How do you find the amplitude and period of $y = \frac{1}{2} sec (3 θ)$ $y=1/2sec(3theta)$ ?

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Answer 1 · 2017-11-03T19:54:34

AMPLITUDE: $\frac{1}{2}$ $1/2$
PERIOD: $\frac{2 π}{3}$ $(2pi)/3$

Standard Form: $y = a sec (b (x - c)) + d$ $y=asec(b(x-c))+d$
$a =$ $a=$ amplitude

So for $y=1/2sec(3θ) -> a=1/2$
Therefore, the amplitude is $1/2$ .

Period of Secant: $(2pi)/|b|$
So for $y=1/2sec(3θ) -> b=3$
$(2pi)/|b|=(2pi)/|3|=(2pi)/3$
Therefore, the period is $(2pi)/3$ .

How do you find the amplitude and period of y=1/2sec(3theta)y=12sec(3θ)y=1/2sec(3theta)?