Thread: Transformations with Trig

Hi,

In the attached file, I do not know how to answer (a)(ii).

I would be grateful for any help,

Thanks.

f(t)=acosb(t-c)+d
maximum value is at t=3 but for maximum value cosb(t-c) should be maximum that is 1 therefor
29=a+d............................................ ...1
minimum value at t=9 but for minimum value cosb(t-c) should be minimum that is -1 therefor
15=-a+d............................................... 2
solving equation 1 and 2 we get
a=7
b=22
now
f(t)=7cosb(t-c)+22
since max value was 29 at t=3 therefor
29=7cosb(3-c)+22 or
7=7cosb(3-c) or
cosb(3-c)=1 or
b(3-c)=0.................................3 (cosx=1 then x=0)
similarly we have
15=7cosb(9-c)+22 or
cosb(9-c)=-1 or
b(9-c)=pi..................................4 (cosx=-1 then x=pi)
solving equation 3 and 4 simultaneously we get
b=pi/6 hence proved