Tuesday, April 27, 2010

Scribe post, 4/27/10

We started with a warm up where we had to
1. Solve on [0,2π)
sin2x-cosx=0
2. Find the exact value of cos105˚
3. Use the figure to find tan2u
















We went over how to figure out if our answer is right when we finally get an answer.
For number 1
sin2x-cosx=0
2sinxcosx-cosx
cosx(2sinx-1)=0
so cosx = 0 and sinx = 1/2


for number 2
cos(105˚)
eq=cos(\frac{1}{2} *210)




eq=-\sqrt{\frac{1+cos210}{2}}




eq=-\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}}



eq=-\sqrt{\frac{\sqrt{2}-\sqrt{3}}{4}}






eq=\frac{2(2)}{1-2^2 }



eq=\frac{4}{-3}


we then went over the Law of Sines which is and
we started deriving the law of sines

then we proved that they are all equal for all triangles


eq=SinA=\frac{h}{c}
eq=sinC=\frac{h}{a}





eq=\frac{sinA}{a}=\frac{sinC}{c}



We also talked about the added bonus which is


eq=A= \frac{1}{2}acsinB


we tried this on a random triangle from GeoGebra where (using AAS therom)
A=59.85˚
C=26.05˚
c=4.78 units

so B =180˚-A-C
B=94.1
eq=\frac{a}{sin59.85}=\frac{4.78}{sin26.05˚}
eq=a=\frac{4.78sin59.85˚}{sin26.05˚}


so a = 9.41
eq=b= \frac{4.78}{sin26.05˚}=\frac{b}{sin94.1˚}

b=10.86
for the last 10 minutes of class we did our homework
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