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 :]
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