# Prove that the two trig identities are equivalent

1. Jun 29, 2013

### QuantumCurt

1. The problem statement, all variables and given/known data

Prove that the two trig identities are equivalent.

$$cos \ x \ -\frac{cos \ x}{1-tan \ x} \ = \ \frac{sin \ x \ cos \ x \ }{sin \ x \ - \ cos \ x}$$

3. The attempt at a solution

My professor recommended that we only work with one side of the equality when we're trying to prove them equivalent, and I've mainly tried using the left side. I've tried this from several different angles, and this seems to be the most viable point I've gotten to so far.

$$cos \ x \ -\frac{cos \ x}{1-tan \ x} \ = \ \frac{sin \ x \ cos \ x \ }{sin \ x \ - \ cos \ x}$$

Working with the left side, I multiplied the cos x by (1-tan x)\(1-tan x) to get them over a common denominator, and I get-

$$\frac{-cos \ x \ tan \ x}{1-tan \ x}$$

Then I rationalized the denominator by multiplying top and bottom by 1+tan x...

$$-\frac{cos \ x \ tan \ x \ -cos \ x \ tan^2 \ x}{1-tan^2 \ x}$$

Then I converted the tangent in the numerator to sine/cosine and cancelled the cosines to get--

$$-\frac{sin \ x \ - \ cos \ x \ tan^2 \ x}{1-tan^2 \ x}$$

And at this point I've hit a brick wall. Should I try using secant and cosecant identities? Should I have started it out differently, or worked with the other side? I've tried this several different ways, and I've tried transforming the right side of the equality into the left side as well, and I can't seem to get anywhere. I also tried converting the tan^2 in the numerator into sin^2/cos^2, and cancelled the cosines to get sin^2/cos, but that didn't seem to help.

What am I doing wrong? Any help is much appreciated. :)

2. Jun 29, 2013

### Simon Bridge

Are you allowed to use $$\tan x = \frac{\sin x}{\cos x} \Rightarrow 1-\tan x = \frac{\cos x - \sin x }{\cos x}$$ ??

Focussing on the LHS seems a good idea - why not start out by putting the LHS over a common denominator more simply

$$\cos x - \frac{\cos x}{1-\tan x} = \frac{(1-\tan x )\cos x - \cos x}{1-\tan x}$$

Aside: in your $\LaTeX$, if you put a backslash before the trig function, it will format correctly ;)

3. Jun 29, 2013

### ehild

Why? You have no reason to do it.

That is a good idea. Do it with all tangents.

ehild

4. Jun 29, 2013

### QuantumCurt

That's an identity that I've never seen before, but I don't think there's really anything we're not allowed to use. We've worked with pretty similar concepts before, so I'm going to give that a try.

And thanks for the tip about the formatting, that'll make things a bit simpler.

Alright, I'll jump back to before I rationalized the denominator, and convert all the tangents.

5. Jun 29, 2013

### QuantumCurt

$$\cos x-\frac{\cos x}{1-\tan x}$$
$$\frac{(1-\tan x)\cos x-\cos x}{1-\tan x}$$
$$\frac{(1-\frac{\sin x}{\cos x})\cos x-\cos x}{\frac{\cos x-\sin x}{\cos x}}$$
$$[(1-\frac{\sin x}{\cos x})\cos x-\cos x] \frac{\cos x}{\cos x-\sin x}$$
$$\frac{(1-\frac{\sin x}{\cos x})\cos^2 x-\cos^2x}{\cos x-\sin x}$$

Was I correct in that step? I wouldn't have to distribute the $\cos x$ into the $(1-\frac{\sin x}{\cos x})$ since it's part of the same term as the $\cos x$ that it's attached to, right?

Then I distributed the $\cos^2 x$, which cancels the cos in the denominator and drops the cos in the numerator to a first degree, and gives me both a positive and a negative $\cos^2 x$ in the numerator which cancel each other out

$$\frac{-\sin x \ \cos x}{\cos x -\sin x}$$

This is incredibly close, but not quite right. Does the negative sign then distribute into the bottom, and make the whole expression positive to leave me with

$$\frac{\sin x \ \cos x}{\sin x-\cos x}$$

or did I do something wrong somewhere?

6. Jun 29, 2013

### Simon Bridge

You can get $$\tan x = \frac{\sin x}{\cos x}$$ by comparing similar right angled triangles ...
one triangle has the hypotenuse = 1, so the adjacent side is cosine and the opposite side is sine; the other has the adjacent side = 1, so the opposite is tangent and the hypotenuse is secant. (these are the definitions of sine, cosine, tangent, and secant!)

Notice that these triangles give you the basic trig identities?
They are all direct consequences of the definition of "angle" as lengths on the unit circle.

Anyway, you can use the identity to "do everything in tangents" as ehild suggests too.

Certainly $$\frac{-a}{b-c}=\frac{a}{c-b}$$ ... so you are there!

7. Jun 29, 2013

### QuantumCurt

Awesome!! It's a good feeling when you get to the bottom of a math problem at 3 am on a Friday night!!...lol

Thanks for the help!!

8. Jun 29, 2013

### QuantumCurt

Now this doesn't seem right. If I plug in a value for x (I used x=2) I get different answers for the 2 different expressions. Did I do something wrong there?

9. Jun 29, 2013

### ehild

-a/(b-c)=a/(c-b) for sure. You miscalculated something.

ehild

10. Jun 29, 2013

### ehild

It is better to eliminate the cosines at that step, after expanding.
$$\frac{\cos x-\tan x \cos x-\cos x}{1-\tan x}=\frac{-\tan x \cos x}{1-\tan x}$$

Replace tanx=sinx/cosx now.

ehild

11. Jun 29, 2013

### QuantumCurt

I'll try to calculate it again. I must have entered something wrong on the calculator.

That does look easier if I eliminate the cosines at that earlier step, I'll give that a try.

Thanks for the help!

12. Jun 29, 2013

### Simon Bridge

x=2 radians is in the second quadrant - some calculators get funny when you leave the first quadrant.

Doing it for both 2 radians and 2 degrees...
Code (Text):
octave:5> s=sin(2)
s =  0.90930
octave:6> c=cos(2)
c = -0.41615
octave:7> s*c/(s-c)
ans = -0.28549
octave:8> -s*c/(c-s)
ans = -0.28549

octave:9> s=sin(2*pi/180)
s =  0.034899
octave:10> c=cos(2*pi/180)
c =  0.99939
octave:11> s*c/(s-c)
ans = -0.036162
octave:12> -s*c/(c-s)
ans = -0.036162

13. Jun 30, 2013

### QuantumCurt

I was entering it in my calculator wrong. I didn't put the denominator in parentheses at first. After I tried it again, with the parentheses around the denominators, they came out to the same answer.

Thanks again guys!