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

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1334897255916:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1334897397459:dw

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0magic happens in between...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1334897856911:dw

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yea, i missed that negative sin up there.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0nice problem @.Sam. .... thankx. :)

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