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April 16, 2014

April 16, 2014

Posted by **Terry** on Wednesday, April 7, 2010 at 7:07pm.

1.)y= 3 sin x + 4 cos x, xE[0, 2pie]

- Calculus -
**Damon**, Wednesday, April 7, 2010 at 7:37pmdy/dx = 3 cos x - 4 sin x

=0 at max or min

3 cos x = 4 sin x

so

tan x = 3/4 at extreme

3,4,5 triangle.

tan is 3/4 in quadrants 1 and 3

In quadrant 1

sin x = 3/5

cos x = 4/5

3 sin x + 4 cos x = 9/5 + 16/5 = 5 (maximum probably)

do similar computation in quadrant 3 and I suspect you will find the minimum.

- Calculus -
**bobpursley**, Wednesday, April 7, 2010 at 7:38pmy'=3cosx-4sinx=0

3cosx=4sinx

tanx=3/4

solve for x. You will get two angles, 180degrees apart.

One is a max, one is a min.

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