When I was in high school, I remember that I had to memorize all trigonometry formulas. I didn’t have a single way to resort to when I was locked.
Fortunately, As I will show here, most trigonometry formulas can be derived via simple algebraic modifications of Euler's Formula.
This way, any time we are locked and can’t remember what the formula was, we can simply derive it.
Euler’s Formula
$$e^{ix} = \cos(x) + i\sin(x)$$
Expressing Sine and Cosine
Since cosine is an even function and sine is an odd function: $$\cos(-x) = \cos(x)$$ $$\sin(-x) = -\sin(x)$$ Therefore: $$e^{-ix} = \cos(x) - i\sin(x)$$ Adding the two Euler’s equations: $$e^{ix} + e^{-ix} = [\cos(x) + i\sin(x)] + [\cos(x) - i\sin(x)]$$ $$e^{ix} + e^{-ix} = 2\cos(x) + i\sin(x) - i\sin(x)$$ $$e^{ix} + e^{-ix} = 2\cos(x)$$
Solving for cosine: $$\cos(x) = \frac{e^{ix} + e^{-ix}}{2}$$
Subtracting the equations: $$e^{ix} - e^{-ix} = [\cos(x) + i\sin(x)] - [\cos(x) - i\sin(x)]$$ $$e^{ix} - e^{-ix} = \cos(x) + i\sin(x) - \cos(x) + i\sin(x)$$ $$e^{ix} - e^{-ix} = 2i\sin(x)$$
Solving for sine: $$\sin(x) = \frac{e^{ix} - e^{-ix}}{2i}$$
So, we now have cosine and sine in terms of imaginary exponential. We can do interesting things from now on.
Double Angle Formulas (Direct)
For cosine: $$\cos(2x) = \frac{e^{i2x} + e^{-i2x}}{2}$$ $$\cos(2x) = \frac{(e^{ix})^2 + (e^{-ix})^2}{2}$$ Substituting Euler’s formula: $$\cos(2x) = \frac{[\cos(x) + i\sin(x)]^2 + [\cos(x) - i\sin(x)]^2}{2}$$Expanding the squares: $$\cos(2x) = \frac{[\cos^2(x) + 2i\sin(x)\cos(x) - \sin^2(x)] + [\cos^2(x) - 2i\sin(x)\cos(x) - \sin^2(x)]}{2}$$ $$\cos(2x) = \frac{2\cos^2(x) - 2\sin^2(x)}{2}$$ $$\cos(2x) = \cos^2(x) - \sin^2(x)$$Using the Pythagorean identity $\cos^2(x) + \sin^2(x) = 1$: $$\cos(2x) = \cos^2(x) - (1 - \cos^2(x))$$ $$\cos(2x) = 2\cos^2(x) - 1$$Alternatively: $$\cos(2x) = 1 - 2\sin^2(x)$$ For sine: $$\sin(2x) = \frac{e^{i2x} - e^{-i2x}}{2i}$$$$\sin(2x) = \frac{(e^{ix})^2 - (e^{-ix})^2}{2i}$$ Substituting Euler’s formula: $$\sin(2x) = \frac{[\cos(x) + i\sin(x)]^2 - [\cos(x) - i\sin(x)]^2}{2i}$$Expanding: $$\sin(2x) = \frac{[\cos^2(x) + 2i\sin(x)\cos(x) - \sin^2(x)] - [\cos^2(x) - 2i\sin(x)\cos(x) - \sin^2(x)]}{2i}$$ $$\sin(2x) = \frac{4i\sin(x)\cos(x)}{2i}$$$$\sin(2x) = 2\sin(x)\cos(x)$$
Triple Angle Formulas (Direct)
Starting with Euler’s formula:
$$e^{ix} = \cos(x) + i\sin(x)$$
For the triple angle, we have:
$$e^{i3x} = \cos(3x) + i\sin(3x)$$
We can also express this as:
$$e^{i3x} = (e^{ix})^3 = (\cos(x) + i\sin(x))^3$$
Expanding using the binomial theorem:
$$(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$$
