Тригонометриядан қысқаша анықтамалық

$\theta<\dfrac{\pi}{2}$ немесе $0^{\circ}<\theta<90^{\circ}$

$\sin\theta=\dfrac{a}{c}=\dfrac{\text{қарсы жатқан катет}}{\text{гипотенуза}}$

$\cos\theta=\dfrac{b}{c}=\dfrac{\text{іргелес жатқан катет}}{\text{гипотенуза}}$

$\tan\theta=\dfrac{a}{b}=\dfrac{\text{қарсы жатқан катет}}{\text{іргелес жатқан катет}}$

$\csc\theta=\dfrac{c}{a}$

$\sec\theta=\dfrac{c}{b}$

$\cot\theta=\dfrac{b}{a}$

Қандай да бір $\theta$ бұрышы үшін:

$\sin\theta=\dfrac{y}{1}=y$

$\cos\theta=\dfrac{x}{1}=x$

$\tan\theta=\dfrac{y}{x}$

$\csc\theta=\dfrac{1}{y}$

$\sec\theta=\dfrac{1}{x}$

$\cot\theta=\dfrac{x}{y}$

$\theta$ аргументінің қабылдай алатын мәндері:

$\sin\theta$: $\theta\in(-\infty;+\infty)$

$\cos\theta$: $\theta\in(-\infty;+\infty)$

$\tan\theta$: $\theta\neq\dfrac{\pi}{2}+\pi n$, $n\in\mathbb{Z}$

$\csc\theta$: $\theta\neq\pi n$, $n\in\mathbb{Z}$

$\sec\theta$: $\theta\neq\dfrac{\pi}{2}+\pi n$, $n\in\mathbb{Z}$

$\cot\theta$: $\theta\neq\pi n$, $n\in\mathbb{Z}$

$-1\leq\sin\theta\leq1$

$-1\leq\cos\theta\leq1$

$-\infty<\tan\theta<+\infty$

$\csc\theta\geq1$ немесе $\csc\theta\leq-1$

$\sec\theta\geq1$ немесе $\sec\theta\leq-1$

$-\infty<\cot\theta<+\infty$

Егер $f(\theta+T)=f(\theta)$ теңдігі орындалса, онда $T$ функцияның периоды деп аталады. $\omega$ — аргументтің алдындағы коэффициент. Коэффициент өзгерген сайын функцияның периоды да өзгереді.

$\sin(\omega\theta)\Rightarrow T=\dfrac{2\pi}{\omega}$

$\cos(\omega\theta)\Rightarrow T=\dfrac{2\pi}{\omega}$

$\tan(\omega\theta)\Rightarrow T=\dfrac{\pi}{\omega}$

$\csc(\omega\theta)\Rightarrow T=\dfrac{2\pi}{\omega}$

$\sec(\omega\theta)\Rightarrow T=\dfrac{2\pi}{\omega}$

$\cot(\omega\theta)\Rightarrow T=\dfrac{\pi}{\omega}$

$\tan\theta=\dfrac{\sin\theta}{\cos\theta}$

$\cot\theta=\dfrac{\cos\theta}{\sin\theta}$

$\csc\theta=\dfrac{1}{\sin\theta}$

$\sec\theta=\dfrac{1}{\cos\theta}$

$\sin\theta=\dfrac{1}{\csc\theta}$

$\cos\theta=\dfrac{1}{\sec\theta}$

$\cot\theta=\dfrac{1}{\tan\theta}$

$\tan\theta=\dfrac{1}{\cot\theta}$

$\sin^2\theta+\cos^2\theta=1$

$\tan^2\theta+1=\dfrac{1}{\cos^2\theta}=\sec^2\theta$

$\cot^2\theta+1=\dfrac{1}{\sin^2\theta}=\csc^2\theta$

$\sin(-\theta)=-\sin\theta$

$\cos(-\theta)=\cos\theta$

$\tan(-\theta)=-\tan\theta$

$\csc(-\theta)=-\csc\theta$

$\sec(-\theta)=\sec\theta$

$\cot(-\theta)=-\cot\theta$

$n\in\mathbb{Z}$ үшін:

$\sin(\theta+2\pi n)=\sin\theta$

$\cos(\theta+2\pi n)=\cos\theta$

$\tan(\theta+\pi n)=\tan\theta$

$\csc(\theta+2\pi n)=\csc\theta$

$\sec(\theta+2\pi n)=\sec\theta$

$\cot(\theta+\pi n)=\cot\theta$

$\sin2\theta=2\sin\theta\cos\theta$

$\cos2\theta=\cos^2\theta-\sin^2\theta=2\cos^2\theta-1=1-2\sin^2\theta$

$\tan2\theta=\dfrac{2\tan\theta}{1-\tan^2\theta}$

Егер $x$ бұрышы градуспен, $t$ бұрышы радианмен өлшенсе:

$\dfrac{\pi}{180}=\dfrac{t}{x}\Rightarrow t=\dfrac{\pi x}{180}$ және $x=\dfrac{180t}{\pi}$

