Pure maths 1 problem

[Freaked12]

I am finding Question 5 part (i) October/November 2008 difficult to understand.

The function f is such that f(x) = a ? b cos x for 0? ? x ? 360?, where a and b are positive constants.
The maximum value of f(x) is 10 and the minimum value is ?2.
(i) Find the values of a and b.

[Freaked12]

a - bcos x for O<=  x <=360, where a and b are positive constants.
The max value of f(x) is 10 and the minimum value is -2

[nid404]

simultaneous equations

a+b=10
a-b=-2

2a=8
a=4

b=10-4=6

[Freaked12]

simultaneous equations

a+b=10
a-b=-2

2a=8
a=4

b=10-4=6

but why is it a+b and a-b
Where did u get the hint from

[nid404]

max value of cos x=1

since max value of f(x)=10

a+b=10 (cos x=1)

least value of cos x=-1

since min value  of f(x)=-2
a-b=-2

[Freaked12]

max value of cos x=1

since max value of f(x)=10

a+b=10 (cos x=1)

least value of cos x=-1

since min value  of f(x)=-2
a-b=-2

Jesus!
feeling embarrased for asking this question.

Thankyou anyways

[nid404]

lol...happens...

You're welcome