for inverse questions for example to find the value of sin(arccos0.75)
Reply 1
Original post by uchihaitachi0023
In core 2 when u had an equation e.g. Sinx=2/3 u could find x and then do 180-x, without plotting graph or using cast diagram, is there a way to do a similar calculation method for inverse questions for example to find the value of sin(arccos0.75)


1. you are finding arcsin(2/3) for x

2. you are finding sin of arccos(0.75)

why and where is the question?
Reply 2
Original post by uchihaitachi0023
In core 2 when u had an equation e.g. Sinx=2/3 u could find x and then do 180-x, without plotting graph or using cast diagram, is there a way to do a similar calculation method for inverse questions for example to find the value of sin(arccos0.75)

Do you mean without a calculator?
Original post by Notnek
Do you mean without a calculator?


Basically what I'm asking is there any way to answer that question without using a quadrant diagram
Reply 4
Original post by uchihaitachi0023
Basically what I'm asking is there any way to answer that question without using a quadrant diagram


Okay.

you are finding sin of arccos(0.75), I think is possible for any type of questions like this. But you have to use cos squared + sin squared = 1.

Y = arccos(0.75), so cos(Y) = 0.75.

You are finding sin(Y), so use cos squared + sin squared = 1or similar trig pthogorean identities.

Though, you still have to use calculator most of the time

Hope this helps.
(edited 6 years ago)
Reply 5
sin[arccos(0.75)] = √(1-0.75^2) = (1/4)√7.

sin[arccos(x)] is symmetric about the x-axis so f(x) = f(-x).
(edited 6 years ago)
why are u revising c3

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