Suppose that 0<c<pi/2. For what value of c is the area of the region enclosed by the curves y = cos(x), y = cos(x-c), and x = 0 equal to the area of the region enclosed by the curves y =...

Suppose that 0<c<pi/2.

For what value of c is the area of the region enclosed by the curves y = cos(x), y = cos(x-c), and x = 0 equal to the area of the region enclosed by the curves y = cos(x-c), x = pi and y = 0?

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to evaluate the area of the region enclosed by the curves y = cos x, y = cos(x - c) and x = 0, such that:

`A_1 = int_0^(c/2) (cos x - cos (x - c)) dx`

`A_1 = int_0^(c/2) (cos x) dx - int_0^(c/2) (cos (x - c)) dx`

`A_1 = sin x|_0^(c/2) - sin (x - c)|_0^(c/2)`

`A_1 = sin (c/2) - sin 0 - sin (c/2 - c) + sin (0 -c)`

`A_1 = sin (c/2) -sin(-c/2) + sin(-c)`

Since sin(-x) = -sin x:

`A = sin (c/2) + sin (c/2) - sin c`

`A = 2 sin (c/2) - sin c`

You need to evaluate the area of the region enclosed by the curves y = 0, y = cos(x - c) and `x = pi` , such that:

`A_2 = int_((pi)/2 + c)^(pi) (0 - cos (x - c)) dx`

`A_2 = -int_((pi)/2 + c)^(pi) cos (x - c) dx`

`A_2 = - sin (x - c)|_((pi)/2 + c)^(pi)`

`A_2 = - sin (pi - c) + sin ((pi)/2 + c - c)`

`A_2 = - sin pi*cos c + sin c*cos pi + sin((pi)/2)`

`A_2 = 0 + sin c*(-1) + 1`

`A_2 = 1 - sin c`

You need to set the two areas equal, such that:

`A_1 = A_2`

`2 sin (c/2) - sin c = 1 - sin c`

You need to add sin c to both sides, such that:

`2 sin (c/2) = 1`

`sin (c/2) = 1/2`

`arcsin(sin (c/2)) = arcsin(1/2)`

Since the sine function has positive values in quadrant 1 and 2, you need to consider the following results, such that:

`c/2 = pi/6` in quadrant 1

`c/2 = pi - pi/6` in quadrant 2

Hence, `c = pi/3` in quadrant 1 and `c = (5pi)/3` in quadrant 2.

Sources:

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