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getuselBest ResponseYou've already chosen the best response.1
\[\frac{ cosx }{ sinx }\]
 one year ago

AtkinsohaBest ResponseYou've already chosen the best response.0
Can you show me how to figure that out?
 one year ago

getuselBest ResponseYou've already chosen the best response.1
because in the picture the question does not seem right
 one year ago

getuselBest ResponseYou've already chosen the best response.1
if this is the question
 one year ago

getuselBest ResponseYou've already chosen the best response.1
\[\sin(\frac{ \pi }{2 }x)=\sin(\frac{ \pi }{ 2 })cosx\cos(\frac{ \pi }{ 2 })sinx=1.cosx0.sinx=cosx\]
 one year ago

getuselBest ResponseYou've already chosen the best response.1
having this in mind, let us insert it in the original expression
 one year ago

getuselBest ResponseYou've already chosen the best response.1
\[\sin(\frac{ \pi }{2 }x)cscx=cosxcscx=cosx.\frac{ 1 }{ sinx }\]
 one year ago

AtkinsohaBest ResponseYou've already chosen the best response.0
Thank you! I have another one that maybe you could help me with?
 one year ago

AtkinsohaBest ResponseYou've already chosen the best response.0
okay, hold on let me get it..
 one year ago

AtkinsohaBest ResponseYou've already chosen the best response.0
If you could do it all in one message that would be great so I can follow along easier.
 one year ago

getuselBest ResponseYou've already chosen the best response.1
\[\frac{ cscAsinA }{ \cot ^{2}A }=\frac{\frac{ 1 }{sinA }sinA }{ \frac{ \cos ^{2}A }{ \sin ^{2}A } }=\frac{ \frac{ 1\sin ^{2}A }{sinA } }{ \frac{ \cos ^{2}A }{\sin ^{2} A} }\] \[\frac{ 1\sin ^{2}A }{sinA }.\frac{ \sin ^{2} A}{ \cos ^{2}A }=\frac{ \cos ^{2}A sinA}{ cosA }=sinAcosA=\frac{ 1 }{2 }\sin(2A)\]
 one year ago

getuselBest ResponseYou've already chosen the best response.1
Here is the final answer
 one year ago

AtkinsohaBest ResponseYou've already chosen the best response.0
So what's the final answer?
 one year ago

getuselBest ResponseYou've already chosen the best response.1
\[\frac{ 1 }{ 2 }\sin(2A)\]
 one year ago

AtkinsohaBest ResponseYou've already chosen the best response.0
uhmm.. yeah sort of. im terrible at this whole simplifying thing. up for a few more?
 one year ago

getuselBest ResponseYou've already chosen the best response.1
If you practice, it is not that hard my friend
 one year ago

AtkinsohaBest ResponseYou've already chosen the best response.0
i got all of the verifying ones but this one..
 one year ago

getuselBest ResponseYou've already chosen the best response.1
I am going for launch. If you stay for 30 minutes i will get back to you.
 one year ago

AtkinsohaBest ResponseYou've already chosen the best response.0
I'll be here all evening. Thank you! Have a good time :)
 one year ago

getuselBest ResponseYou've already chosen the best response.1
\[\frac{ \cos \alpha\cos \beta }{ \sin \alpha+\sin \beta }+\frac{ \sin \alpha\sin \beta }{ \cos \alpha+\cos \beta}=\] \[\frac{ \cos ^{2}\alpha\cos ^{2}\beta+\sin ^{2}\alpha\sin^2\beta }{(\sin \alpha+\sin \beta)(\cos \alpha +\cos \beta)}\] =\[\frac{ \cos^2 \alpha+\sin^2\alpha(\sin^2\beta +\cos^2\beta) }{(\sin \alpha+\sin \beta)(\cos \alpha+\cos \beta) }=\frac{ 11 }{(\sin \alpha+\sin \beta) (\cos \alpha+\cos \beta) }=0\]
 one year ago

AtkinsohaBest ResponseYou've already chosen the best response.0
how is the denominator verified? nothing was done to it?
 one year ago

getuselBest ResponseYou've already chosen the best response.1
since it does not have any significance to simplify the denominator, we can leave it as it is.
 one year ago

AtkinsohaBest ResponseYou've already chosen the best response.0
How do you know if it has significance to simplify it or not?
 one year ago

getuselBest ResponseYou've already chosen the best response.1
In this particular problem, the purpose is to simplify the right hand side to zero. That tells us the numerator of the RHS has to be siplified to zero. our focus should be mostly on working out the numerator.
 one year ago
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