Question

wat it the maximum of \[a \cos x+b \sin x\]

Answer 1

without calculus

let \[c=\sqrt{a^2+b^2}\]
\[c( \frac{a}{c}\cos x+\frac{b}{c}\sin x)=c(\sin y\cos x+\cos y\sin x)=c\sin(x+y)\]
so max is c
but with calculus
\[f(x)=a \cos x+b \sin x\\f'(x)=0\\0=-a\sin x+b \cos x\\ \tan x=b/a \]
wich meanc cos x=a/c,sin x=b/c
so
\[f(x)=a(a/c)+b(b/c)=c\]

Answer 2

Did you just solve your own question? Nice. Lol

Answer 3

jus concerned if both methods are correct n also to help my problem solving sklls...

Answer 4

n if theres any

Answer 5

bsinx + acosx, it can be converted to k cos(x - θ)
with k = sqrt(b^2 + a^2), and we knowed that the maximum value of cos(x- θ) is 1, so the maximum value would be k * 1 = k = sqrt(b^2 + a^2) = sqrt(a^2 + b^2)

Answer 6

lol @abb0t thanx @RadEn

Answer 7

you're welcome