use the graph of y=sin2θ to find the value of sin 2θ for θ= pi/4 radians A. -1 B. 0 C. 0.5 D. 1
for theta = pi/4 substitute pi/4 for tehta \[y=\sin(2\cdot \frac{\pi}{4})\] simplify the fraction
and then graph the equation https://www.desmos.com/calculator what is y when x=pi/4
1?
how would you find the period ?
I honestly don't know. I am so confused by this lesson
looks like you're just guessing well here is the general equation \[\large\rm y=A \sin(Bx-C)+D\] \[\left| A \right|\] is amplitude \[P=\frac{2\pi}{B}\] 2pi because the wave repeats every 2 units (As you can see on the unit circle) and C= horizontal shift D= vertical shift
P=period use that equation to find the value of `B` when p=4pi
\[4\pi=\frac{2\pi}{B}\] solve for B (B=the number at front of x variable )
I can't figure out what to put for B. I don't see a number before the x
so would it be B? since there is a 4 in it?
use the equation i gave to find B
we know that Period = 2pi/b so since the given is period which is 4pi we can solve for b
\[ \color{Red}{P}=\frac{2\pi}{B}\] \[ \color{Red}{4\pi}=\frac{2\pi}{B}\] solve for B
y=2*sin(2*π-4)+π y=π-2*sin(4)
no solve this equation for B \[ \color{Red}{P}=\frac{2\pi}{B}\] \[ \color{Red}{4\pi}=\frac{2\pi}{B}\] what is b = ?
given is Amplitude (which is 2) and the period (which is 4pi) we don't have to deal with C and D
((4*π)/(2*π)) 2 ((2*π)/(4*π)) 1/2
that's correct B=1/2
btw in this example x is same thing as theta
so if B =1/2 how would you write the equation with A=2 and B =1/2
y=0.5sin4θ
well amplitude is 2 for sure right ?? bec that is give so just replace A with 2 and we just found out that b =1/2 so replace b with 1/2
\[\huge\rm y=Asin b \theta\] without vertical and horizontal shift
y=2sinθ/2?
yes
thank you so much
np
Join our real-time social learning platform and learn together with your friends!