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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

getusel
 one year ago
Best ResponseYou've already chosen the best response.1In 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.
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