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Find an equation of the tangent line to $$y = \cos(x)+3\sin(x)$$ at $x=\pi/3$.

This is what I have done...

Find $y$, $y= \cos(\pi/3) + 3\sin(\pi/3)$
this equals $1 + \sqrt 3/2$

Next

Find $f'(x) = \sin(\pi/3) + 3\cos(\pi/3)$
this equals $3+ 3\sqrt3/2$

Next

Plug into point slope form

$(y-1+\sqrt 3/2) = (3 +3\sqrt3/2) (x - \pi/3)$

$y = \left(\frac{3+3\sqrt 3}{2}\right)(x-\pi/3)+(1+3\sqrt 3/2)$

Am I doing something wrong? Thanks.

Em.
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pewpew
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  • Yes, there is a mistake. When you put in point slope form, there needs to be a minus, not a plus in front of that radical (Distributive property) – imranfat Mar 04 '16 at 03:02
  • That edit makes things lots easier for the eye...:) – imranfat Mar 04 '16 at 03:03
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    Use $f'(x) = \sin(\pi/3) + 3\cos(\pi/3)$ to show $f'(x) = \sin(\pi/3) + 3\cos(\pi/3)$. Formatting tips here. – Em. Mar 04 '16 at 03:04
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    I believe your evaluations of the trigonometric terms are incorrect: please check your results for the value of the function and its derivative. It looks like you've swapped sine and cosine values, at the least. Also, what is the derivative of $ \ \cos x \ $ ? – colormegone Mar 04 '16 at 03:08
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    Note that $f'(x)=-\sin x+3\cos x$. There are other errors, and several missing parentheses. The general strategy is fine. – André Nicolas Mar 04 '16 at 03:12
  • ok thanks I will work on it some more. – pewpew Mar 04 '16 at 03:20

1 Answers1

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While your strategy is correct, you incorrectly evaluated the sine and cosine functions at $x = \pi/3$ and incorrectly took the derivative.

Let $f(x) = \cos x + 3\sin x$. Since $$\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$$ and $$\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$$ when we evaluate $f(x)$ at $\pi/3$ we obtain

$$f\left(\frac{\pi}{3}\right) = \cos\left(\frac{\pi}{3}\right) + 3\sin\left(\frac{\pi}{3}\right) = \frac{1}{2} + \frac{3\sqrt{3}}{2} = \frac{1 + 3\sqrt{3}}{2}$$

Since the derivative of $g(x) = \cos x$ is $g'(x) = -\sin x$ and the derivative of $h(x) = \sin x$ is $h'(x) = \cos x$, the derivative of $f(x)$ is

$$f'(x) = -\sin x + 3\cos x$$

Thus, the derivative of $f(x)$ evaluated at $\pi/3$ is

$$f'\left(\frac{\pi}{3}\right) = -\sin\left(\frac{\pi}{3}\right) + 3\cos\left(\frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2} + \frac{3}{2} = \frac{3 - \sqrt{3}}{2}$$

Hence, the equation of the tangent line is

$$y - \frac{1 + 3\sqrt{3}}{2} = \frac{3 - \sqrt{3}}{2}\left(x - \frac{\pi}{3}\right)$$

N. F. Taussig
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