With $a = \cos(x)$ and $b = i\sin(x)$:
$$(\cos(x) + i\sin(x))^3 = \cos^3(x) + 3\cos^2(x)(i\sin(x)) + 3\cos(x)(i\sin(x))^2 + (i\sin(x))^3$$
Simplifying, noting that $(i\sin(x))^2 = -\sin^2(x)$ and $(i\sin(x))^3 = -i\sin^3(x)$:
$$(\cos(x) + i\sin(x))^3 = \cos^3(x) + 3i\cos^2(x)\sin(x) - 3\cos(x)\sin^2(x) - i\sin^3(x)$$
Rearranging:
$$(\cos(x) + i\sin(x))^3 = [\cos^3(x) - 3\cos(x)\sin^2(x)] + i[3\cos^2(x)\sin(x) - \sin^3(x)]$$
Comparing this with $e^{i3x} = \cos(3x) + i\sin(3x)$, we get:
$$\cos(3x) = \cos^3(x) - 3\cos(x)\sin^2(x)$$ $$\sin(3x) = 3\cos^2(x)\sin(x) - \sin^3(x)$$
Pythagorean Identity with Steps
From Euler’s formula:
$$e^{ix} = \cos(x) + i\sin(x)$$
The modulus of a complex number squared equals the sum of squares of real and imaginary parts:
$$|e^{ix}|^2 = |\cos(x) + i\sin(x)|^2 = \cos^2(x) + \sin^2(x)$$
Since $|e^{ix}| = |e^i|^{|x|} = 1^{|x|} = 1$ (+):
$$\cos^2(x) + \sin^2(x) = 1$$
Sum and Difference Formulas
For cosine of sum:
$$\cos(a+b) = \frac{e^{i(a+b)} + e^{-i(a+b)}}{2}$$ $$\cos(a+b) = \frac{e^{ia}e^{ib} + e^{-ia}e^{-ib}}{2}$$
Substituting Euler’s formula:
$$\cos(a+b) = \frac{[\cos(a) + i\sin(a)][\cos(b) + i\sin(b)] + [\cos(a) - i\sin(a)][\cos(b) - i\sin(b)]}{2}$$
Multiplying the terms:
$$\cos(a+b) = \frac{[\cos(a)\cos(b) - \sin(a)\sin(b) + i(\sin(a)\cos(b) + \cos(a)\sin(b))] + [\cos(a)\cos(b) - \sin(a)\sin(b) - i(\sin(a)\cos(b) + \cos(a)\sin(b))]}{2}$$ $$\cos(a+b) = \frac{2[\cos(a)\cos(b) - \sin(a)\sin(b)]}{2}$$ $$\cos(a+b) = \cos(a)\cos(b) - \sin(a)\sin(b)$$
For cosine of difference, replace $b$ with $-b$:
$$\cos(a-b) = \cos(a)\cos(-b) - \sin(a)\sin(-b)$$ $$\cos(a-b) = \cos(a)\cos(b) - \sin(a)(-\sin(b))$$ $$\cos(a-b) = \cos(a)\cos(b) + \sin(a)\sin(b)$$
For sine of sum:
$$\sin(a+b) = \frac{e^{i(a+b)} - e^{-i(a+b)}}{2i}$$ $$\sin(a+b) = \frac{e^{ia}e^{ib} - e^{-ia}e^{-ib}}{2i}$$
Substituting Euler’s formula:
$$\sin(a+b) = \frac{[\cos(a) + i\sin(a)][\cos(b) + i\sin(b)] - [\cos(a) - i\sin(a)][\cos(b) - i\sin(b)]}{2i}$$
Expanding:
$$\sin(a+b) = \frac{[\cos(a)\cos(b) - \sin(a)\sin(b) + i(\sin(a)\cos(b) + \cos(a)\sin(b))] - [\cos(a)\cos(b) - \sin(a)\sin(b) - i(\sin(a)\cos(b) + \cos(a)\sin(b))]}{2i}$$ $$\sin(a+b) = \frac{2i[\sin(a)\cos(b) + \cos(a)\sin(b)]}{2i}$$ $$\sin(a+b) = \sin(a)\cos(b) + \cos(a)\sin(b)$$
For sine of difference, replace $b$ with $-b$:
$$\sin(a-b) = \sin(a)\cos(-b) + \cos(a)\sin(-b)$$ $$\sin(a-b) = \sin(a)\cos(b) + \cos(a)(-\sin(b))$$ $$\sin(a-b) = \sin(a)\cos(b) - \cos(a)\sin(b)$$
Half Angle Formulas
Half-angles can be derived from results of above equations. I couldn’t find a way to simply extract it from Euler’s formula.
