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trig identity: sin @ / (1-cos @) - (1 + cos @) / sin @ = 0
- Thread starter sonyanale
- Start date
Re: Math identity --Please help!
Hello, sonyanale!
Another approach . . .
\(\displaystyle \text{Multiply the first fraction by }\,\frac{1+\cos\theta}{1+\cos\theta}\)
\(\displaystyle \frac{\sin\theta}{1-\cos\theta}\cdot\frac{1+\cos\theta}{1+\cos\theta} \;=\;\frac{\sin\theta(1+\cos\theta)}{1 - \cos^2\!\theta} \;=\;\frac{\sin\theta(1+\cos\theta)}{\sin^2\!\theta} \;=\;\frac{1+\cos\theta}{\sin\theta}\)
\(\displaystyle \text{And the problem becomes: }\:\frac{1+\cos\theta}{\sin\theta} - \frac{1+\cos\theta}{\sin\theta} \quad\hdots\;\text{ which equals 0.}\)
Hello, sonyanale!
Another approach . . .
\(\displaystyle \text{Prove: }\:\frac{\sin\theta}{1- \cos\theta} - \frac{1+ \cos\theta}{\sin\theta} \:=\:0\)
\(\displaystyle \text{Multiply the first fraction by }\,\frac{1+\cos\theta}{1+\cos\theta}\)
\(\displaystyle \frac{\sin\theta}{1-\cos\theta}\cdot\frac{1+\cos\theta}{1+\cos\theta} \;=\;\frac{\sin\theta(1+\cos\theta)}{1 - \cos^2\!\theta} \;=\;\frac{\sin\theta(1+\cos\theta)}{\sin^2\!\theta} \;=\;\frac{1+\cos\theta}{\sin\theta}\)
\(\displaystyle \text{And the problem becomes: }\:\frac{1+\cos\theta}{\sin\theta} - \frac{1+\cos\theta}{\sin\theta} \quad\hdots\;\text{ which equals 0.}\)
Sonya, with respect, if you're that lost you need to see your instructor. Soroban gave about as much detail as can be shown. I just showed a simple identity
\(\displaystyle 1-cos^{2}(t)=sin^{2}(t)\).
That leaves 0 in the numerator and it's finished.
\(\displaystyle sin^{2}(t)-(1-cos^{2}(t))=0\)
\(\displaystyle 1-cos^{2}(t)=sin^{2}(t)\).
That leaves 0 in the numerator and it's finished.
\(\displaystyle sin^{2}(t)-(1-cos^{2}(t))=0\)
I thought you did not know what you were doing when you said to cross multiply because the book says not to assume that both sides are equal. It also says that you cannot prove an identity by first assuming that both sides are equal. Sorry. I just wanted more explanations with each step on the other reply.
Thanks anyway
Thanks anyway