Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (anonymous):
OpenStudy (anonymous):
@shaydoe Can you help?
OpenStudy (anonymous):
yep
OpenStudy (anonymous):
im not sure exactly sorry
OpenStudy (anonymous):
@zepdrix Can you help me?
Still Need Help?
Join the QuestionCove community and study together with friends!
Sign Up
myininaya (myininaya):
so we have f(x)=cos(x)
and we know f(-x)=cos(-x)=cos(x) since cos(x) is an even function
myininaya (myininaya):
so you know f(x)=f(-x) for all x
myininaya (myininaya):
that means you know f(-a) since you are given f(a)
myininaya (myininaya):
for example cos(pi/4)=sqrt(2)/2 so cos(-pi/4) also equal sqrt(2)/2
myininaya (myininaya):
so if cos(a)=2/7 then cos(-a)=?
Still Need Help?
Join the QuestionCove community and study together with friends!
Sign Up
myininaya (myininaya):
well if you come back and I'm gone
a hint for the second question is cos has period 2pi
OpenStudy (anonymous):
IM SO SORRY!!!!! @myininaya
OpenStudy (anonymous):
cos(-a)= 2/7 since it's an even function
myininaya (myininaya):
yep
OpenStudy (anonymous):
For the second one what's changing is the period? So (a+ 2pi) means having a period of 4pi?
Still Need Help?
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (anonymous):
I'm confused
myininaya (myininaya):
do you know that cos(pi/4)=sqrt(2)/2
and that cos(pi/4+2pi)=sqrt(2)/2
and that cos(pi/4+4pi)=sqrt(2)/2
and that cos(pi/4+2npi)=sqrt(2)/2
where n is an integer
myininaya (myininaya):
so if cos(a)=2/7
then cos(a+2pi)=?
OpenStudy (anonymous):
2/7. ohhhh so its like a reference angle
myininaya (myininaya):
yeah
and cos(a+4pi) still equals
Still Need Help?
Join the QuestionCove community and study together with friends!
Sign Up
myininaya (myininaya):
like it is like we are at a
and go all the way around to get back to a
so cos(a) is the same cos(a+2*pi*n) for an integer n
OpenStudy (anonymous):
2/7 + 2/7 + 2/7 = 6/7?
myininaya (myininaya):
yeah
OpenStudy (anonymous):
I understand what your saying. Thanks a lot :) youre the best