*I Feel Like I'm Running in Circles*
lol.
Hi guys! :D It's Aya.
So let's recall what we did in class today. First, we had our normal "everyday quiz" about previous lessons. Today's quiz was about Transformation of Trig Functions. *I personally didn't do well on mine -.- but oh well*. We did the checking and discussed a few points with our partners. Next, we learned about the Unit Circle.
So let's recall what we did in class today. First, we had our normal "everyday quiz" about previous lessons. Today's quiz was about Transformation of Trig Functions. *I personally didn't do well on mine -.- but oh well*. We did the checking and discussed a few points with our partners. Next, we learned about the Unit Circle.
*The Unit Circle
In our handouts, the () were blank and in order to solve that, we had to look for x and y: cosϴ being the x and sinϴ being the y.
REVIEW:
The special triangles that were written on the blackboard helped us look for the x and y.
The given triangles were:
Just remember:
COS = adjacent/hypotenuse
SIN = opposite/hypotenuse
Okay, let's move on. The next thing we did is answer another hand out about Reference Angles.
This is basically the continuation of what we did yesterday.
I'm not really sure how this goes, but this is what we did:
*FIND THE REFERENCE ANGLE*
Solution:
390° - 360° = 30°
-Disregard the negative sign (The reason why it's negative is because the rotation moved clockwise.)
-Notice that the terminal arm is at Quadrant IV, and the round line went one whole round on the cartesian plane. Therefore, we subtract the given to 360°. Try looking at the Unit Circle.
Solution:
525° - 360° = 165°
180° - 165° = 15°
-The given is not negative because the rotation moved counterclockwise.
-This time, line went round the plane in more than 360° , therefore we need to subtract this two times. First, with 360 "525° -360° =165° ", then because the terminal line is on Quadrant II, we will subtract the answer to 180, so "180° -165° =15° "
So yeah, that's pretty much all. ;)
The handout contains the answers so you could check there if you want.
Sorry for the bad quality (pictures). It was really hard to make them manually. Except for the unit circle cuz I just copied that. LOL. Okay, I'll shh now. :)
PS. Don't forget, the deadline for the Principia Mathematica Project Outline is on March 14. :D
Okay. :D Godbless.
In our handouts, the () were blank and in order to solve that, we had to look for x and y: cosϴ being the x and sinϴ being the y.REVIEW:
The special triangles that were written on the blackboard helped us look for the x and y.
The given triangles were:
Just remember:
COS = adjacent/hypotenuse
SIN = opposite/hypotenuse
Okay, let's move on. The next thing we did is answer another hand out about Reference Angles.
This is basically the continuation of what we did yesterday.
I'm not really sure how this goes, but this is what we did:
*FIND THE REFERENCE ANGLE*
1.
Solution:390° - 360° = 30°
-Disregard the negative sign (The reason why it's negative is because the rotation moved clockwise.)
-Notice that the terminal arm is at Quadrant IV, and the round line went one whole round on the cartesian plane. Therefore, we subtract the given to 360°. Try looking at the Unit Circle.
Solution:525° - 360° = 165°
180° - 165° = 15°
-The given is not negative because the rotation moved counterclockwise.
-This time, line went round the plane in more than 360° , therefore we need to subtract this two times. First, with 360 "525° -360° =165° ", then because the terminal line is on Quadrant II, we will subtract the answer to 180, so "180° -165° =15° "
So yeah, that's pretty much all. ;)
The handout contains the answers so you could check there if you want.
Sorry for the bad quality (pictures). It was really hard to make them manually. Except for the unit circle cuz I just copied that. LOL. Okay, I'll shh now. :)
PS. Don't forget, the deadline for the Principia Mathematica Project Outline is on March 14. :D
Okay. :D Godbless.
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