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Proving easy one,
Area of sector POB is equal to twich the area of the shaded region, show that
\[3\theta=2(pisin( pi\theta))\]
 2 years ago
 2 years ago
Proving easy one, Area of sector POB is equal to twich the area of the shaded region, show that \[3\theta=2(pisin( pi\theta))\]
 2 years ago
 2 years ago

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.Sam.Best ResponseYou've already chosen the best response.1
Actually you can simplify a little bit by \[A(shaded)=\frac{1}{2}r^2(pi\theta)\frac{1}{2}(r^2)(\sin(pi\theta))\] \[A(shaded)=\frac{1}{2}(r^2)(pi\theta\sin(pi\theta))\]
 2 years ago

dpaIncBest ResponseYou've already chosen the best response.1
magic happens in between...
 2 years ago

.Sam.Best ResponseYou've already chosen the best response.1
POB=2AP \[\frac{1}{2}r^2 \theta=2(\frac{1}{2}r^2(pi\theta\sin(pi\theta))\] \[\theta=2(pi\theta\sin(pi\theta))\] \[\theta=2pi2\theta2\sin(pi\theta))\] \[3\theta=2(pi\sin(pi\theta))\]
 2 years ago

dpaIncBest ResponseYou've already chosen the best response.1
yea, i missed that negative sin up there.
 2 years ago

dpaIncBest ResponseYou've already chosen the best response.1
nice problem @.Sam. .... thankx. :)
 2 years ago

.Sam.Best ResponseYou've already chosen the best response.1
I'll post another one but a bit mind boggling :D
 2 years ago
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