Trig help - How do i know cos^-1(-sqrt3/2) = 5pi/6?

[Alcsaar]

I understand how to figure out that it represents a pi/6 or 30 degree angle since its sqrt3/2 and the 30-60-90 triangle can tell me its a 30 degree angle as a result, but how do I know its 5pi/6? What do I do to solve the problem after I know what angle its based off? I'm like missing a step here and its driving me crazy for an exam I have in about an hour.


Edit* This should probably be moved to the "Fun Stuff" offtopic forum /endsarcasm

[Independent voter]

wolframalpha.com says it's true but you probably already knew that.

[Elorii]

Because it is negative, it must be in quadrant II or III. So it can't be pi/6, so 5pi/6 is next.

[Deleted]

I understand how to figure out that it represents a pi/6 or 30 degree angle since its sqrt3/2 and the 30-60-90 triangle can tell me its a 30 degree angle as a result, but how do I know its 5pi/6? What do I do to solve the problem after I know what angle its based off? I'm like missing a step here and its driving me crazy for an exam I have in about an hour.


Edit* This should probably be moved to the "Fun Stuff" offtopic forum /endsarcasm

Well, if you already know cos^-1(sqrt(3)/2) is pi/6, then you know that cos^-1(-sqrt(3)/2) is pi-pi/6=5pi/6, because cos(pi-alpha)=-cos(alpha). It was that or are you missing something else?

[StayTuned]

And then people wonder why children find no fun in math. Boy am I happy that I don't need this to survive. Sorry OP, just looking at this gave me headaches. I hope you had good luck with your exam

[the game]

The answer is 42

[fengosa]

And then people wonder why children find no fun in math. Boy am I happy that I don't need this to survive. Sorry OP, just looking at this gave me headaches. I hope you had good luck with your exam

This isn't that complicated. If you literally had to learn it for your own safe being you'd probably pick it up in half an hour.

[Kruncholyo]

And then people wonder why children find no fun in math. Boy am I happy that I don't need this to survive. Sorry OP, just looking at this gave me headaches. I hope you had good luck with your exam

This gets more fun if you solve it using the exponential expansion for trig functions:

http://www.phy.duke.edu/~rgb/Class/p...y51/img135.png

And
cosx=Re{e^ix}=[e^ix+e^−ix]/2

Euler was Awesome!

[Xargoth6634]

This isn't that complicated. If you literally had to learn it for your own safe being you'd probably pick it up in half an hour.

It's not complicated if math comes easily to you. Anything beyond algebra or basic states I just can't do, and not from lack of trying. Not everyone's built the same way.

[Dendrek]

Sorry I wasn't here to answer this when you needed it, but here's the reason (which Elorii already discussed):

Let x be the angle we're looking for.

cos^-1(-sqrt(3)/2) = x

The value for x must be related to pi/6
You've already pointed this out. But let's clarify what that means.

If the 30-60-90 triangle is in the:

• First quadrant, x = pi/6
• Second quadrant, x = 5pi/6
• Third quadrant, x = 7pi/6, or x = -5pi/6
• Forth quadrant, x = 11pi/6, or x = -pi/6

I assume you know each of these rules, but if you don't say so and I'll explain further.

So which quadrant is our problem in?
To answer that, there's a technicality for inverse cosine you need to know: it can only affect two of the quadrants, not all four. The reason it can't exist in all 4 is because it has to be a function (if you know this or if you don't want to get too technical, you can skip to the next bolded part below)

function (in math) definition:
For any input value (x), there can be, at most, one output value (y)

In the case of inverse cosine, the input values are lengths and the output values are angles. Let's backtrack to regular cosine (non inverse cosine). In order to get the output sqrt(3)/2, there are two inputs you can use: pi/6 or 11pi/6. So for the inverse cosine, pi/6 and 11pi/6 are possible answers to the input sqrt(3)/2. But it's supposed to be a function, so there can only be one possible answer. You can't have both pi/6 or 11pi/6 be answers or else inverse cosine is not a function. So we need to limit the output values we can use. For inverse cosine, the output values are limited to the range [0,pi]. In other words, any value between the angle 0 and the angle pi are allowed. Anything bigger or smaller are not.

Why limit the range to [0,pi] and not some other arbitrary set of angles? Because it's the nicest, smallest set of angles that account for all possible input values. In other words, it accounts for all input values between cos^(-1)(-1) and cos^(-1)(1). With a range of output values being [0,pi], that means the quadrants inverse cosine exists in are the first and second. And that means cos^(-1)(-sqrt(3)/2) has to be either pi/6 or 5pi/6.

So why 5pi/6?
You plugged in -sqrt(3)/2. For cosine, negative outputs only occur in the second or third quadrant. (Again, I assume you know this, but if not I can explain in more detail.) But since inverse cosine only exists in the first or second quadrant, we're stuck with the second quadrant answer.

For reference:
inverse sine only exists in quadrants 1 and 4 (and the answers can be between -pi/2 to pi/2), quadrant 4 contains the negative inputs
Inverse tangent only exists in quadrants 1 and 4 (and the answers can be between -pi/2 and pi/2), quadrant 4 contains the negative inputs
Inverse cosine only exists in quadrants 1 and 2 (and the answers can be between 0 and pi), quadrant 2 contains the negative inputs