# Thread: The Square root in Trigonometry equations.

1. ## The Square root in Trigonometry equations.

Hi everyone, okay so basically i want a little help with this example from my book.

tanx=\sqrt{3}

so to start that it is a special ratio, so we are to use special triangles. And I get that part of it. however this is where I get confused.

x=tan^-1(\sqrt{3}

x= \pi /3 ,4 \pi/3

I don't understand where the -1 comes from. Also what do i do with a square root of 3 in an equation? I have another similar question that I'm working on for an assignment.

sinx=\sqrt{3} cosx

I'm at lost of how to solve it. So i was using youtube videos to help figure out but unfortunately none explain how to work it out.

sinx=\sqrt{3} cosx
\sqrt{3} +sinx=0

not sure if i can just move everything and make it equal to 0.

How come the Math tag doesn't work anymore? I hope you can all understand the question. :S

2. If I read your question right, you're wondering where the eff the -1 comes from in the equation x = \tan ^{-1}(\sqrt{3}) which results from the equation $\tan x = \sqrt{3}$. (I don't know why LaTeX won't let me enter the first thing as math text, but it won't.) That -1 is just a symbol which means the inverse of tangent. It's not actually like a number which is performing an operation. What we mean by tan^-1(x) is just that function which is the opposite of tan(x). Thus for the equation tan(x) = \sqrt{3} we take tan^-1 of both sides yielding tan^-1(tan(x)) = tan^-1(\sqrt{3}). Because tan^-1 is the opposite of tan, they cancel each other and leave just x.

3. To answer the other questions, what you do with the root-3. You enter it into the tan^-1 function and see what spits out. If you want to avoid using a calculator, you can figure it out because tan^-1 is what does the opposite of tan. Thus, find out what value you can put into tan to get out root-3 and this will be equal to what you get out of tan^-1 when you put in root-3. (But you have to do all of this within the interval (-pi/2, pi/2).)

To solve your other problem, divide both sides by cosx. When you do you'll get (sinx)/(cosx) = root-3. But sine over cosine is tangent. So this is the same thing as tanx = 0. Take tan^-1 of both sides and get x = tan^-1(root-3). You've already calculated tan^-1(root-3).

For the second one, subtract root-3 from both sides and get sinx = -root-3. Take sin^-1 of both sides and get x = sin^-1(-root-3). This one doesn't have a nice, easy answer. Get a calculator and enter sin^-1(-root-3).

4. Originally Posted by sara213
\sqrt{3} +sinx=0
This reduces to solving sin(x) = -sqrt(3). As sqrt(3) > 1 we are looking for an angle such that the sine of that angle is greater than 1. This is impossible: -1 <= sin(x) <= 1 for all x.

-Dan

5. Originally Posted by ragnar
If I read your question right, you're wondering where the eff the -1 comes from in the equation x = \tan ^{-1}(\sqrt{3}) which results from the equation $\tan x = \sqrt{3}$. (I don't know why LaTeX won't let me enter the first thing as math text, but it won't.) That -1 is just a symbol which means the inverse of tangent. It's not actually like a number which is performing an operation. What we mean by tan^-1(x) is just that function which is the opposite of tan(x). Thus for the equation tan(x) = \sqrt{3} we take tan^-1 of both sides yielding tan^-1(tan(x)) = tan^-1(\sqrt{3}). Because tan^-1 is the opposite of tan, they cancel each other and leave just x.
okay that makes sense...lol actually i knew that but I didn't really relate it to being tan.
I punched it into the calculator and it came to 60 degrees which would be pi/3 and 4pi/3.

Now, as for the second question I'm looking at what you wrote there. Dividing cosx by both sides is that a standard thing to do? I guess I can look at it differently when solving the questions...like algebra wise I suppose? I know i have seen that way to solve problems but it can be confusing too.

@_@

6. Originally Posted by ragnar
To answer the other questions, what you do with the root-3. You enter it into the tan^-1 function and see what spits out. If you want to avoid using a calculator, you can figure it out because tan^-1 is what does the opposite of tan. Thus, find out what value you can put into tan to get out root-3 and this will be equal to what you get out of tan^-1 when you put in root-3. (But you have to do all of this within the interval (-pi/2, pi/2).)

