The functions f(theta) and g(theta) are sine functions, where f(0) = g(0) = 0. The amplitude of f(theta) is twice the amplitude of g(theta). The period of f(theta) is one-half the period of g(theta). If g(theta) has a period of 2pi, and f(pi/4) = 4, write the function rule for g(theta).
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Can you help?
that hurts my brain ..
Yeah, tell me about it..
canyou organize the data in a more convenient manner?
what do we know pull it apart and use t not theta ... the less letters the better The functions f(t) and g(t) are sine functions, where f(0) = g(0) = 0. The amplitude of f(theta) is twice the amplitude of g(theta). f(t) = 2 g(t) The period of f(t) is one-half the period of g(t). f is twice as fast as g f(2t) = 2 g(t) If g(t) has a period of 2pi, then g is a normal sin function and f(pi/4) = 4, write the function rule for g(theta). 2t = pi/4 when t = pi/8
okay
so when 2 g(pi/8) = f(pi/4) = 4
how'd you get that, and why?
i wrote it all out .... what part do you get lost on?
this part: "The period of f(t) is one-half the period of g(t). f is twice as fast as g f(2t) = 2 g(t) If g(t) has a period of 2pi, then g is a normal sin function and f(pi/4) = 4, write the function rule for g(theta). 2t = pi/4 when t = pi/8"
if we cover the same distance in half the time, are we going faster or slower?
faster
how'd you get \(f(2\theta)= ~2~ g(\theta)\)?
f completes a cycle in say a time of 1, g completes a cycle in a time of 2 f is 2 times as fast as g. f(2t) for a given t value is relative to g(t) sin(2t) is twice as fast (takes half the time) or rather it completes 2 cycles when sin(t) completes 1 cycle
do we agree that the arguements are sin(2t) and sin(t) ?
sin(pi/2) = 1 sin(2(pi/4)) = 1 pi/4 is half of pi/2
okay, I get that..
|dw:1431559321100:dw| half the period is twice as fast.
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