From the double angle formula $\cos(2x) = 2\cos^2(x) - 1$, we get:
$$\cos^2(x) = \frac{1 + \cos(2x)}{2}$$
So, $$\cos\left(\frac{x}{2}\right) = \pm\sqrt{\frac{1 + \cos(x)}{2}}$$
The sign depends on the quadrant of $\frac{x}{2}$.
Similarly, from $\cos(2x) = 1 - 2\sin^2(x)$:
$$\sin^2(x) = \frac{1 - \cos(2x)}{2}$$ So, $$\sin\left(\frac{x}{2}\right) = \pm\sqrt{\frac{1 - \cos(x)}{2}}$$
Using the tangent identity $\tan(x) = \frac{\sin(x)}{\cos(x)}$:
$$\tan\left(\frac{x}{2}\right) = \frac{\sin\left(\frac{x}{2}\right)}{\cos\left(\frac{x}{2}\right)} = \frac{\pm\sqrt{\frac{1 - \cos(x)}{2}}}{\pm\sqrt{\frac{1 + \cos(x)}{2}}}$$
Simplifying:
$$\tan\left(\frac{x}{2}\right) = \sqrt{\frac{1 - \cos(x)}{1 + \cos(x)}}$$
Using the identity $1 - \cos(x) = 2\sin^2\left(\frac{x}{2}\right)$:
$$\tan\left(\frac{x}{2}\right) = \frac{\sin(x)}{1 + \cos(x)}$$
Alternatively:
$$\tan\left(\frac{x}{2}\right) = \frac{1 - \cos(x)}{\sin(x)}$$
Solving Some Challenging Trigonometry Formulas
Problem 1: Sum of Cosines
Prove that $\sum_{k=0}^{n-1} \cos\left(x + \frac{2\pi k}{n}\right) = 0$ for any integer $n > 1$ and any angle $x$.
Solution:
Using Euler’s formula, we have: $$\cos\left(x + \frac{2\pi k}{n}\right) = \frac{e^{i(x + \frac{2\pi k}{n})} + e^{-i(x + \frac{2\pi k}{n})}}{2}$$
The sum becomes: $$\sum_{k=0}^{n-1} \cos\left(x + \frac{2\pi k}{n}\right) = \frac{1}{2}\sum_{k=0}^{n-1} \left(e^{i(x + \frac{2\pi k}{n})} + e^{-i(x + \frac{2\pi k}{n})}\right)$$ $$= \frac{1}{2}\sum_{k=0}^{n-1} \left(e^{ix} \cdot e^{i\frac{2\pi k}{n}} + e^{-ix} \cdot e^{-i\frac{2\pi k}{n}}\right)$$ $$= \frac{e^{ix}}{2}\sum_{k=0}^{n-1} e^{i\frac{2\pi k}{n}} + \frac{e^{-ix}}{2}\sum_{k=0}^{n-1} e^{-i\frac{2\pi k}{n}}$$
Both sums are geometric series with ratio $r = e^{i\frac{2\pi}{n}}$ and $r = e^{-i\frac{2\pi}{n}}$ respectively.
For the first sum, using the formula for geometric series: $$\sum_{k=0}^{n-1} e^{i\frac{2\pi k}{n}} = \frac{1 - e^{i2\pi}}{1 - e^{i\frac{2\pi}{n}}} = \frac{1 - 1}{1 - e^{i\frac{2\pi}{n}}} = 0$$
Since $e^{i2\pi} = 1$ (one full rotation in the complex plane).
Similarly, the second sum equals zero.
Therefore, $\sum_{k=0}^{n-1} \cos\left(x + \frac{2\pi k}{n}\right) = 0$
Problem 2: Products to Powers
Express $\cos^n(x)$ in terms of cosines of multiple angles.
