Pythagoras' theorem
In the unit circle, the radius 1 is equivalent to the hypotenuse and the sine and cosine ratios are the legs of the right-angled triangle.
So,
$$ \sin^2 {\theta} + \cos^2 {\theta} \equiv 1 $$
tangent
If the sides of the triangle are divided by $\cos {\theta} $, we get a new ratio
$$\tan \theta \equiv \frac {\sin {\theta}}{\cos {\theta}} $$
Dividing the Pythagorean identity by $\cos^2 {\theta}$, we get following new identity. (see diagram)
$$ \tan^2 {\theta} + 1 \equiv \frac {1}{\cos^2 {\theta}} \equiv \sec^2 {\theta} $$
Similarly, by dividing by $\sin^2 {\theta}$
$$ 1 + \cot^2 {\theta} \equiv \frac {1}{\sin^2 {\theta}} \equiv \csc^2 {\theta} $$
ratios of compound angles
$$\begin{align} \sin{(A + B)} & = \frac{RT}{OR} \newline & = \frac{RS + ST}{OR} \newline & = \frac{RS}{OR} + \frac{ST}{OR} \newline & = \frac{RS}{OR} + \frac{PQ}{OR} \newline & = \frac{RS}{RQ} \cdot \frac{RQ}{OR} + \frac{PQ}{OQ} \cdot \frac{OQ}{OR} \newline & = \cos B \cdot \sin A + \sin B \cdot \cos A \end{align} $$
Hence,
$$ \sin {(A+B)} \equiv \sin A \cos B + \cos A \sin B $$
Can you now derive the identity for $ \cos{(A + B)}$ ?
alternative proof beginning with cos(A+B)
recently saw this elegant proof for the compound angle of cosine.
- draw two chords on a unit circle.
- one starting at A(1,0) to B(cos(A+B),sin(A+B)) and the other beginning at D(cos(-B), sin(-B)) to C(cos(A), sin(A))
- arc AB = arc CD
- AB = CD
- apply pythagoras to the coordinates
- square and expand
- simplify
TODO diagram
double angles
sin(2A)
Set B = A, then
$$\begin{align} \sin {(2A)} & = \sin{(A+A)} \newline & =\sin A \cos A + \cos A \sin A \newline & = 2 \sin {A}\cos{A}\end{align}$$
cos(2A)
For $ \cos{(2A)}$ see here
half-angle identities
$\sin{x}$
From $\begin{align} \sin {(2A)} & = 2 \sin {A}\cos{A}\end{align}$ we get
$$\begin{align} \sin {x} & = 2 \sin {\frac{x}{2}}\cos{\frac{x}{2}}\end{align}$$
$\cos {x}$
And similarly,
$$ \cos {x} \equiv \cos^2 {\frac{x}{2}} - \sin^2 {\frac{x}{2}} \equiv 2\cos^2{\frac{x}{2}} - 1 \equiv 1 - 2\sin^2{\frac{x}{2}} $$
A useful result we can derive from this is
$$ \frac{1}{2}(\cos {x} - 1) = - \sin^2{\frac{x}{2}} $$
This is used when finding the limit of
$$ \frac{1}{2} \frac{(\cos {x} - 1)}{x} = - \frac{\sin{\frac{x}{2}}}{x} \cdot \sin {\frac{x}{2}} $$
dividing by $\frac{1}{2}$
$$ \frac{(\cos {x} - 1)}{x} = - \frac{\sin{\frac{x}{2}}}{\frac{x}{2}} \cdot \sin {\frac{x}{2}} $$
Thus,
$$ \lim_{x \to 0} \frac{(\cos {x} - 1)}{x} = - \lim_{x \to 0} \left( \frac{\sin{\frac{x}{2}}}{\frac{x}{2}}\right) \cdot \lim_{x \to 0} \sin {\frac{x}{2}} = -1 \cdot 0 = 0$$
other sites on trigonometry
interesting alternative view using jsxgraph
Ptolemy’s theorem and addition theorems
Another teacher’s site on trigonometry Wumbo.net