Re: Finding Period and Amplitude for y = 1/2 cos(2x/3)
hello, this is my first question, for i just signed up for this help & it is greatly appreciated =) my alg2/trig teacher never explained this to us but decided to give us problems anyway. so heres my question..warning i dont kno this at all!
Sketch and find (theta) and (theta one)
a) sin (theta)R = -.85;cos <0
b) tan(theta)degrees = -8.3;sin >0
Hello, Moe Moe,
You need to start a “New Topic” rather than append your question to someone else’s problem. I recommend you also see the “Read before posting” info here:
viewtopic.php?f=10&t=12146
You might find that helpful.
To answer your first question, “sin (theta)R = -.85;cos <0”, notice that the sine of some angle is going to equal a negative number (-.85). Notice also that we are told that “cos<0”. This means that both sine and cosine are negative. This only occurs in the third quadrant. The reason we need to figure out which quadrant we’re talking about is that “sin (theta)R = -.85” has more than one solution, and we need to narrow it down to the correct one.
I suggest you draw a triangle in the third quadrant to get a visual representation of what you are solving.
To solve for theta, set the mode in your calculator to radians.
Next we take the inverse sine (or arcsine) of both sides of the equation and get
Theta = arcsin(-.85) = -1.016 radians (approx.)
This isn’t much help to us though because -1.016 radians is in the fourth quadrant. However, the triangle we would make with this –1.016 radian angle would be the mirror image of the triangle we want in the third quadrant. Therefore, the angle we want is pi + 1.016. The “pi” part covers the first two quadrants, and the “+1.016” part extends the angle into the third quadrant.
Pi + 1.016 = 4.158 radians (approx.)
Hope that helps.
