arcsin(y) + arccos(y)

Original post by oniisanitstoobig

The question is:

Let y = cos(x)
Evaluate arcsin(y) + arccos(y)

I used the trig identity cos(x) = sin(x + pi/2) and the answer came out to be 2x + pi/2. However that was apparently wrong - the mark scheme used the identity cos(x) = sin(pi/2 - x) instead and got the answer pi/2.
I couldn't seem to find anything wrong with my method at the moment, wondering if anyone could help identify why my answer came out wrong.

Think about the range of arcsin and arccos.

Spoiler

(edited 8 years ago)

Reply 2

OP

I'm sorry for being so dense but I'm still a bit confused as to why the two identities makes a difference when considering the range, could you possibly elaborate a bit on that?

(edited 8 years ago)

Reply 3

Original post by oniisanitstoobig

I'm sorry for being so dense but I'm still a bit confused as to why the two identities makes a difference when considering the range, could you possibly elaborate a bit on that?

There are many solutions to sinx=0.6 (for example) and the function arcsin simply chooses one of them, namely the one between -0.5pi and 0.5pi. Thus arcsin(sin(x+pi/2)) is not equal to x+pi/2. Think about the graph of y=arcsin(x) and y=arccos(x) and the transformation that takes one to the other (think about symmetry here, one is the reflection of the other).

(edited 8 years ago)

Reply 4

OP

I do get what you mean now (I think),
arcsin(x) = -arccos(x) + pi/2;
so arcsin(x) + arccos(x) = pi/2

But how comes arcsin(sin(x+pi/2)) != x+pi/2 whilst arcsin(sin(pi/2-x)) = pi/2-x, (according to the mark scheme)? Wouldn't both of them have values of x which brings it out of range of the arcsin function?

Reply 5

ian.slater

12

Original post by oniisanitstoobig

But how comes arcsin(sin(x+pi/2)) != x+pi/2 whilst arcsin(sin(pi/2-x)) = pi/2-x, (according to the mark scheme)? Wouldn't both of them have values of x which brings it out of range of the arcsin function?

I like the way you put this question, and shuddered a bit when I saw your mark scheme quote.

Renzhi is quite right ... you have to starting by thinking what arcsin means. If it were me, I would start by figuring out what happens when 0 < arcsin(y) < pi/2 and draw a triangle. Then I'd work out what happens for other values of y. Just jumping in with some identity seems a bit too risky.

(edited 8 years ago)

Reply 6

OP

I went through this question again and I'm wondering if I got the right idea:
since arcsin(y)'s range is limited to -pi/2 and pi/2, when sin(x+pi/2) = cos(x) = y is subbed in, x will be limited to the range of -pi and 0. This would mean arccos(cos(x)) is undefined and so the identity sin(x+pi/2) can't be used. I'm sorry if that was phrased horribly, but I hope I'm somewhat on the right track.

Thank for your time Renzhi and ian.slater, and I do apologise for my stubbornness!

(edited 8 years ago)

Reply 7

Original post by oniisanitstoobig

I went through this question again and I'm wondering if I got the right idea:
since arcsin(y) and arccos(y)'s shared range is from 0 to pi/2, both functions must remain in that shared range for the addition arcsin(y) + arccos(y) and hence the reason the trig identity sin(x+pi/2) can't be used. I'm sorry if that was phrased horribly, but I hope I'm somewhat on the right track.

Thank for your time Renzhi and ian.slater, and I do apologise for my stubbornness!

The reason the identity cos(x)=sin(pi/2+x) can't be used is simply because

arcsin(sin(x))\neq x

for some x. The same is true for arccos. It's not really anything to do with a shared range as you call it, and in fact, arcsin(y)+arccos(y) doesn't have to remain in that range.

And no problem, this is why TSR exists!

(edited 8 years ago)

Reply 8

OP

Sorry, I edited my post right before yours after reading back through it and realising it made 0 sense.

