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Monday, November 13, 2006

TEST HELP CENTER
OPEN 6 PM - 9 PM
Trig Lessons 1.1 - 1.4

Don't forget your 1.4 homework check is tomorrow before the test.

posted by Mrs. S. at 4:13 PM

8 Comments:

  • Can someone tell me why the period of the sine function is 2pi?

    By Blogger Mrs. S., at 4:32 PM  

  • is it because you go around the unit circle and the sine equal 0

    By Anonymous Anonymous, at 5:56 PM  

  • This comment has been removed by a blog administrator.

    By Anonymous Anonymous, at 5:58 PM  

  • can you see if i done this right?
    its number 31. sec 3pi/2
    cos=0 sec=-1
    sin=-1 csc=-1
    tan=-1/0= undefined cot=0/1

    By Anonymous Anonymous, at 5:59 PM  

  • Josh
    Regarding problem 31. It asks for the secant of the angle 3pi/2.

    answer: At 3pi/2, the ordered pair is (0,-1). Secant is the reciprocal of cosine. Since cos (3pi/2) = 0, sec(3pi/2) is 1/0 or undefined.

    Hope this helps.

    By Blogger Mrs. S., at 6:17 PM  

  • Josh...
    The period of the sine function is 2pi because when you complete one revolution around the circle, the ordered pairs began to repeat themselves. One revolution equals 2pi radians i.e. the period.

    Mrs. S

    By Blogger Mrs. S., at 6:19 PM  

  • Step 1: Think about the quadrants in which sin theta is positive: I and II. So.. sin theta must be negative in quadrants III and IV.

    Step 2: Think about the quadrants in which cos theta is positive: I and IV. So...cos theta must be negative in quadrants II and III.

    Conclusion: Since sin theta and cos theta are negative in Quad. III, theta lies in Quad. III.

    If this doesn't help, let me know before nine:)
    Mrs. S

    By Blogger Mrs. S., at 6:32 PM  

  • Monday December 4, 2006

    Solving Oblique Triangles

    -An oblique Triangle is a triangle with NO right angles.

    -To solve an oblique triangle you have to know the measure of atleast one side. you can have any other two parts of the triangle as long as you have one side.

    -There are four cases of oblique triangles:
    Angle,Angle,Side(AAS);Angle,Side,Angle(ASA);Side,Side,Angle(ambiguous case); Side,Side,Side(SSS);Side,Angle,Side(SAS)

    We only learned about AAS and ASA today. These are the only ones you can solve using The Law of Sines also.

    The Law of Sines:
    a/SinA = b/SinB = c/SinC

    If you are given a Triangle ABC where A=24 degrees and a=14 and C=62 degrees Find the remaining sides and angles.

    1st you have to find the other angle, so you will add the two angles you have been given( 24 + 62 = 86) and subtract that answer from 180 degrees(180 - 86 = 94).This number is your other angle. A=24 B=94 C=62

    2nd you pick one of the sides you are solving for (b or c). using the law of sines you plug in your numbers

    you were not given side c, so take c/Sin62(C's angle)and set it equal to something you have both the angle and side for. You have angle A and side a. So your equation will be c/Sin62 = 14/Sin24

    Multiply both sides by Sin62, it cancles on the left side.So you have c = 14(Sin62)/Sin24

    Now type it in your calculater as (14Sin(62))/Sin(24)

    you should get 30 (rounded to the nearest whole number)

    You repeat this same process to get the other side (b).

    By Anonymous Anonymous, at 3:34 PM  

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