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
since 3 radians is in QII, sin is positive, and cos is negative. 3 is almost pi: 171.9 degrees
Hi Steve, thank you for the reply. I think I got it.
Hello Constantine,
To find the values of sin(3) and cos(3), we need to use a calculator or a mathematical software that can handle trigonometric functions.
First, let's calculate sin(3):
sin(3) ≈ 0.1411
Similarly, let's calculate cos(3):
cos(3) ≈ -0.9900
These values can be obtained by using a scientific calculator or a software that can evaluate trigonometric functions. It's important to note that these values are approximate, as they are rounded to a certain number of decimal places.
In this case, the value of sin(3) is approximately 0.1411, and the value of cos(3) is approximately -0.9900. These values are then used in the expression for the derivative to find the slope of the tangent line at x = 3.
I hope this explanation helps! Let me know if you have any further questions.