Analytic Trigonometry

Reciprocal:  $\csc\theta=\dfrac{1}{\sin\theta}$,  $\sec\theta=\dfrac{1}{\cos\theta}$,  $\cot\theta=\dfrac{1}{\tan\theta}$.

Quotient:  $\tan\theta=\dfrac{\sin\theta}{\cos\theta}$,  $\cot\theta=\dfrac{\cos\theta}{\sin\theta}$.

Pythagorean:  $\sin^2\theta+\cos^2\theta=1$,  $1+\tan^2\theta=\sec^2\theta$,  $1+\cot^2\theta=\csc^2\theta$.

Even/Odd:  $\sin(-\theta)=-\sin\theta$,  $\cos(-\theta)=\cos\theta$,  $\tan(-\theta)=-\tan\theta$.

Cofunction:  $\sin\!\left(\tfrac{\pi}{2}-\theta\right)=\cos\theta$,  $\tan\!\left(\tfrac{\pi}{2}-\theta\right)=\cot\theta$.

Work one side only (usually the messier one) until it matches the other. Do not move terms across the $=$ sign as if solving an equation. Useful moves:

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• Convert to $\sin$ and $\cos$.
• Get a common denominator.
• Use a Pythagorean identity, or multiply by a conjugate such as $\dfrac{1+\sin\theta}{1+\sin\theta}$.
• Factor, or split a fraction into separate terms.

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• Isolate a single trig function, or move all terms to one side and factor.
• Quadratic type ($a\,u^2+b\,u+c=0$ with $u=\sin x$, etc.): factor or use the quadratic formula.
• Use identities to reduce to one function before factoring.
• On $[0,2\pi)$: list only the angles in that interval. All solutions: add the period: $+2\pi n$ for $\sin,\cos$; $+\pi n$ for $\tan$ (here $n$ is any integer).
• If you squared or divided, check for extraneous solutions in the original equation.

Watch the sign flip: $\cos$ and $\tan$ use the opposite sign of the one in $(A\pm B)$.

Double-angle:

Power-reducing:  $\sin^2\theta=\dfrac{1-\cos2\theta}{2}$,  $\cos^2\theta=\dfrac{1+\cos2\theta}{2}$.

Half-angle:  $\sin\dfrac{\theta}{2}=\pm\sqrt{\dfrac{1-\cos\theta}{2}}$,  $\cos\dfrac{\theta}{2}=\pm\sqrt{\dfrac{1+\cos\theta}{2}}$,  $\tan\dfrac{\theta}{2}=\dfrac{1-\cos\theta}{\sin\theta}=\dfrac{\sin\theta}{1+\cos\theta}$.

The $\pm$ is chosen by the quadrant of $\theta/2$.

Product-to-sum:

Sum-to-product:

To evaluate something like $\sin(\arccos u)$, let $\theta=\arccos u$, so $\cos\theta=u$ with $\theta$ in the range of arccosine ($[0,\pi]$). Draw a right triangle (or use a Pythagorean identity) to find the other functions, respecting the sign forced by the range. The same idea handles $\sin(2\arctan x)$, $\cos(\arcsin a+\arccos b)$, etc.

Repeatedly using $\sin 2\alpha=2\sin\alpha\cos\alpha$ collapses a whole product of doubling-angle cosines into a single ratio of sines. For any $\theta$ with $\sin\theta\neq0$,

The trick: multiply the product by $2^n\sin\theta$ and peel off one factor at a time. This single identity powers a surprising number of “impossible-looking” exact values.

Opposite, adjacent and hypotenuse are named from the angle.

Not every “nice” angle needs a sum/difference split --- some fall out of a polynomial. Let $\theta=36^\circ$, so $5\theta=180^\circ$, giving $2\theta=180^\circ-3\theta$ and hence $\cos2\theta=-\cos3\theta$. With $x=\cos\theta$ and the identities $\cos2\theta=2x^2-1$, $\cos3\theta=4x^3-3x$:

Since $x=\cos36^\circ>0$, discard $x=-1$ and solve $4x^2-2x-1=0$:

This is the golden-ratio value $\tfrac{\varphi}{2}$, where $\varphi=\tfrac{1+\sqrt5}{2}$.

For an arithmetic progression of angles $a,\,a+d,\,a+2d,\dots$, the telescoping idea from sum-to-product gives closed forms. For $\sin\tfrac{d}{2}\neq0$,

Why it works: multiply each term by $2\sin\tfrac{d}{2}$ and apply $2\sin\tfrac{d}{2}\cos(a+kd)=\sin\!\big(a+kd+\tfrac{d}{2}\big)-\sin\!\big(a+kd-\tfrac{d}{2}\big)$. Consecutive terms cancel in pairs, leaving only the first and last.

The addition law, read backward, adds arctangents:

valid outright when $ab<1$. This yields elegant collapses such as

Chaining the rule gives Machin's formula, historically used to compute $\pi$: