anonymous
  • anonymous
Proving Trigonometric Identities
Mathematics
schrodinger
  • schrodinger
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jdoe0001
  • jdoe0001
posting specifics helps
zzr0ck3r
  • zzr0ck3r
and go
anonymous
  • anonymous
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UsukiDoll
  • UsukiDoll
\[(\frac{\tan(\theta)\sin(\theta)}{1-\cos(\theta)}) \equiv 1 + \frac{1}{\cos(\theta)} \]
zzr0ck3r
  • zzr0ck3r
\(\dfrac{\tan\theta \sin \theta}{1-\cos\theta} = \dfrac{\frac{\sin\theta}{\cos\theta}\sin\theta}{1-cos\theta}=\dfrac{\frac{\sin^2\theta}{\cos\theta}}{1-\cos\theta}=\dfrac{\frac{1-\cos^2\theta}{\cos\theta}}{1-\cos\theta}=\frac{1-\cos^2\theta}{\cos\theta}*\frac{1}{1-\cos\theta}=\\ \frac{(1-\cos\theta)(1+\cos\theta)}{\cos\theta}\frac{1}{1-\cos\theta}=\frac{1+\cos\theta}{\cos\theta}=\frac{1}{\cos\theta}+\frac{\cos\theta}{\cos\theta}=1+\frac{1}{\cos\theta}\)
zzr0ck3r
  • zzr0ck3r
there is prob an easier way but I just started and went until I got there...
UsukiDoll
  • UsukiDoll
I did the same thing in my mind with the tangent being rewritten and a trig identity being used.
anonymous
  • anonymous
that format is confusing tho @zzr0ck3r @UsukiDoll
UsukiDoll
  • UsukiDoll
ok... start from the left side of the equation.. and we use manipulations or techniques so we can have the right side of the equation
anonymous
  • anonymous
i mean this (\dfrac{\tan\theta \sin \theta}{1-\cos\theta} = \dfrac{\frac{\sin\theta}{\cos\theta}\sin\theta}{1-cos\theta}=\dfrac{\frac{\sin^2\theta}{\cos\theta}}{1-\cos\theta}=\dfrac{\frac{1-\cos^2\theta}{\cos\theta}}{1-\cos\theta}=\frac{1-\cos^2\theta}{\cos\theta}*\frac{1}{1-\cos\theta}=\\ \frac{(1-\cos\theta)(1+\cos\theta)}{\cos\theta}\frac{1}{1-\cos\theta}=\frac{1+\cos\theta}{\cos\theta}=\frac{1}{\cos\theta}+\frac{\cos\theta}{\cos\theta}=1+\frac{1}{\cos\theta}\)
UsukiDoll
  • UsukiDoll
\[\tan \theta = \frac{\sin(\theta)}{\cos(\theta)}\] so we use this fact first...
jdoe0001
  • jdoe0001
hmmm
anonymous
  • anonymous
the curly brackets and stuff.. i got to go..........parents....but plz finish helping and ill medal/fan or whatever when i get back online
anonymous
  • anonymous
Thanks
UsukiDoll
  • UsukiDoll
then after you use the tangent identity, we have another identity to use \[\sin^2(\theta) = 1-\cos^2(\theta)\] flip the second fraction and something will cancel.
UsukiDoll
  • UsukiDoll
we can also use the right hand side of the equation as a guideline too. 1/cosine is actually secant... so we need to obtain 1 + secant.
jdoe0001
  • jdoe0001
\(\bf \cfrac{tan(\theta)sin(\theta)}{1-cos(\theta)}\implies \cfrac{\frac{sin(\theta)}{cos(\theta)}sin(\theta)}{1-cos(\theta)}\implies \cfrac{\frac{sin^2(\theta)}{cos(\theta)}}{1-cos(\theta)} \\ \quad \\ \cfrac{sin^2(\theta)}{cos(\theta)}\cdot \cfrac{1}{1-cos(\theta)}\implies \cfrac{sin^2(\theta)}{cos(\theta)[1-cos(\theta)]} \\ \quad \\ recall\implies sin^2(\theta)+cos^2(\theta)=1\implies sin^2(\theta)={\color{brown}{ 1-cos^2(\theta) }}\qquad thus \\ \quad \\ \cfrac{sin^2(\theta)}{cos(\theta)[1-cos(\theta)]}\implies \cfrac{{\color{brown}{ 1-cos^2(\theta) }}}{cos(\theta)[1-cos(\theta)]} \\ \quad \\ \textit{now keep in mind that }1^2, 1^3, 1^4 = 1\qquad thus \\ \quad \\ \cfrac{{\color{brown}{ 1-cos^2(\theta) }}}{cos(\theta)[1-cos(\theta)]}\implies \cfrac{{\color{brown}{ 1^2-cos^2(\theta) }}}{cos(\theta)[1-cos(\theta)]} \\ \quad \\ \textit{now using difference of square }a^2-b^2 = (a-b)(a+b) \\ \quad \\ \cfrac{{\color{brown}{ \cancel{[1-cos(\theta)]}[1+cos(\theta)] }}}{cos(\theta)\cancel{[1-cos(\theta)]}}\implies \cfrac{1+cos(\theta)}{cos(\theta)} \\ \quad \\ \cfrac{1}{cos(\theta)}+\cfrac{cos(\theta)}{cos(\theta)}\implies ?\)
Mertsj
  • Mertsj
\[\frac{\tan \theta \sin \theta}{1-\cos \theta}=\frac{\frac{\sin \theta}{\cos \theta}\times \sin \theta}{1-\cos \theta}\times\frac{\cos \theta}{\cos \theta}=\frac{\sin ^2\theta}{\cos \theta(1-\cos \theta)=}\] \[\frac{1-\cos ^2\theta}{\cos \theta(1-\cos \theta)}=\frac{(1-\cos \theta)(1+\cos \theta)}{\cos \theta(1-\cos \theta)}=\frac{1+\cos \theta}{\cos \theta}\]
Loser66
  • Loser66
Ha!! it is fun, I want to contribute my way also, I replace \(\theta = t\), from the right hand side \(1+\dfrac{1}{cos (t)}=\dfrac{cos (t)}{cos(t)}+\dfrac{1}{cos(t)}=\dfrac{1+cos(t)}{cos(t)}\) multiple both numerator and denominator by (1-cos(t)) we get \(\dfrac{(1+cos(t))(1-cos(t))}{cos(t)(1-cos(t))}=\dfrac{1-cos^2(t)}{cos(t)(1-cos(t))}\) and \(1-cos^2(t) = sin^2(t) = sin(t)*sin(t)\) hence, it becomes \(\dfrac{sin(t)}{cos(t)}*\dfrac{sin(t)}{1-cos(t)}= \dfrac{tan(t)*sin(t)}{1-cos(t)}\) this is the left hand side.
anonymous
  • anonymous
wow guys.....thanks alot!!

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