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Trigonometric differentiation question

Hey guys, I’m working on this question here

The gradient of the curve y=cos2x when x=15 is:

The options are

-pi/180
-1
pi(sqrt(3))
1

I differentiated it, substituted 15 in and I got -1. For some reason it’s saying this is incorrect but I don’t see how. The question specifies that this is in degrees and not radians btw.
Reply 1
Original post by ramell
Hey guys, I’m working on this question here

The gradient of the curve y=cos2x when x=15 is:

The options are

-pi/180
-1
pi(sqrt(3))
1

I differentiated it, substituted 15 in and I got -1. For some reason it’s saying this is incorrect but I don’t see how. The question specifies that this is in degrees and not radians btw.

When you do calculus with trig, you always work in radians (i.e. the argument to sin, cos etc must be in radians). That means you need to convert your argument from degrees to radians.
Reply 2
Original post by DFranklin
When you do calculus with trig, you always work in radians (i.e. the argument to sin, cos etc must be in radians). That means you need to convert your argument from degrees to radians.

I know that, but the question has a degrees sign after y=cos2x. Also, working it out on radians gives me an answer that is not one of the 4 options
Reply 3
Original post by ramell
I know that, but the question has a degrees sign after y=cos2x. Also, working it out on radians gives me an answer that is not one of the 4 options

The derivative of sinθ is equal to π/180cosθ if θ is measured in degrees.
Reply 4
Original post by ramell
I know that, but the question has a degrees sign after y=cos2x. Also, working it out on radians gives me an answer that is not one of the 4 options

If it helps, think of a degree as pi/180 radians.

So, for example, sin3xo=sin(3xπ180)\sin 3x^o = \sin(3x \frac{\pi}{180})

You can then differentiate as normal.
(edited 5 months ago)

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