# Math (Calculus)

Hello,

I have a difficulty with the following question. The question is asking to find the slope of the tangent at x=3 for this function: f(x)=4x^3/sinx. X is measured in radians.

The is the derivative of f(x):

f'(x)=(12x^2)(sinx)^-1 + (4x^3)(-1)(sinx)^-2(cosx)

When x=3 is plugged in:

=f'(3)=(12(3)^2)(1/sin3) - 4(3)^3((cos3)/(sin3)^2)
=108(1/0.1411) - 108(-0.9900/(0.1411)^2)=6135.8

What I don't understand is why sin3 equals to 0.1411 and why cos3 is -0.9900.

I would be very grateful for your help.

Constantine

1. 👍
2. 👎
3. 👁
1. since 3 radians is in QII, sin is positive, and cos is negative. 3 is almost pi: 171.9 degrees

1. 👍
2. 👎
2. Hi Steve, thank you for the reply. I think I got it.

1. 👍
2. 👎

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