Question

What is the absolute maximum value taken by the function f(x) = 5sin(x)-2cos(x) on [0, pi/2]?

Answer 1

are u stuck at 5cos x +2 sin x =0 ?

Answer 2

yeah... i'm trying to find the critical values again... but i can't seem to find any... does that mean there arn't any?

Answer 3

i think there are...
divide the entire equation by
\(\sqrt{5^2+2^2}\)
then put 5/sqrt[...] as sin a
and 2/ sqrt[..] as cos a

Answer 4

and u get the form
sin (x+a) =0

Answer 5

getting this ?

Answer 6

not really... is it possible you could work it out so i can see how it works?

Answer 7

yeah..will take some time...

Answer 8

\(5cos x +2 sin x =0\)
\(\huge \frac{5cosx}{\sqrt{5^2+2^2}}+\frac{2sin x}{\sqrt{5^2+2^2}}=0\)
put sin a = 5/sqrt[29], then 2/sqrt[29] will be cos a
\(\sin a\cos x+\cos a\sin x=0 \\ sin[x+a]=0\)
ok till here ?

Answer 9

where a is \(\tan^{-1}(5/2)\) or \(\sin^{-1}(5/\sqrt{29})\)

Answer 10

I understand the first part, but where did the a's come from?

Answer 11

a is just a constant.
i denoted 5/sqrt 29 as sin a
to get the form sin a cos b +cos a sin b
to write that as sin(a+b)

Answer 12

ok...i think i get it so far...

Answer 13

sin (x+a) =0
so x+a =.... ?

Answer 14

sin(x+a) = 0
since, sin 0 = 0
u get
x+a = 0
x=-a
now since a is positive,
x is negative, which is not in the range of [0,pi/2]
so again to find absolute maxima, substitute end-points
put x=0 and x=pi/2
whichever is greater is your maxima.

Answer 15

Alright! I think i pretty much get what you're saying.... i just need to stare at it a little yet! Thanks so much again! :D

Answer 16

welcome ^_^