# Verify trig identity

## Homework Statement

Use a graphing calculator to test whether the following is an identity. If it is an identity, verify it. If it is not an identity, find a value of x for which both sides are defined but not equal.

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

None

## The Attempt at a Solution

Ok, plug in the left side for y1, right side for y2, obviously not an identity. The second part where it ask for a x value is where Im having trouble. I thought maybe simplify the left hand side and find a value for whatever that is that equals 1..

Cos(-x)/sin(X)(1/-tan(X))
Cos(x)/sin(x)(-cosx/sinx)
Cos(x)/-cos(x)
Cos(x)/1-sin(x)
Cos(x)-Sin(x)cos(x)
Factor out to get
1-sin(x)
ok so now I have 1-sin(x)=1
-sin(x)=0
so find the value where -sin(x)=0??

The teachers answer is - Not an identity, x=$$\pi/4$$

Im clueless about the second part I guess..

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so, you simplified it to the form of Cos(x)/-cos(x). This just equals -1. (your next step was incorrect, though... how did you get from -cos(x) to 1-sin(x)?)
So it obviously is not equal to one. The problem statement asks for "a value of x for which both sides are defined but not equal". We have just shown that whenever the expression is defined, the equation doesn't hold. So you just need to find a value where the expression is defined.

so, you simplified it to the form of Cos(x)/-cos(x). This just equals -1. (your next step was incorrect, though... how did you get from -cos(x) to 1-sin(x)?)
So it obviously is not equal to one. The problem statement asks for "a value of x for which both sides are defined but not equal". We have just shown that whenever the expression is defined, the equation doesn't hold. So you just need to find a value where the expression is defined.
Yeah your right should be -1, I'm not quite sure how I would go about finding a value where the expression is defined.

Anyone? That was my original question, Im not sure how to find that value.

Last edited:
It's defined almost everywhere. Since cot(-x)=-cos(x)/sin(x), you need to make sure sin(x)=/=0.

Then you also need to make sure that the simplified expression, cos(x)/-cos(x) is defined, i.e. cos(x)=/=0.

That's all. There are plenty of angles for which this holds. pi/4 is one, but you can also have pi/3, 2*pi/3, etc..

Mentallic
Homework Helper
Stevo6754, since you're able to simplify the expression and make it $$\frac{cos(-x)}{sinxcot(-x)}=-1$$, this is telling you that for all defined values of x (they are undefined where the denominator=0) it is not equal to 1. So like grief has said, the values you're looking for are not just $\pi/4$ but any other value that is defined.

Oh and while $cos^2x=1-sin^2x$ this does not mean $cosx=1-sinx$ since, if you square both sides you'll get $cos^2x=(1-sinx)^2=1-2sinx+sin^2x\neq 1-sin^2x$