$\sin\dfrac{\theta}{2}=\pm\sqrt{\dfrac{1-\cos\theta}{2}}$

$\cos\dfrac{\theta}{2}=\pm\sqrt{\dfrac{1+\cos\theta}{2}}$

$\tan\dfrac{\theta}{2}=\pm\sqrt{\dfrac{1-\cos\theta}{1+\cos\theta}}$

$\sin^2\theta=\dfrac{1-\cos2\theta}{2}$

$\cos^2\theta=\dfrac{1+\cos2\theta}{2}$

$\tan^2\theta=\dfrac{1-\cos2\theta}{1+\cos2\theta}$

$\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta$

$\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta$

$\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta$

$\tan(\alpha+\beta)=\dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$

$\tan(\alpha-\beta)=\dfrac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}$

$\sin\alpha\sin\beta=\dfrac{1}{2}\left(\cos(\alpha-\beta)-\cos(\alpha+\beta)\right)$

$\cos\alpha\cos\beta=\dfrac{1}{2}\left(\cos(\alpha-\beta)+\cos(\alpha+\beta)\right)$

$\sin\alpha\cos\beta=\dfrac{1}{2}\left(\sin(\alpha+\beta)+\sin(\alpha-\beta)\right)$

$\cos\alpha\sin\beta=\dfrac{1}{2}\left(\sin(\alpha+\beta)-\sin(\alpha-\beta)\right)$

$\sin x+\sin y=2\sin\dfrac{x+y}{2}\cos\dfrac{x-y}{2}$

$\sin x-\sin y=2\sin\dfrac{x-y}{2}\cos\dfrac{x+y}{2}$

$\cos x+\cos y=2\cos\dfrac{x+y}{2}\cos\dfrac{x-y}{2}$

$\cos x-\cos y=-2\sin\dfrac{x+y}{2}\sin\dfrac{x-y}{2}$

$\tan x+\tan y=\dfrac{\sin(x+y)}{\cos x\cos y}$

$\tan x-\tan y=\dfrac{\sin(x-y)}{\cos x\cos y}$

$\cot x+\cot y=\dfrac{\sin(x+y)}{\sin x\sin y}$

$\cot x-\cot y=\dfrac{\sin(y-x)}{\sin x\sin y}$

$\sin\left(\dfrac{\pi}{2}-\theta\right)=\cos\theta$

$\cos\left(\dfrac{\pi}{2}-\theta\right)=\sin\theta$

$\csc\left(\dfrac{\pi}{2}-\theta\right)=\sec\theta$

$\sec\left(\dfrac{\pi}{2}-\theta\right)=\csc\theta$

$\tan\left(\dfrac{\pi}{2}-\theta\right)=\cot\theta$

$\cot\left(\dfrac{\pi}{2}-\theta\right)=\tan\theta$

Шеңбер бойындағы кез келген $(x;y)$ нүктесінің координаталарына $\cos\theta=x$ және $\sin\theta=y$ мәндері сәйкес келеді.

Мысалы: $\cos\dfrac{5\pi}{3}=\dfrac{1}{2}$ және $\sin\dfrac{5\pi}{3}=-\dfrac{\sqrt{3}}{2}$.

$y=\arcsin x$ функциясы $x=\sin y$ теңдеуіне мәндес.

$y=\arccos x$ функциясы $x=\cos y$ теңдеуіне мәндес.

$y=\arctan x$ функциясы $x=\tan y$ теңдеуіне мәндес.

$y=\arcsin x$: $-1\leq x\leq1$, $-\dfrac{\pi}{2}\leq y\leq\dfrac{\pi}{2}$.

$y=\arccos x$: $-1\leq x\leq1$, $0\leq y\leq\pi$.

$y=\arctan x$: $-\infty<x<+\infty$, $-\dfrac{\pi}{2}<y<\dfrac{\pi}{2}$.

$\cos(\arccos x)=x$

$\sin(\arcsin x)=x$

$\tan(\arctan x)=x$

$\arccos(\cos\theta)=\theta$

$\arcsin(\sin\theta)=\theta$

$\arctan(\tan\theta)=\theta$

$\sin^{-1}x=\arcsin x$

$\cos^{-1}x=\arccos x$

$\tan^{-1}x=\arctan x$

$\dfrac{\sin\alpha}{a}=\dfrac{\sin\beta}{b}=\dfrac{\sin\gamma}{c}$

$a^2=b^2+c^2-2bc\cos\alpha$

$b^2=a^2+c^2-2ac\cos\beta$

$c^2=a^2+b^2-2ab\cos\gamma$

$\dfrac{a+b}{c}=\dfrac{\cos\dfrac{\alpha-\beta}{2}}{\sin\dfrac{\gamma}{2}}$

$\dfrac{a-b}{a+b}=\dfrac{\tan\dfrac{\alpha-\beta}{2}}{\tan\dfrac{\alpha+\beta}{2}}$

$\dfrac{b-c}{b+c}=\dfrac{\tan\dfrac{\beta-\gamma}{2}}{\tan\dfrac{\beta+\gamma}{2}}$

$\dfrac{a-c}{a+c}=\dfrac{\tan\dfrac{\alpha-\gamma}{2}}{\tan\dfrac{\alpha+\gamma}{2}}$