Reply 9

ian.slater

12

Original post by oniisanitstoobig

I went through this question again and I'm wondering if I got the right idea:
since arcsin(y) and arccos(y)'s shared range is from 0 to pi/2, both functions must remain in that shared range for the addition arcsin(y) + arccos(y) and hence the reason the trig identity sin(x+pi/2) can't be used. I'm sorry if that was phrased horribly, but I hope I'm somewhat on the right track.

Thank for your time Renzhi and ian.slater, and I do apologise for my stubbornness!

You owe no apology ... you just want to understand.

Having thought about this a bit more, I think the most reliable method is to graph both functions. When you've got the sketch you can use arguments about symmetry. Just using trig identities just seems to get messy if you want to do it properly. I'm a bit curious as to why they use y ... is this part of a bigger question by some chance?

Reply 10

Buses

i got 0.5pi. Just use your compound trig identies to prove sinx = cos(0.5pi-x). That's all there is to it.

May I ask which book are you using? :smile:

(edited 8 years ago)

Reply 11

OP

Ah, both of you replied right after I edited #7. I rephrased what I gathered so far a few minutes ago in an edit, hopefully I made a bit more sense there.

Original post by ian.slater

I'm a bit curious as to why they use y ... is this part of a bigger question by some chance?

yes, this is a part of a larger question but this one seems pretty standalone relatively (as in it's done without needing to do the parts before). Here's the rest of the question:

Original post by Buses

i got 0.5pi. Just use your compound trig identies to prove sinx = cos(0.5pi-x). That's all there is to it.May I ask which book are you using? :smile:

Yes, that identity is the correct way to do it. But my question was that since cos(x) is also sin(x+pi/2), why does subbing in that identity give the wrong answer?
I'm using Cambridge Maths HL Exam Prep Guide.

(edited 8 years ago)

Reply 12

ian.slater

12

Original post by Buses

i got 0.5pi. Just use your compound trig identies to prove sinx = cos(0.5pi-x). That's all there is to it.

May I ask which book are you using? :smile:

The trouble is that OP got the wrong answer by starting from a different trig identity.

btw - I suggest you write pi/2 rather than a decimal. Otherwise pi/2 = pi/3 + pi/6 starts to get messy. Decimals are best reserved for applied maths where some approximation is implied.

Reply 13

Original post by oniisanitstoobig

Sorry, I edited my post right before yours after reading back through it and realising it made 0 sense.

Lol, no worries. You have to realise that arccos(cos(x)) is never undefined, and I suspect that that is why they said y=cos(x) at the beginning, since arccos(y) is defined far all -1<=y<=1. As i said before, it's about the range of arccos and arcsin. In fact, if you wanted to use your identity, then you would have to say

arcsin(sin(x))=x+2k\pi, \pi-x+2k\pi

, for some k, such that one of the two lies between -0.5pi and 0.5pi. But things get a bit complicated at that point, it's probably best to just consider the symmetry of arccos and arcsin.

Reply 14

Original post by Buses

i got 0.5pi. Just use your compound trig identies to prove sinx = cos(0.5pi-x). That's all there is to it.

May I ask which book are you using? :smile:

Not really, since

arcsin(sin(x))\not\equiv x

and

arccos(cos(x))\not\equiv x

Reply 15

Buses

Original post by Renzhi10122

Not really, since

arcsin(sin(x))\not\equiv x

and

arccos(cos(x))\not\equiv x

well yeah no need for the technicalities when answering these C2 a level maths kind of questions.

Reply 16

Buses

Original post by ian.slater

The trouble is that OP got the wrong answer by starting from a different trig identity.

btw - I suggest you write pi/2 rather than a decimal. Otherwise pi/2 = pi/3 + pi/6 starts to get messy. Decimals are best reserved for applied maths where some approximation is implied.

ah i see, cheers :smile:

Reply 17

OP

My argument there was that since from arccos(sin(x+pi/2)) the range of x has been limited, then for the same x, in arccos(cos(x)) the values for arccos(cos(x)) becomes undefined if the same domain for x is used.

Anyways, thanks again. I realised you probably spent a while replying to all my questions, have a nice evening & good luck on your results day!

(edited 8 years ago)

Reply 18

atsruser