Trigonometry

Proof of sum and product of sine and cosine
Recall $e^{ix}=\cos x+i\sin x$
$e^{i(x+y)}=\cos (x+y)+i\sin (x+y)$
$e^{ix}\cdot e^{iy}\\
=(\cos x+i\sin x)(\cos y+i\sin y)\\
=\cos x \cos y -\sin x \sin y+i(\sin x \cos y+\sin y \cos x)\\
\cos x \cos y -\sin x \sin y+i(\sin x \cos y+\sin y \cos x)=\cos (x+y)+i\sin (x+y)$

Comparing the real and imaginary part respectively,
$\cos (x+y)=\cos x \cos y -\sin x \sin y\\
\sin (x+y)=\sin x \cos y+\sin y \cos x\\$
Similarly,
$\cos (x-y)=\cos x \cos y +\sin x \sin y\\
\sin (x-y)=\sin x \cos y-\sin y \cos x$

Then,
$\cos (x+y)+\cos (x-y)=2\cos x \cos y\\
\sin (x+y)+\sin (x-y)=2\sin x \cos y\\
\cos (x+y)-\cos (x-y)=-2\sin x \sin y\\
\sin (x+y)-\sin (x-y)=2\sin y \cos x$,

or equivalently

[Let A = x+y, B = x-y; x = $\frac{A+B}{2}$, y = $\frac{A-B}{2}$]

$\cos A+\cos B=2\cos \frac{A+B}{2} \cos \frac{A-B}{2}\\
\sin A+\sin B=2\sin \frac{A+B}{2} \cos \frac{A-B}{2}\\
\cos A-\cos B=-2\sin \frac{A+B}{2} \sin \frac{A-B}{2}\\
\sin A-\sin B=2\sin \frac{A-B}{2} \cos \frac{A+B}{2}$

Proving trigonometric identities
Example:
Prove $\Large \frac{\sec \theta+\csc \theta}{1+\cot^2 \theta}=\frac{\sec \theta+2\sin \theta}{1+\cot \theta}$.



The proof from LHS to RHS is similar. A third way is to prove LHS - RHS = 0.

Geometric/graphical proof
[later]

When to reject solutions
If $\csc \theta+7\cot \theta=4$, where $0<\theta<\pi$, find $\theta$.
$\csc \theta+7\cot \theta=4$
$7\cot \theta=4-\csc \theta$
$49\cot^2 \theta=16-8\csc \theta+\csc^2 \theta$
$49(1-\csc^2 \theta)=16-8\csc \theta+\csc^2 \theta$
$48\csc^2 \theta+8\csc \theta-65=0$
$\Large \csc \theta=\frac{13}{12}, \frac{-5}{4}$ (rejected because sin > 0 when $0<\theta<\pi$)
$\sin \theta = \frac{12}{13}$
$\theta$ = 1.18 or 1.97 (rejected because $\cot \theta > 0$, meaning $\theta$ is in quadrant I.)

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It is given that $\sin \theta+\cos \theta=\frac{7}{13}$, where $\frac{\pi}{2} < \theta < \pi$.
Then $\sin \theta-\cos \theta=\frac{17}{13}$
Find  $\sin \theta$ and  $\cos \theta$.

Method I:
$\sin \theta=\frac{1}{2}[(\sin \theta+\cos \theta)+(\sin \theta-\cos \theta)]=\frac{12}{13}$
$\cos \theta=\frac{-5}{13}$

Method II:
$\sin \theta - \cos \theta=\frac{17}{13}$
$\sin^2 \theta = (\frac{17}{13}+\cos \theta)^2$
...
$\cos \theta = \frac{-5}{13} or \frac{-12}{13}$
when $\cos \theta=\frac{-5}{13}, \sin \theta=\frac{12}{13}$
when $\cos \theta=\frac{-12}{13}, \sin \theta=\frac{5}{13}$ (rejected because $\sin \theta+\cos \theta=\frac{7}{13}$, which implies $\sin \theta >1$.)

Moral of the story: Always check the solutions!

[pending] area, eq, s=rθ, a = ½ r^2θ, prove formulas, geometry, application