How to derive some trigonometric formulas?
I was reading a book. Where I found some equations.
$$\sin (\alpha+\beta)=\sin \alpha \cos \beta + \cos \alpha \sin \beta$$
$$\sin (\alpha+\beta)=\cos \alpha \cos \beta - \sin \alpha \sin \beta$$
$$\sin (\alpha-\beta)=\sin \alpha \cos \beta - \cos \alpha \sin \beta$$
$$\sin (\alpha-\beta)=\cos \alpha \cos \beta + \sin \alpha \sin \beta$$
I could derive double angle formula from them.
$$\sin (2\alpha)=\sin \alpha \cos \alpha+\cos \alpha\sin \alpha$$ $$=2\sin\cos\alpha$$
Same way I had derived for $\cos$ also. But, I didn't find how to prove that.
While, $$\sin (\alpha+\beta)\ne \sin \alpha + \sin \beta$$
than, I can't use algebra to derive them. Even, some old trigonometric formulas was derived using a simple triangle. But, here they have given two angle that's why I can't understand how to solve it.
2 answers
You are accessing this answer with a direct link, so it's being shown above all other answers regardless of its score. You can return to the normal view.
Memorizing or using actual trigonometry to re-derive trigonometric formulas is a waste of time and mental resources. (This isn't to say it isn't useful to go through the trigonometric approach once and to understand how they relate.) Instead, there is just one relatively simple formula that one needs from which all of this follows from basic algebra. Namely, Euler's formula: $$e^{\theta i} = \cos\theta + i\sin\theta$$ where $i$ is the imaginary unit, i.e. $i = \sqrt{-1}$. Given this, all of the above formulas follow readily from normal properties of exponents and then straightforward complex algebra.
Angle addition corresponds to the fact that, in general, $x^ax^b = x^{a+b}$ and so concretely, $e^{\alpha i}e^{\beta i} = e^{(\alpha + \beta)i}$. Expanding each exponential with Euler's formula and multiplying the complex number on the left hand side gives:
$$\begin{align} e^{\alpha i}e^{\beta i} & = (\cos\alpha + i\sin\alpha)(\cos\beta + i\sin\beta) \\ & = \cos\alpha\cos\beta + i\sin\alpha\cos\beta + i\cos\alpha\sin\beta + i^2\sin\alpha\sin\beta \\ & = (\cos\alpha\cos\beta - \sin\alpha\sin\beta) + i(\cos\alpha\sin\beta + \sin\alpha\cos\beta) \\ & = \cos(\alpha+\beta) + i\sin(\alpha + \beta) \\ & = e^{(\alpha + \beta)i} \end{align}$$
Equating the components of the complex numbers gives both of the angle addition formulas. Angle subtraction is just as easy. One can derive the double angle formula via $e^{(\alpha + \alpha)i}$ or equivalently $(e^{\alpha})^2$.
More challengingly but still a straightforward exercise in algebra is the half angle formula. Here, we use an implicit formula $(e^{(\alpha/2)i})^2 = e^{\alpha i}$. Write $a$ and $b$ for the real and imaginary components of $e^{(\alpha/2)i}$, i.e. $e^{(\alpha/2)i} = a + bi$. Squaring, we get $(a+bi)^2 = (a^2 - b^2) + 2abi$. Equating this to $e^{\alpha i}$ gives us the following system of three equations: $$\cos\alpha = a^2 - b^2 \qquad \sin\alpha = 2ab \qquad a^2 + b^2 = 1$$ the last because $a = \cos(\alpha/2)$ and $b = \sin(\alpha/2)$ and $\cos^2\theta + \sin^2\theta = 1$ corresponding to $|e^{\theta i}| = 1$ for any $\theta$.
With this we can quickly compute that $a^2 = (1 + \cos\alpha)/2$ and thus $b^2 = (1 - \cos\alpha)/2$. That gives $a = \pm\sqrt{(1 + \cos\alpha)/2}$ and $b = \pm\sqrt{(1 - \cos\alpha)/2}$ where the signs are correlated based on the $\sin\alpha = 2ab$ equation, but we still end up with two possibilities corresponding to the fact that, e.g., a $90^\circ$ angle can be split into two $45^\circ$ angles or two $135^\circ$ angles.
0 comment threads
Can't use algebra, you say? Nonsense! All you need is Euler's formula $$ e^{i\theta} = \cos\theta + i\sin\theta $$ and these identities just fall out of the algebra. For example:
\begin{align} \sin (\alpha + \beta) &= \frac1{2i} \left(e^{i(\alpha + \beta)} - e^{-i(\alpha + \beta)}\right)\\ &= \frac1{2i} \left(e^{i\alpha}e^{i\beta} - e^{-i\alpha}e^{-i\beta}\right)\\ &= \frac1{2i} \left((\cos\alpha + i\sin\alpha)(\cos\beta + i\sin\beta) - (\cos\alpha - i\sin\alpha)(\cos\beta - i\sin\beta)\right)\\ &= \frac1{2i} \left(\cos\alpha\cos\beta + i\sin\alpha\cos\beta + i\cos\alpha\sin\beta - \sin\alpha\sin\beta\right.\\ &\left.\qquad\qquad- \cos\alpha\cos\beta + i\sin\alpha\cos\beta + i\cos\alpha\sin\beta + \sin\alpha\sin\beta\right)\\ &= \frac1{2i} \left(2i\sin\alpha\cos\beta + 2i\cos\alpha\sin\beta\right)\\ &= \sin\alpha\cos\beta + \cos\alpha\sin\beta \end{align}
0 comment threads
2 comment threads