Precal Blog

Today in class, we began by going over the claims for sections 2-2 and 2-3.

Ali presented the first problem and it asked for the refence angle if theta=-2746 degrees.

What he did was divide -2746 by 360, and the answer ends up being -7.62777778.  THis number means that theta makes 7 full rotations in the negative direction and a partial rotation of .62777778.  Net he multiplied -.62777778 by 360, and that answer would give him the measure of theta so he could plot it on a uv axis without spiraling seven times around the origin.  The value for theta is -226 degrees, making it in the 3rd quardrant in the negative direction.  Then to find the reference of theta,  Ali completed the rotation to the nearest horizontal axis, keeping in mind that it must always go in the positive direction.  It would be drawn to 180 degrees, and 226-180= 46 degrees.

Jared presented the next problem, and it asked to find the reference angle for an angle measuring 352 degrees 16′ 44″.

 The first step was to draw the angle in the fourth quadrant, because it was between 270 and 360.  Then the reference angle can be drawn to the nearest horizontal axis.  But since the initial angle is in degrees minutes and seconds, you need to subtract it from 360 degrees 0′ 0″ properly.  The way to do it is to just simply borrow from the previous number.  The answer comes out to be 7 degrees 43′ 16″.

For this problem, Pat was asked to write g(x) when the pre-image is f(x).

 Visually you can see that it is dilated in both the x and y direction.  To find out the value for the vertical dilation, you take the y point of one of the humps of the pre-image and compare it to the y coordinate of one of the humps of the image.  In this case, the peak oif the pre-image hump is 3, and the peak of the image hump is 9, making it a dilation by 3.  g(x)=3f(x).  For the horizontal dilation, count the horizontal distance between the humps of the pre-image and and compare.  The pre-image is twice as small as the image, meaning that the image got two times as big.  g(x)=3f(1/2x).  It is 1/2x inside the parenthesis because it is always inverse in the x direction.

In this problem presented by Streim, he was asked to find the reference angle for 300 degrees and calculate the sine and cosine of the initial angle and reference angle.  It is the same as the first two problems that were presented in that you need to find the reference angle.  The reference angle ends up being 60 degrees.  We have the sin and cosine value memorized for 60 (sq. root(3)/2=sin60 and 1/2=cos60) so we know that since 60 is 300’s reference angle, their sines are opposite and their cosines are equal.

Justin’s problem gave him a point (-7,-24) and it told him to find the sine and cosine of its angle.  First he needs to draw the triangle, knowing the veritcal side is -7 and the horizontal side is -24.  Then use the Pythagorean theorem to find the hypotenuse, and use sine and cosine functions to solve for it.  Answer: sin=-7/25, cos=-24/25

For this one, Jocelyn needed to graph the given equation and tell its transformations.  4cos tells you it is dilated vertically by 4 and (theta+60) means it is translated 60 degrees to the left.  By the way the drawing was just superb for this one hahaha.

After claims, Mr. Bieniek talked abou the difference between inverse and reciprocal.  The notation is very slightly different but is critical.  For inverse, the notation is cos^-1(theta).  For reciprocal the notation is cos(theta)^-1.  He also talked about how theta and the inverse function (sine, cosine, tangent) of theta are the same. 

The next scribe is Stinz Leerix, who is droppin’ a new mixtape ‘Tha Letdown”.  Coming soon to stores near you