Tuesday, October 30, 2007

Hello, everyone. Today we had two math classes, in the morning and in the afternoon. The morning class was spent on a pre-test and the afternoon there was a quiz...



The pre-test contain eight questions..


1) Simplify: sin(2x+π)

A. sin2x
B. cos2x
C. -sin2x
D. -cos2x


The answer is C because sin(2x+π) is similar to sin(α+β) and the equation for this is simple, just do the sine dance!

sin(α+β)=sinαcosβ+cosαsinβ .. now you just plug sin(2x+π) into the equation

sin(2x+π)=sin2xcosπ+cos2xsinπ
sin(2x+π)=sin2x(-1)+cos2x(0)
sin(2x+π)=-sin2x+0
sin(2x+π)=-sin2x

cos π= -1
sinπ= 0 anything multiplied by 0 equals 0 so the right side reduces..


2) The terminal arm of an angle Ө in standard position passes through point (m,n) where m>0, n>0. Determine the value of sin(π+Ө).

A. ( -n)/√(m^2+n^2)
B. ( -m)/√(m^2+n^2)
C. ( n)/√(m^2+n^2)
D. ( m)/√(m^2+n^2)

The answer is A.
The question says that m and n are both positive so you know the coordinates are in the first quadrant.
m=x coordinates
n=y coordinates
Ө= angle

Once you draw a diagram you can use the pythagorean theorem to find the hypotunse, which is √(m^2+n^2).
Since sin=Opposite/adjacent then..
n/√(m^2+n^2)

BUT... it asks for the value of sin(π+Ө) so you substitute it in the sin(α+β)=sinαcosβ+cosαsinβ when you finish you get -sinӨ. Soo.. you multiply everything by -1.

And your answer is
-n/√(m^2+n^2).

Question# 3.




In this slide mr.k drew a diagram of cos A and you realize that the sin of this triangle is the sin of sin B... this means that both A and B are the same angle. The question is asking for the value of cos (A-B) and since A and B are the same the answer will be zero!

cos(A-B) = cos 0
= 1 < the answer is one because 1 is the raduis always 1 in the unit circle.

Question #4.



In this slide instead of having 2cos^2x+cosx-1=0 we let cosx=A so it's easier to read...
after we factored it to find the values of A.

- cosx = 1/2 is not a solution because it is in quadrants 1 and 4. So we move onto the next value which is x=π. This value works because it's greater than π/2 and smaller than 3π/2.



Okay, this is the last question of our pre-test. I was a bit confused when 1+cos2x changed to 1+2cos^2x-1 but you need to remember the identities because cos2x=2cos^2x-1. You also need to remember that sec^2x-1=tan^2.

When you repeatedly do the questions it's easier to remember all the identities...


* sorry if i didn't use colour or anything extra, i was using an older computer, at the time, and i was limited on tools. (: if i didn't anything wrong please comment & i'll correct it. The next scribe for wednesday is Bryan..

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