Understanding the Simple Proof of sin^2θ = 1-cos^2θ in Trigonometry

[claytonh4]

Hi everyone.
I'm in high school still so please bear with me on the simplicity of this problem. I'm working through a calculus book and the trig intro section has a proof I know to be true, but I have a hard time following in one spot. Here it is:

sin^2θ=(y^2/r^2)=([r^2-x^2]/r^2)=1-(x/r)^2=1-cos^2θ

The part where I get confused is how ([r^2-x^2]/r^2)=1-(x/r)^2
I would have thought the r^2/r^2 would make 1, leaving a negative x^2; thus rather than it being a 1-(x/r)^2, it would be 1-x^2. If someone could explain where my mistake is i would appreciate it. And sorry it's so messy, I couldn't figure out how to superscript so I used the ^ symbol.

 

[Mark44]

Hi everyone.
I'm in high school still so please bear with me on the simplicity of this problem. I'm working through a calculus book and the trig intro section has a proof I know to be true, but I have a hard time following in one spot. Here it is:

sin^2θ=(y^2/r^2)=([r^2-x^2]/r^2)=1-(x/r)^2=1-cos^2θ

The part where I get confused is how ([r^2-x^2]/r^2)=1-(x/r)^2
I would have thought the r^2/r^2 would make 1, leaving a negative x^2; thus rather than it being a 1-(x/r)^2, it would be 1-x^2. If someone could explain where my mistake is i would appreciate it. And sorry it's so messy, I couldn't figure out how to superscript so I used the ^ symbol.

Your mistake is thinking that $\frac{r^2 - x^2}{r^2} = 1 - x^2$

In other words, that the r² term in the numerator "cancels" with the r² term in the denominator. Fractions and rational expressions don't work that way. The only time you get cancellation is when the same factor appears in both numerator and denominator. The problem in what you did is that the numerator is not factored.

This is what you could have done:
$$\frac{r^2 - x^2}{r^2} = \frac{r^2(1 - (x^2/r^2))}{r^2} = \frac{r^2}{r^2}\cdot (1 - (x^2/r^2)) = 1 - (x^2/r^2)$$

The second expression, above, has a factor of r² in both numerator and denominator, so it cancels.

 

[azizlwl]

Homework Statement

Hi everyone.
I'm in high school still so please bear with me on the simplicity of this problem. I'm working through a calculus book and the trig intro section has a proof I know to be true, but I have a hard time following in one spot. Here it is:

Homework Equations

sin^2θ=(y^2/r^2)=([r^2-x^2]/r^2)=1-(x/r)^2=1-cos^2θ

The Attempt at a Solution

The part where I get confused is how ([r^2-x^2]/r^2)=1-(x/r)^2
I would have thought the r^2/r^2 would make 1, leaving a negative x^2; thus rather than it being a 1-(x/r)^2, it would be 1-x^2. If someone could explain where my mistake is i would appreciate it. And sorry it's so messy, I couldn't figure out how to superscript so I used the ^ symbol.

r²=y²+x²
You have to divide both r² and x² by r²
$(r^2-x^2)\frac{1}{r^2}$

 

[claytonh4]

Your mistake is thinking that $\frac{r^2 - x^2}{r^2} = 1 - x^2$

In other words, that the r² term in the numerator "cancels" with the r² term in the denominator. Fractions and rational expressions don't work that way. The only time you get cancellation is when the same factor appears in both numerator and denominator. The problem in what you did is that the numerator is not factored.

This is what you could have done:
$$\frac{r^2 - x^2}{r^2} = \frac{r^2(1 - (x^2/r^2))}{r^2} = \frac{r^2}{r^2}\cdot (1 - (x^2/r^2)) = 1 - (x^2/r^2)$$

The second expression, above, has a factor of r² in both numerator and denominator, so it cancels.

Oh ok I see. So if I want to employ that train of thought I need to expand the problem so it will work? Thank you for your help, that makes more sense now.

 

[claytonh4]

r²=y²+x²
You have to divide both r² and x² by r²
$(r^2-x^2)\frac{1}{r^2}$

Ok that makes more sense. I think I was looking at it too simplistically. When I saw the 1 in the equation, I wanted to try to cancel, but I didn't account for dividing there. Thanks for your help.

 

[Simon Bridge]

Neat - I'd like to add to what Mark44 said with:

$$\frac{a+b}{c} = (a+b)\div c = \frac{1}{c}(a+b) = \frac{a}{c} + \frac{b}{c}$$

 

[claytonh4]

Neat - I'd like to add to what Mark44 said with:

$$\frac{a+b}{c} = (a+b)\div c = \frac{1}{c}(a+b) = \frac{a}{c} + \frac{b}{c}$$

Thank you! I think that's an excellent example of how that dividing property works.

 

[Simon Bridge]

As the Heart of Gold doors' say, "pleased to be of service."