# Secrets of the Mathematical Ninja: Sines and cosines near 45º

This is the one area where I’m better with degrees than with radians - and I suspect that’s only because I don’t particularly notice when radian angles are close to $\frac{\pi}{4}$, but I do when degree angles are close to 45º.

This one’s a trickier one than we’ve been looking at, but I’m sure you can manage; there’s also a simple version: for angles very near 45º, 0.7 is a good guess for both sine and cosine. (In fact, $\sin(45º) = \cos(45º) = \frac{\sqrt{2}}{2} \simeq 0.707$, which is about 1% more than 0.7.)

However, such a rough answer for, say, $\sin(43º)$, would be unacceptable for the true mathematical ninja. So how can we adjust our answer?

Well… unless you’re doing Further Maths, Taylor series probably don’t concern you. Even if you are, you probably only get to see the more specific Maclaurin series. However, it’s a useful little feller here:

$f(X + x) \simeq f(X) + x f^\prime(X)$.

Who to the what now? Well, what it means is, if you move a small distance from a known value, you can use the derivative to figure out a guess for the new answer. In this case, we have $X = 45º$, or - since we’re doing calculus - $\frac{\pi}{4}$ radians.

When we’re x radians away from π/4, this becomes: $\sin(\frac{\pi}{4} + x) \simeq 0.707 ( 1 + x )$ $\cos(\frac{\pi}{4} + x) \simeq 0.707 ( 1 - x )$

Now we’re in business - because we know that you can convert degrees to radians using the magic number $\frac{7}{400}$ - which is 1.75%. Can you see where this is going yet?

To work out $\sin(43º)$, you say ‘it’s two degrees below 45, so I need to minus 3.5% from sin(45º), which is 1% more than 0.7. So, it’s 0.7 minus 2.5%, and 2.5% of 0.7 is about 0.02, making 0.68. (It’s actually 0.682 - if I’d thought through the details a bit better, I’d have got that).

How about $\cos(40º)$? That’s 5º below 45, so I need to add $5 \times 1.75% = 8.75%$ on to 0.707, or 9.75% onto 0.7. That’ll be close enough to 0.77 as makes no odds - in fact, it’s 0.766. Again, less lazy calculation would have got me closer, but not exactly there.

As an exercise, you can try applying the Taylor series near other known values of $\sin(x)$ and $\cos(x)$, even $\tan(x)$ if you’re feeling brave ((Yes, it’s in the formula book, it’s $\sec^2(x)$)).