Proving an equation

[shelarson]

The equation is

sin?/(1-cos? )-(1+cos?)/sin? = 0

sin?/(1-cos? )*(1+cos?)/(1+cos? )=(sin? (1+cos?))/(1-cos^2? )

(sin? (1+cos?))/sin^2? =(1+cos?)/sin?

(1+cos?)/sin? -(1+cos?)/sin? =0


The part that I am confused about is the *(1+cos?)/(1+cos? ). I need to justify that 1+cos? does not equal 0, because you can not divide by 0. What is the proof that goes with that part of the equation.

Thanks, Shelly

 

[chrisr]

Hi Shelly,

here's how you should approach this....

\(\displaystyle \frac{sin\theta}{1-cos\theta}-\frac{1+cos\theta}{sin\theta}=0\)

\(\displaystyle \frac{sin\theta}{sin\theta}\ \frac{sin\theta}{1-cos\theta}-\frac{1-cos\theta}{1-cos\theta}\ \frac{1+cos\theta}{sin\theta}=0\)

\(\displaystyle The\ numerator\ must\ be\ zero, so\)

\(\displaystyle sin^2\theta=(1-cos\theta)(1+cos\theta)\)

\(\displaystyle sin^2\theta=1+cos\theta-cos\theta-cos^2\theta\)

\(\displaystyle sin^2\theta+cos^2\theta=1....\ true\)

 

[soroban]

Hello, shelarson!

\(\displaystyle \text{Prove that: }\;\frac{\sin\theta}{1-\cos\theta} - \frac{1+\cos\theta}{\sin\theta} \:=\:0\)

\(\displaystyle \text{Multiply the first fraction by }\frac{1+\cos\theta}{1+\cos\theta}\)

.\(\displaystyle \frac{\sin\theta}{1-\cos\theta}\cdot\frac{1+\cos\theta}{1+\cos\theta} \;=\;\frac{\sin\theta(1+\cos\theta)}{1-\cos^2\!\theta} \;=\;\frac{\sin\theta(1+\cos\theta)}{\sin^2\theta} \;=\;\frac{1+\cos\theta}{\sin\theta}\)

\(\displaystyle \text{Therefore, the problem becomes: }\;\frac{1+\cos\theta}{\sin\theta} - \frac{1+\cos\theta}{\sin\theta} \;=\;0\)



. . \(\displaystyle \sim\;\sim\;\sim \quad \text{EDIT}\;\sim \;\sim\;\sim\)

As chrisr pointed out, the original denominators must not be zero.

\(\displaystyle \text{Then: }\:\sin\theta \,\neq\,0 \quad\Rightarrow\quad \theta \,\neq\,\pi\)

\(\displaystyle \text{If }\theta \,\neq\,\pi\,\text{ then: }1 + \cos\theta\,\neq\,0\)

\(\displaystyle \text{Hence, we }can\text{ freely multiply by }\frac{1+\cos\theta}{1+\cos\theta}\)

 

[chrisr]

\(\displaystyle I'm\ sorry,\ Shelly,\)

\(\displaystyle I\ did\ not\ read\ your\ question\ properly.\)

\(\displaystyle 1+cos\theta\ is\ zero\ for\ cos\theta=-1,\ so\ \theta=\pi\ radians.\)

 

[chrisr]

\(\displaystyle Using\ the\ same\ logic,\ the\ original\ equation\ has\ denominators\ 1-cos\theta\ and\ sin\theta,\)
\(\displaystyle which\ ought\ not\ be\ zero.\)