How do you find the amplitude and period of a function #y = sin(pi/2Ө)#?

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Answer 1 · 2016-01-25T07:43:45

Amplitude#=1#

Period #=4#

from the given #y=sin (pi/2* theta)#

it should be taken into account that the above is the same as

#y=1*sin (pi/2* theta)#

Take the absolute value of any number you see as numerical coefficient of sine or cosine expression and you get the Amplitude.

Amplitude #=abs1=1#

The Period can be obtain using #P=(2pi)/a# where #a# is the number beside #theta#

so that #P=(2 pi)/abs(pi/2)=4#

Learn from the following examples,

#y=-7*sin 3theta#
Amplitude#=abs-7=7# and Period#=(2pi)/abs(3)=(2pi)/3#

#y=-3/4*cos ((pitheta)/6)#

Amplitude#=abs(-3/4)=3/4#and Period#=(2pi)/abs(pi/6)=12#

#y=12*sin (theta)#

Amplitude#=abs(12)=12#and Period#=(2pi)/abs(1)=2pi#

I hope the additional examples help..

Have a nice day from the Philippines ....