Question

Show (sin x+cos x)^2<2

Answer 1

Check the maxima of the function.

Answer 2

actually, (cos^2(x) + sin^2(x))^2 = 2 when x = pi/4

perhaps you meant (cos^2(x) + sin^2(x)) <= 2

expanding gives sin^2x + cos^2x + 2cosx sinx

using trig identities, this equals to
1 + sin(2x)

we know that
-1 <= sin(2x) <= 1

0 <= 1 + sin(2x) <= 2

Answer 3

I meant (cosx + sinx)^2 = 2 when x = pi/4

Answer 4

No, actually it's |sin x+cos x|< √ ̅ 2

Answer 5

well when x = pi/4, then the expression equals sqrt(2)

Answer 6

and what about |a*sin x+b*cos x|<√ ̅ (a^2+b^2)

Answer 7

what if a=b=0 ?

Answer 8

I think |a*sin x+b*cos x|<=√ ̅ (a^2+b^2)

Answer 9

but it doesn't matter because |sin x+cos x|< √ ̅ 2 is false anyway.

but |sin x+cos x|<= √ ̅ 2 is true

Answer 10

yes. it comes <= .I forgot to put it

Answer 11

actually if a=b=0 then a*sinx and b*cos x is simpliying so that's not the answer.. anyway

Answer 12

so you want to show that |sin x+cos x|<= √ ̅ 2 ?

Answer 13

that the first.. then i have to show |a*sin x+b*cos x|<=√ ̅ (a^2+b^2)

Answer 14

just hint :
a*sin x+b*cos x can be converted to k cos(x-θ), with k = sqrt(a^2 + b^2)

Answer 15

I give up :|

Answer 16

the 1st and the 2nd problem are exactly the samethink.
see how to change sin x+cos x into kcos(x-θ)
here a = 1 and b = 1, so k = sqrt(1^1+1^2) = sqrt(2)
and tanθ = a/b = 1/1 = 1 (here θ in the 1st quadrant)
get θ = 45 degrees

therefore,
|sin x+cos x|<= √ ̅ 2
|sqrt(2) cos(x-θ)} <= √ ̅ 2

and we knowed that -1 <= cos(x-θ) <= 1
and absolut sign makes everthink be positive and zero. so, obvious the LHS would be less than or equal the RHS