Hi there,
Can someone help me with the question:
"Prove that the value of 7sinx +3cos2x cannot be greater than 5/1/24 for all values of x between 0 degrees and 360 degrees"
Thanks lots. =)
I've already simplified 7sinx + 3cos2x into (2sinx-3)(3sinx+1).
I do not agree with your factoring
7 sin x +3 (1 - 2sin^2x)
7 sin x + 3 - 6 sin^2 x
-[6 sin^2 x - 7sin x -3 )
to find the maximum or minimum take the derivative and set to 0
-[ 12 sin x cos x - 7 cos x ] = 0
sin x = 7/12
then
sin^2 x = 49/144
7 sin x = 49/12
2 sin^2 x = 98/144
so function = 49/12 +3 (1 - 98/144)
= 588/144 +432/144 -98*3/144
= 726/144 = 5.041666...
1/24 = .04166666..
so
5 1/24 is right
Oh, I do agree with the factoring but did not do it.
Here without using calculus
let sin x = z
-y = 6z^2-7z -3
that is a parabola
complete the square to find the vertex
-y/6 = z^2 -7/6 z -1/2
-y/6 +1/2 = z^2 -7/6 z
add (7/12)^2 or 49/144 to both sides
-y/6 +72/144 +49/144 = z^2 -7/6 z + 49/144
-y/6 + 121/144 = (z-7/12)^2
so
vertex is where z = sin x = 7/12
then continue as I did with the calculus way
Thanks for the help =) Clear and detailed explanation.
Certainly! To prove that the value of 7sinx + 3cos2x cannot be greater than 5/24 for all values of x between 0 degrees and 360 degrees, we can take the following steps:
1. Recall that the maximum value of sinx is 1 and the maximum value of cos2x is 1 as well. Therefore, the maximum possible value of 7sinx + 3cos2x is 7(1) + 3(1) = 10.
2. Now, we need to find the minimum value of the expression (2sinx - 3)(3sinx + 1) to determine if it can be greater than 5/24.
3. To find the minimum value, we analyze the critical points where the expression is either zero or undefined. In this case, the critical points occur when (2sinx - 3) = 0 or (3sinx + 1) = 0.
a) When (2sinx - 3) = 0, we solve for x:
2sinx - 3 = 0
2sinx = 3
sinx = 3/2 (which is not possible)
Since sinx cannot be greater than 1, there are no critical points here.
b) When (3sinx + 1) = 0, we solve for x:
3sinx + 1 = 0
3sinx = -1
sinx = -1/3
Since sinx cannot be less than -1 or greater than 1, the only critical point is x = arcsin(-1/3) ≈ -19.47 degrees.
4. Now, since we are considering values of x between 0 degrees and 360 degrees, the expression (2sinx - 3)(3sinx + 1) is continuous within this interval except at x = -19.47 degrees.
5. To prove that the value of 7sinx + 3cos2x cannot be greater than 5/24 for all values of x between 0 degrees and 360 degrees, we need to evaluate the expression at both endpoints of the interval: x = 0 degrees and x = 360 degrees.
a) When x = 0 degrees:
(2sin(0) - 3)(3sin(0) + 1)
(0 - 3)(0 + 1)
(-3)(1)
-3
b) When x = 360 degrees:
(2sin(360) - 3)(3sin(360) + 1)
(0 - 3)(0 + 1)
(-3)(1)
-3
6. As both endpoints yield the result -3, which is less than 5/24, we can conclude that the value of 7sinx + 3cos2x cannot be greater than 5/24 for all values of x between 0 degrees and 360 degrees.
Therefore, the proof is complete.