Solution:
Using Euler’s formula: $$\cos(x) = \frac{e^{ix} + e^{-ix}}{2}$$
Therefore: $$\cos^n(x) = \left(\frac{e^{ix} + e^{-ix}}{2}\right)^n$$
Using the binomial theorem: $$\cos^n(x) = \frac{1}{2^n}\sum_{k=0}^{n} \binom{n}{k} (e^{ix})^k (e^{-ix})^{n-k}$$ $$= \frac{1}{2^n}\sum_{k=0}^{n} \binom{n}{k} e^{ix(2k-n)}$$
Using Euler’s formula again: $$e^{ix(2k-n)} = \cos(x(2k-n)) + i\sin(x(2k-n))$$
Since we’re finding $\cos^n(x)$ which is real, we take only the real part: $$\cos^n(x) = \frac{1}{2^n}\sum_{k=0}^{n} \binom{n}{k} \cos(x(2k-n))$$
For example, when $n=3$: $$\cos^3(x) = \frac{1}{2^3}[\binom{3}{0}\cos(-3x) + \binom{3}{1}\cos(-x) + \binom{3}{2}\cos(x) + \binom{3}{3}\cos(3x)]$$ $$= \frac{1}{8}[\cos(3x) + 3\cos(x) + 3\cos(-x) + \cos(-3x)]$$ $$= \frac{1}{8}[4\cos(3x) + 3\cos(x) + 3\cos(x)]$$ (since cosine is even) $$= \frac{1}{4}\cos(3x) + \frac{3}{4}\cos(x)$$
Problem 3: Prove a Complex Identity
Prove that $\cos^4(x) - \sin^4(x) = \cos(2x)$.
Solution:
Using Euler’s formula: $$\cos(x) = \frac{e^{ix} + e^{-ix}}{2}$$ $$\sin(x) = \frac{e^{ix} - e^{-ix}}{2i}$$
Calculating $\cos^4(x)$: $$\cos^4(x) = \left(\frac{e^{ix} + e^{-ix}}{2}\right)^4 = \frac{(e^{ix} + e^{-ix})^4}{16}$$
Expanding using the binomial theorem: $$\cos^4(x) = \frac{1}{16}[e^{4ix} + 4e^{2ix} + 6 + 4e^{-2ix} + e^{-4ix}]$$
Similarly for $\sin^4(x)$: $$\sin^4(x) = \left(\frac{e^{ix} - e^{-ix}}{2i}\right)^4 = \frac{(e^{ix} - e^{-ix})^4}{16i^4} = \frac{(e^{ix} - e^{-ix})^4}{16 \cdot 1}$$
Expanding: $$\sin^4(x) = \frac{1}{16}[e^{4ix} - 4e^{2ix} + 6 - 4e^{-2ix} + e^{-4ix}]$$
Now calculating $\cos^4(x) - \sin^4(x)$: $$\cos^4(x) - \sin^4(x) = \frac{1}{16}[8e^{2ix} + 8e^{-2ix}] = \frac{1}{2}[e^{2ix} + e^{-2ix}]$$
Using Euler’s formula again, this equals: $$\cos^4(x) - \sin^4(x) = \cos(2x)$$
Which completes the proof.
Problem 4: Sum of Sines Series
Find the sum: $S = \sum_{k=1}^{n} \sin(kx)$ for $x \neq 2\pi m$ where $m$ is an integer. Solution:
Using Euler’s formula: $$\sin(kx) = \frac{e^{ikx} - e^{-ikx}}{2i}$$
The sum becomes: $$S = \sum_{k=1}^{n} \sin(kx) = \frac{1}{2i}\sum_{k=1}^{n} (e^{ikx} - e^{-ikx})$$ $$= \frac{1}{2i}\left[\sum_{k=1}^{n} e^{ikx} - \sum_{k=1}^{n} e^{-ikx}\right]$$
Each sum is a geometric series: $$\sum_{k=1}^{n} e^{ikx} = e^{ix} \cdot \frac{1 - e^{inx}}{1 - e^{ix}}$$ $$\sum_{k=1}^{n} e^{-ikx} = e^{-ix} \cdot \frac{1 - e^{-inx}}{1 - e^{-ix}}$$
Substituting: $$S = \frac{1}{2i}\left[e^{ix} \cdot \frac{1 - e^{inx}}{1 - e^{ix}} - e^{-ix} \cdot \frac{1 - e^{-inx}}{1 - e^{-ix}}\right]$$
Multiplying numerator and denominator of the second fraction by $e^{ix}$: $$S = \frac{1}{2i}\left[\frac{e^{ix}(1 - e^{inx})}{1 - e^{ix}} - \frac{1 - e^{-inx}}{e^{ix} - 1}\right]$$ $$= \frac{1}{2i}\left[\frac{e^{ix}(1 - e^{inx})}{1 - e^{ix}} + \frac{1 - e^{-inx}}{1 - e^{ix}}\right]$$
Combining fractions: $$S = \frac{1}{2i(1 - e^{ix})}\left[e^{ix}(1 - e^{inx}) + (1 - e^{-inx})\right]$$ $$= \frac{1}{2i(1 - e^{ix})}\left[e^{ix} - e^{ix}e^{inx} + 1 - e^{-inx}\right]$$ $$= \frac{1}{2i(1 - e^{ix})}\left[e^{ix} - e^{i(n+1)x} + 1 - e^{-inx}\right]$$
Using Euler’s formula, we can express this in terms of sine and cosine: $$S = \frac{\sin(\frac{(n+1)x}{2})\sin(\frac{nx}{2})}{\sin(\frac{x}{2})}$$
This is the closed form for the sum of the first $n$ sines.
