Simple Calculus Help.?

I'm pretty sure I understand why the function f(pi/2)= 20 cos (pi/2 ) equates to 0


Why the hell does f(pi/2 + 0.1) = 20 cos (pi/2 + 0.1) equate to approximately -1.9?

I have no idea where the negative is coming from and have attempted to reverse engineer the problem using other math applications-- I still can't figure it out.


To be clear-- Applications like Symbolab etc just generate the answer without showing me how it generated a -1.9.

Before when I did the pi/2 function it at least justified why the result was 0

Update 2:

I THINK I figured it out-- but the answer is still slightly off from what Symbolab specifies, I'd appreciate any input.

2 Answers

  • 1 month ago
    Favorite Answer

    If you look at the unit circle, you can see that cosine is positive in quadrant 1. It is zero when exactly at π/2. Once you switch over to quadrant 2, cosine is negative.

    So as you increase the angle beyond 90° (aka π/2 radians), the cosine is negative.

    Easier still, just look at the graph of the cosine function. You can see the graph is going down. When you reach π/2 the graph is at zero. And then it goes negative from there (until it turns around, goes back up and crosses zero at 3π/2).

    By the way, it's worth noticing that cos(π/2 + 0.1) ≈ -0.1 and so when you multiply it by 20 you get 20 cos(π/2 + 0.1) ≈ 20 * -0.1 ≈ -2.0 (or -1.9966683...)

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  • Amy
    Lv 7
    1 month ago

    Not quite sure what you're asking for.

    20 cos (pi/2 + 0.1) is approximately -1.997, as you can find by typing it into any calculator.

    So you're asking why some particular approximation gives a less accurate result? But you haven't told us what algorithm you're using to make the approximation.

    Perhaps your program simply truncates (rather than rounds) the answer to some number of digits.

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