Stephanie M. answered • 06/10/15

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(a)

I'm just going to switch to a and b for this.

Solve the first expression for b:

a + b = (2/3)π

b = (2π)/3 - a

Substitute that into the second expression:

sin(a) = 2sin(b)

sin(a) = 2sin(2π/3 - a)

Apply the angle difference identity:

sin(a) = 2sin(2π/3 - a)

sin(a) = 2(sin(2π/3)cos(a) - sin(a)cos(2π/3))

sin(a) = 2(√(3)/2)cos(a) - 2(-1/2)sin(a)

sin(a) = √(3)cos(a) + sin(a)

0 = √(3)cos(a)

0 = cos(a)

cos

^{-1}(0) = aCosine is 0 at π/2, so a = π/2.

Substitute that into the first expression:

a + b = 2π/3

π/2 + b = 2π/3

b = 2π/3 - π/2

b = 4π/6 - 3π/6

b = π/6

Finally, plug into the third expression:

tan(a-b) = tan(π/2 - π/6)

tan(a-b) = tan(3π/6 - π/6)

tan(a-b) = tan(2π/6)

tan(a-b) = tan(π/3)

tan(a-b) = sin(π/3) / cos(π/3)

tan(a-b) = (√(3)/2) / (1/2)

tan(a-b) = √(3)

To find a function's extrema, take its derivative:

f(x) = 4x

^{3}- 18x^{2}+ 15x - 20f'(x) = 12x

^{2}- 36x + 15Find the zeros of the derivative:

0 = 12x

^{2}- 36x + 150 = 4x

^{2}- 12x + 50 = 4x

^{2}- 2x - 10x + 50 = 2x(2x - 1) - 5(2x - 1)

0 = (2x - 5)(2x - 1)

0 = 2x - 5 OR 0 = 2x - 1

5 = 2x OR 1 = 2x

5/2 = x OR 1/2 = x

So, x = 5/2 or x = 1/2.

(b)

To figure out which one of those points is the maximum, divide the derivative of the function into pieces and test to see whether the derivative is positive or negative for each piece:

<---[0]----(1/2)---[1]----(5/2)---[3]---->

f'(0) = 12(0)

^{2}- 36(0) + 15f'(0) = 12(0) - 0 + 15

f'(0) = 0 - 0 + 15

f'(0) = 15 (POSITIVE)

f'(1) = 12(1)

^{2}- 36(1) + 15f'(1) = 12(1) - 36 + 15

f'(1) = 12 - 36 + 15

f'(1) = -24 + 15

f'(1) = -9 (NEGATIVE)

f'(3) = 12(3)

^{2}- 36(3) + 15f'(3) = 12(9) - 108 + 15

f'(3) = 108 - 108 + 15

f'(3) = 0 + 15

f'(3) = 15 (POSITIVE)

<---[POS]----(1/2)---[NEG]----(5/2)---[POS]---->

This means the function increases until x = 1/2, then decreases until x = 5/2, then increases again afterwards. So, x = 1/2 is a local maximum.