My solution:

After adding up both equalities we have:

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 :]

Printable View

- October 14th 2008, 05:45 AMachacyProve this inequality

My solution:

After adding up both equalities we have:

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 :] - October 16th 2008, 05:08 AMADARSH...
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