To solve your other problem, divide both sides by cosx. When you do you'll get (sinx)/(cosx) = root-3. But sine over cosine is tangent. So this is the same thing as tanx = 0. Take tan^-1 of both sides and get x = tan^-1(root-3). You've already calculated tan^-1(root-3).

For the second one, subtract root-3 from both sides and get sinx = -root-3. Take sin^-1 of both sides and get x = sin^-1(-root-3). This one doesn't have a nice, easy answer. Get a calculator and enter sin^-1(-root-3).
I don't understand.

I divided cosx on both sides to get: sinx/cosx=root3

then i have to subtract root 3 from both sides so it would be sinx/cosx-root3=0? or do i have it wrong?

but then where does the cosx go? how do i end up with sin^-1(root3).

7. Well, dividing by cosine in that problem is sorta standard and sorta not. Not standard because you don't always just divide by cosine. Take the problem sinx(cos^2(x) + sin^2(x)) = 0. If you go dividing like a mad(wo)man you'll probably end up with the wrong answer. Dividing by cosine will make everything ugly, and dividing by sine will make everything wrong. To do this problem, notice that (cos^2(x) + sin^2(x)) = 1. Thus this is really just the problem sinx = 0 and so x = 0. If you had tried, at the beginning, to divide by sinx you'd have gotten that (cos^2(x) + sin^2(x)) = 0 but as we said (cos^2(x) + sin^2(x)) = 1, so you'd have screwed yourself. (As a general note, don't always assume that you can divide, when there's some threat that the thing you're dividing by could be 0. Case in point, as we saw, sinx = 0 thus dividing by sine in the original problem is invalid and is what leads to the contradiction we find here. By that token, when doing the problem that you gave in the original post, you should have first checked that cosx could not be zero before dividing by it. This is easy to check, because if cosine were zero then the right-hand side of the equation would be zero. But it would also mean that x = pi/2 or some crap. But then the left-hand side of the equation would be sin(pi/2) which is not zero. But the left-hand side needs to equal the right-hand side. Thus cosine cannot be zero in this problem, and thus you are permitted to divide by it.)

But on the other hand, it's kind of standard for this problem to divide by cosine, because you should be training yourself to readily recognize that sine/cosine = tangent, so it should just be one of those things that you look for. You definitely have to apply your algebra skills to solving trigonometric problems, and be very comfortable with your collection of trigonometric identities.

8. Originally Posted by sara213
I don't understand.

I divided cosx on both sides to get: sinx/cosx=root3

then i have to subtract root 3 from both sides so it would be sinx/cosx-root3=0? or do i have it wrong?

but then where does the cosx go? how do i end up with sin^-1(root3).
No, no, no. Don't subtract root-3. Leave root-3. Change sinx/cosx to tanx, giving tanx = root-3. Now take tan^-1 of both sides, giving x = tan^-1(root-3). That's for the problem which starts as sinx=\sqrt{3} cosx.

You have a second, very different problem, \sqrt{3} +sinx=0. But as Mr. Top pointed out, this problem has no solution.

9. I need a lot of practice on this stuff for sure. It is very new for me at the moment so that is why...i have a hard time. however thanks for you help.
I would ask my tutor but I won't be able to go and see her this weekend because it is Easter and family stuff have some obligation...@_@

10. Originally Posted by ragnar
No, no, no. Don't subtract root-3. Leave root-3. Change sinx/cosx to tanx, giving tanx = root-3. Now take tan^-1 of both sides, giving x = tan^-1(root-3). That's for the problem which starts as sinx=\sqrt{3} cosx.

You have a second, very different problem, \sqrt{3} +sinx=0. But as Mr. Top pointed out, this problem has no solution.

Oh, no....i think i discovered a typo...i actually meant to put...cosx (root 3) + sinx=0 i was trying to show what i put for the question of sinx=(root 3) cosx. My bad so sorry

11. Sal good.

12. Originally Posted by ragnar
...But as Mr. Top pointed out, this problem has no solution.
"Mr. Top?"

Just kidding. It's cool.

-Dan