Sinusoidal Functions (I )

[mathuravasant]

Find the first two positive x-intercepts for y= -2cos(3(x-25°)) +1



(Can someone help me for this)

 

[topsquark]

Find the first two positive x-intercepts for y= -2cos(3(x-25°)) +1



(Can someone help me for this)

Well, some basic algebra first:
\(\displaystyle 0 = -2 ~ cos(3 (x - 25) ) + 1\)

\(\displaystyle -1 = -2 ~ cos(3 (x - 25) )\)

\(\displaystyle \dfrac{1}{2} = cos(3 (x - 25) )\)

Where are the first two places you find cos(y) = 1/2?

Can you finish?

-Dan

 

[mathuravasant]

im not sure if I did it right but would the answer be 1452.4 degree +k*120 degree , 1532.4 degree +k*120 degree ??

 

[topsquark]

No. Even if the numbers were right where are you getting the k's from?

Start here: Where are the first two places you find cos(y) = 1/2? If you need to use your calculator! Graph it! Do something. I have no idea how you got your answer so I can't address what you might have done wrong. Please tell us how you are doing your calculations.

The first two places you find cos(y) = 1/2 are y = 30 and y = 300 degrees. So solve 3(x - 25) = 30 and 3(x - 25) = 300.

-Dan

 

[HOI]

Cosine is "near side over hypotenuse". So if cos(y)= 1/2 we can represent it as a right triangle with hypotenuse of length 2 and one leg of length 1. Two of those right triangles, placed together gives a triangle with two sides of length 2 and the third side of length 1+ 1= 2. That's an equilateral triangle! Its three angles, and in particular the one whose cosine is 1/2, are all 60 degrees.

Since cos(3(x-25))= 1/2, 3(x- 25)= 3x- 75= 60 so 3x= 60+ 75= 135, x= 45 degrees.
Since cos(180- x)= COS(x), cos(180- 60)= cos(120)= 1/2 so we can also have 3x- 75= 120, 3x= 195, x= 65 degrees.