trigonometry exam question

y=arccosx.
cosy=x
siny=??

I'm doing part i, and I dont know where to start. can someone explain this question to me please? thanks

should I start with drawing the arccosx graph?

This was one of the worst answered Edexcel pure questions of all time by the way. Nasty question.

yeah, I dont get it.

I see that cos y = x

then I dont know where to go from there?

Do you know an identity that links cos and sin (not cos^2 and sin^2)? It’s a less used identity but can be seen by observing that sin/cos are translations of each other.

yeah, I dont get it.

I see that cos y = x

then I dont know where to go from there?

Do you know an identity that links cos and sin (not cos^2 and sin^2)? It’s a less used identity but can be seen by observing that sin/cos are translations of each other.

Well tbh just arcsin both sides and say done

thats not the right answer though

Well tbh just arcsin both sides and say done

Do you know an identity that links cos and sin (not cos^2 and sin^2)? It’s a less used identity but can be seen by observing that sin/cos are translations of each other.

so, I'm looking at the arcs and arcsin graphs.

for arcsin to = arcos (y), it must reflect in the y axis (-y) and go up pi/2?

which means arcsinx = pi/2 - y ????

is that right way of thinking and doing it?

That's correct. The more standard approach would be to consider sin/cos. The identities you need are

$\cos x = \sin\left(\frac{\pi}{2}-x\right)$

$\sin x = \cos\left(\frac{\pi}{2}-x\right)$

This is why e.g. sin(60) = cos(30) and cos(80) = sin(10) in other words the sin/cos angles add up to 90 or $\frac{\pi}{2}$ radians. You need to know these identities for the exam (although they're used rarely).

Back to the question:

$y=arccos(x)$
$\Rightarrow \cos(y) = x$
$\Rightarrow \sin\left(\frac{\pi}{2}-y\right) = x$

Can you finish it off to find $\arcsin(x)$ in terms of $y$?

thanks!! no i can't, i don't understand how you got the answer. what did you do?