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anonymous
 one year ago
Proving Trigonometric Identities
anonymous
 one year ago
Proving Trigonometric Identities

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0posting specifics helps

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1436486070798:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0\[(\frac{\tan(\theta)\sin(\theta)}{1\cos(\theta)}) \equiv 1 + \frac{1}{\cos(\theta)} \]

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1\(\dfrac{\tan\theta \sin \theta}{1\cos\theta} = \dfrac{\frac{\sin\theta}{\cos\theta}\sin\theta}{1cos\theta}=\dfrac{\frac{\sin^2\theta}{\cos\theta}}{1\cos\theta}=\dfrac{\frac{1\cos^2\theta}{\cos\theta}}{1\cos\theta}=\frac{1\cos^2\theta}{\cos\theta}*\frac{1}{1\cos\theta}=\\ \frac{(1\cos\theta)(1+\cos\theta)}{\cos\theta}\frac{1}{1\cos\theta}=\frac{1+\cos\theta}{\cos\theta}=\frac{1}{\cos\theta}+\frac{\cos\theta}{\cos\theta}=1+\frac{1}{\cos\theta}\)

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1there is prob an easier way but I just started and went until I got there...

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0I did the same thing in my mind with the tangent being rewritten and a trig identity being used.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that format is confusing tho @zzr0ck3r @UsukiDoll

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0ok... 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
 one year ago
Best ResponseYou've already chosen the best response.0i mean this (\dfrac{\tan\theta \sin \theta}{1\cos\theta} = \dfrac{\frac{\sin\theta}{\cos\theta}\sin\theta}{1cos\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
 one year ago
Best ResponseYou've already chosen the best response.0\[\tan \theta = \frac{\sin(\theta)}{\cos(\theta)}\] so we use this fact first...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the curly brackets and stuff.. i got to go..........parents....but plz finish helping and ill medal/fan or whatever when i get back online

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0then 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
 one year ago
Best ResponseYou've already chosen the best response.0we 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(\bf \cfrac{tan(\theta)sin(\theta)}{1cos(\theta)}\implies \cfrac{\frac{sin(\theta)}{cos(\theta)}sin(\theta)}{1cos(\theta)}\implies \cfrac{\frac{sin^2(\theta)}{cos(\theta)}}{1cos(\theta)} \\ \quad \\ \cfrac{sin^2(\theta)}{cos(\theta)}\cdot \cfrac{1}{1cos(\theta)}\implies \cfrac{sin^2(\theta)}{cos(\theta)[1cos(\theta)]} \\ \quad \\ recall\implies sin^2(\theta)+cos^2(\theta)=1\implies sin^2(\theta)={\color{brown}{ 1cos^2(\theta) }}\qquad thus \\ \quad \\ \cfrac{sin^2(\theta)}{cos(\theta)[1cos(\theta)]}\implies \cfrac{{\color{brown}{ 1cos^2(\theta) }}}{cos(\theta)[1cos(\theta)]} \\ \quad \\ \textit{now keep in mind that }1^2, 1^3, 1^4 = 1\qquad thus \\ \quad \\ \cfrac{{\color{brown}{ 1cos^2(\theta) }}}{cos(\theta)[1cos(\theta)]}\implies \cfrac{{\color{brown}{ 1^2cos^2(\theta) }}}{cos(\theta)[1cos(\theta)]} \\ \quad \\ \textit{now using difference of square }a^2b^2 = (ab)(a+b) \\ \quad \\ \cfrac{{\color{brown}{ \cancel{[1cos(\theta)]}[1+cos(\theta)] }}}{cos(\theta)\cancel{[1cos(\theta)]}}\implies \cfrac{1+cos(\theta)}{cos(\theta)} \\ \quad \\ \cfrac{1}{cos(\theta)}+\cfrac{cos(\theta)}{cos(\theta)}\implies ?\)

Mertsj
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{\tan \theta \sin \theta}{1\cos \theta}=\frac{\frac{\sin \theta}{\cos \theta}\times \sin \theta}{1\cos \theta}\times\frac{\cos \theta}{\cos \theta}=\frac{\sin ^2\theta}{\cos \theta(1\cos \theta)=}\] \[\frac{1\cos ^2\theta}{\cos \theta(1\cos \theta)}=\frac{(1\cos \theta)(1+\cos \theta)}{\cos \theta(1\cos \theta)}=\frac{1+\cos \theta}{\cos \theta}\]

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Ha!! 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 (1cos(t)) we get \(\dfrac{(1+cos(t))(1cos(t))}{cos(t)(1cos(t))}=\dfrac{1cos^2(t)}{cos(t)(1cos(t))}\) and \(1cos^2(t) = sin^2(t) = sin(t)*sin(t)\) hence, it becomes \(\dfrac{sin(t)}{cos(t)}*\dfrac{sin(t)}{1cos(t)}= \dfrac{tan(t)*sin(t)}{1cos(t)}\) this is the left hand side.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wow guys.....thanks alot!!
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