2 ms·
My best trick for being able to derive trig identies from scratch is to use the complex analysis identity: exp{i \theta} = cos(\theta) + i sin{\theta} where i
by hackerblues 17y ago
My best trick for being able to derive trig identies from scratch is to use the complex analysis identity:
exp{i \theta} = cos(\theta) + i sin{\theta}
where i^2 = -1.
As an example, lets derive the double angle formula:
cos(2\theta) + i sin{2\theta} = exp{i 2\theta}
= (exp{i \theta})^2
= (cos(\theta) + i sin{\theta})^2
= cos(\theta)^2 + i 2 sin(\theta) cos(\theta) + i^2 sin{\theta}^2
= cos(\theta)^2 - sin{\theta}^2 + i 2 sin(\theta) cos(\theta)
Since the real part must equal the real part and the imaginary part equal the imaginary part we conclude that:
cos(2\theta) = cos(\theta)^2 - sin{\theta}^2
and
sin{2*\theta} = 2 sin(\theta) cos(\theta)
~~~
The general procedure is:
cos(complicated) = ???
1) cos(complicated) + i sin(complicated) = exp{i complicated}
2) Break it up: exp{i complicated} = exp{i simple1 + ... + i simple4} = exp{i simple1}...exp{i simple4}
3) Translate back to sins and cos: (cos(simple1) + i sin(simple1))...(cos(simple4) + i sin(simple4))
4) Then multiply them out and choose the real part to get cos(complicated) or the imaginary part to get sin(complicated)
where the simple1,...,simple4 was an arbitrarily chosen number of simple bits.