Problem 5: Trigonometric Equation (Euler’s Formula Only)
Solve the equation $\sin(3x) + \sin(5x) = 0$ for $x \in [0, 2\pi)$.
Solution:
I’ll solve this using only Euler’s formula:
$$e^{ix} = \cos(x) + i\sin(x)$$
From this formula, we can express sine in terms of complex exponentials:
$$\sin(x) = \frac{e^{ix} - e^{-ix}}{2i}$$
Applying this to our equation:
$$\sin(3x) + \sin(5x) = 0$$
$$\frac{e^{i3x} - e^{-i3x}}{2i} + \frac{e^{i5x} - e^{-i5x}}{2i} = 0$$
Multiplying both sides by $2i$:
$$e^{i3x} - e^{-i3x} + e^{i5x} - e^{-i5x} = 0$$
Rearranging terms:
$$e^{i3x} + e^{i5x} = e^{-i3x} + e^{-i5x}$$
Let $z = e^{ix}$. Then our equation becomes:
$$z^3 + z^5 = z^{-3} + z^{-5}$$
Multiplying both sides by $z^5$:
$$z^8 + z^{10} = z^2 + 1$$
Rearranging:
$$z^{10} + z^8 - z^2 - 1 = 0$$
This can be factored as:
$$z^{10} + z^8 - z^2 - 1 = z^8(z^2 + 1) - (z^2 + 1) = (z^8 - 1)(z^2 + 1) = 0$$
Therefore, either $z^8 - 1 = 0$ or $z^2 + 1 = 0$.
From $z^8 - 1 = 0$, we get: $$z^8 = 1$$
This means $z$ is an 8th root of unity. In the complex plane, the 8th roots of unity are equally spaced around the unit circle and can be written as:
$$z = e^{i\frac{2\pi k}{8}} = e^{i\frac{\pi k}{4}}$$ for $k = 0, 1, 2, …, 7$
Since we defined $z = e^{ix}$, we can find $x$ by comparing the exponents:
$$e^{ix} = e^{i\frac{\pi k}{4}}$$
This gives us: $$x = \frac{\pi k}{4}$$ for $k = 0, 1, 2, …, 7$
In the interval $[0, 2\pi)$, these solutions are: $$x = 0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, \pi, \frac{5\pi}{4}, \frac{3\pi}{2}, \frac{7\pi}{4}$$
From $z^2 + 1 = 0$, we get: $$z^2 = -1$$
This gives us: $$z = \pm i = e^{i\frac{\pi}{2}}$$ or $$z = e^{i\frac{3\pi}{2}}$$
Which corresponds to: $$x = \frac{\pi}{2}$$ or $$x = \frac{3\pi}{2}$$
These values are already included in our solutions from $z^8 = 1$ (specifically when $k = 2$ and $k = 6$).
Therefore, the complete set of solutions in $[0, 2\pi)$ is: $$x = 0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, \pi, \frac{5\pi}{4}, \frac{3\pi}{2}, \frac{7\pi}{4}$$