Prove this inequality

achacy · #1 $Old$ October 14th, 2008, 07:45 AM

$- \sqrt{a ^{2}+b ^{2} } \leqslant a \cdot cos \alpha + b \cdot sin \alpha \leqslant \sqrt{a ^{2}+b ^{2} }$

My solution:

$\begin{cases} - \sqrt{ a^{2}+b^{2} } \leqslant a \cdot cos \alpha +b \cdot sin \alpha \\a \cdot cos \alpha +b \cdot sin \alpha \leqslant \sqrt{ a^{2}+b^{2} }\end{cases}$

After adding up both equalities we have:

$a \cdot cos \alpha +b \cdot sin \alpha-\sqrt{ a^{2}+b^{2} } \leqslant \sqrt{ a^{2}+b^{2} } + a \cdot cos \alpha +b \cdot sin \alpha$

$2\sqrt{ a^{2}+b^{2} } \geqslant 0$

My question is as follows:
Is this solution correct, or is it after adding up the two equalities still obvious to write two?
I hope you understand what I mean :]

ADARSH · #2 $Old$ October 16th, 2008, 07:08 AM

u can try this
LHS=[acos(x)+bsin(x)]
dividing andd multiplying with

take [a^2+b^2]^(1/2)=k
[acos(x)+bsin(x)]*k/k..(1)

when we take
a/[a^2+b^2]^(1/2)=sin(y)
we get
b/[a^2+b^2]^(1/2)=cos(y)
equation (1) than becomes
k*(sin(x+y))
here
-1=< sin(x+y) <=1
so
-k<=LHS<=k
thus proved