Trigonometric Intersections

First, find the points of intersection:

$\cos(x)=\tan(x)$

$\cos(x)=\frac{\sin(x)}{\cos(x)}$

$\cos^2(x)=\sin(x)$

$1-\sin^2(x) = \sin(x)$

Solve the quadratic to find:

$\sin(x)=\frac{1-\sqrt{5}}{2}$

Happily, we don't need to find the co-ordinates. Differentiating:

$d/dx(\cos(x))=-\sin(x)$

$d/dx(\tan(x))=-\sec^2(x)$

So, at the intersection, the gradient of $\cos(x)$ is:

$-\sin(x)=-\frac{1-\sqrt{5}}{2}$

Using the standard labels for a right-angled triangle (O,A,H), we know:

$\sin(x)=\frac{1-\sqrt{5}}{2}=\frac{O}{H}$

$H(\frac{1-\sqrt{5}}{2})=O$ - (1)

We know from the problem:

$\frac{A}{H}=\frac{O}{A}$

$A^2=OH$

Substitute equation (1):

$H^2(\frac{1-\sqrt{5}}{2})=A^2$

$\frac{H^2}{A^2}=\sec^2(x)=\frac{2}{1-\sqrt{5}}$

So, we see that $\tan(x)$ and $\cos(x)$ are perpendicular at the intersection points, giving us an answer of $\pi/2$ for the angle between them.

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Now we know that the tangents are perpendicular, we can prove it in a very succinct way:

$f(x)=\cos(x)$

$g(x)=\tan(x)$

$f'(x)=-\sin(x)$

$g'(x)=\sec^2(x)$

At the points of intersection, $f(x)=g(x)$

$\cos(x)=\frac{\sin(x)}{\cos(x)}$

$1=\frac{\sin(x)}{\cos^2(x)}$

So we have

$f'(x)g'(x)=-1$

Which implies the tangents at this point are perpendicular.