finding ratio of area of two rectangles in a graphed trigonometic function.

[lethal raptor]

question 80, im not looking for an answer since this was a question that my math teacher gave me as a challenge, i just want to know if my steps are right and why if they aren’t

first i got the area of rectangle xyzl by constructing a perpendicular line(say the point of intersection with YZ is F and the point at the range is M) from the minimum value (-1) on YZ, then i got the length of that perpendicular which is in the graph (0.5) , i was planning to solve this by drawing a triangle then using pythagoras theorem, anyways i got the diagonal distance between Y and M using distance formula which was 2.281, then used pythagoras theorem to get YF then multiplied YF by 2 to get YZ, then got the area by multiplying YZ and XY.

i was gonna solve for the other triangle with the same way.

[void bison]

You don't need F or M.

[lethal raptor]

i just solved it by constructing so i said F and M

but i think i get what u mean by i dont need them since maybe constructing was unnecessary

[void bison]

How did you find the distance YM? did you need coordinates of Y?

[lethal raptor]

i got the coordinate of Y which is (pi, -1/2) and M which is (3/4pi , -1) then got the distance using the distance formula

[void bison]

hmmm

if you plug pi into the function, you get 0.

so the x-value of Y has to be bigger than pi

[lethal raptor]

can you elaborate? i don’t think i understand this part

[void bison]

the point Y is on the curve y=sin(x)

but sin(pi) = 0

you need to find sin(x) = -1/2

[void bison]

and the solution to this is larger than pi

[lethal raptor]

i see but how will this help to solving it ?

[void bison]

solve this: sin(x) = -1/2

and then you can find the length XL

[lethal raptor]

x is -30 but how should it help me in finding XL?

[short mica]

@lethal raptor I was just reading your work: the x-coordinate of point M and Y is incorrect.
So your final answer for area of XYZL is incorrect (just a calculation error).
Even though your reasoning is a "harder" way to solve this question, good job for your work!

Let's say that there are much easier ways to solve this, but your way does indeed work.

[void bison]

x = -30 is one possible solution to that equation, but it's not the right one here

[lethal raptor]

oh you mean i get the reference angle?

[void bison]

notice that Y does not have a negative x-value

[lethal raptor]

yeah i usually have a tendency to overcomplicate things by unnecessarily constructing, but why is the coordinate for M and Y incorrect?

[void bison]

yes, -30 is the reference angle
(though your other work is in radians. Both are fine here, but you should pick one.)

We already went over why your coordinate is wrong for Y

oh, the x-coordinate of M is halfway between pi and 2pi, but was accidentally written as 3pi/4

[lethal raptor]

but isnt it supposed to be 3/4pi ?

[short mica]

Well they're incorrect. For example sin(3pi/4) is not equal to 1. So the x-coordinate of M is incorrect. Same for point Y.

[lethal raptor]

i got the average value which is 1+1/2 / 2

[short mica]

Just try to find the correct coordinates for X and Y and other than that using your method everything is right.

[lethal raptor]

ah okay then thank youu

[void bison]

(You don't need this for this question, but halfways between pi and 2pi is (pi+2pi)/2 = 3pi/2. )

[lethal raptor]

ohh i just got the halfway point between pi and 2pi so its the x coordinate of point M but i think its wrong

[void bison]

3pi/2 is the correct x-coordinate for M. Though you don't need that for this question.

[lethal raptor]

yeah because there are more simple ways to solve this but i needed the coordinate for M since i overcomplicated it by constructing