Question

Prove that the max and min values of asinX + bcosX are respectively sqrt.(a^2+b^2) and -sqrt(a^2+b^2) .

Answer 1

See

-1<= sinx <=1 ==> -a<=asinx<=a
-1<=cosx<=1 ==> -b<=bcosx <=b

Now add them up
-a-b<= asinx+bcosx <= a+b

Answer 2

use compound angle formulae

Answer 3

using derivatives: \[\Large a \cdot \cos(X) -b \cdot \sin(X)=0\]\[\huge \therefore \tan(X)=\frac ab\]

Answer 4

\[ asinX + bcosX = \sqrt{a^2 + b^2} \cos (X - \tan^-1(b/a))\]

Answer 5

put \( a = A\sin\phi \) and \( b = A\cos\phi \)

Answer 6

i personally like @experimentX 's as it doesn't assume calculus

Answer 7

but both are valid and nice :)

Answer 8

lol ... not sure if this is called compound formula!!

Answer 9

@eigenschmeigen ,@PaxPolaris ,@experimentX ,would you mind explaining why method won't work ??

Answer 10

*my method

Answer 11

so, at max or min:
\[y={a^2 \over \sqrt {a^2+b^2}} +{b^2 \over \sqrt {a^2+b^2}}\]

Answer 12

well, @shivam_bhalla you assumed that x are different for sin and cos
but they are same ... try for x=0 and x=pi/2

Answer 13

** that should be \(\pm \sqrt{a^2+b^2}\) for the denominators

Answer 14

Ok. getting it now @experimentX , Thanks a lot. I am still a noob in maths. :P

Answer 15

thanks everybody.. I got it...