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Тригонометрія

$$ \sin^2{(x)} + \cos^2{(x)} = 1 $$ $$ {\rm tg}{(x)} \cdot {\rm ctg}{(x)} = 1 $$

$$ {\rm tg}(x) = \frac{\sin (x)}{\cos (x)} $$ $$ {\rm ctg}(x) = \frac{\cos (x)}{\sin (x)} $$

$$ 1 + {\rm tg}^2(x) = \frac{1}{\cos^2 (x)} $$ $$ 1 + {\rm ctg}^2(x) = \frac{1}{\sin^2(x)} $$

$$ 2 \sin^2(x) = 1 − \cos(2x) $$ $$ 2 \cos^2(x) = 1 + \cos(2x) $$

$$ 4 \sin^3(x) = 3 \sin(x) − \sin(3x) $$ $$ 4 \cos^3(x) = 3 \cos(x) + \cos(3x) $$

$$ 8 \sin^4(x) = 3 − 4 \cos(2x) + \cos(4x) $$ $$ 8 \cos^4(x) = 3 + 4 \cos(2x) + \cos(4x) $$

$$ \sin (a+b) = \sin(a) \cdot \cos(b) + \cos(a) \cdot \sin(b) $$ $$ \cos(a+b) = \cos(a) \cdot \cos(b) − \sin(a) \cdot \sin(b) $$

$$ \sin(a−b) = \sin(a) \cdot \cos(b) − \cos(a) \cdot \sin(b) $$ $$ \cos(a−b) = \cos(a) \cdot \cos(b) + \sin(a) \cdot \sin(b) $$

$$ {\rm tg}(a+b) = \frac{ {\rm tg}(a) + {\rm tg}(b) }{ 1 − {\rm tg}(a) \cdot {\rm tg(b)} } $$ $$ {\rm ctg}(a+b) = \frac{ {\rm ctg}(a) \cdot {\rm ctg}(b) − 1 }{ {\rm ctg}(a) + {\rm ctg}(b) } $$

$$ {\rm tg}(a−b) = \frac{{\rm tg}(a) − {\rm tg}(b) }{ 1 + {\rm tg(a)} \cdot {\rm tg}(b) } $$ $$ {\rm ctg}(a-b) = \frac{ {\rm ctg}(a) \cdot {\rm ctg}(b) + 1 }{ {\rm ctg}(a) - {\rm ctg}(b) } $$

$$ \sin(2x) = 2 \sin(x) \cdot \cos(x) = \frac{ 2 {\rm tg}(x) }{ 1 + {\rm tg}^2(x) } $$ $$ \cos(2x) = \cos^2(x) − \sin^2(x) = 2 \cos^2(x) − 1 = 1 − 2 \sin^2(x) = $$ $$ = \frac{ 1 − {\rm tg}^2(x) }{ 1 + {\rm tg}^2(x) } $$

$$ {\rm tg}(2x) = \frac{ 2 {\rm tg}(x) }{ 1 − {\rm tg}^2(x) } = \frac{ 2 }{ {\rm ctg}(x) − {\rm tg}(x) } $$ $$ {\rm ctg}(2x) = \frac{ {\rm ctg}^2(x) − 1 }{ 2 {\rm ctg}(x) } = \frac{ {\rm ctg}(x) − {\rm tg}(x) }{ 2 } $$

$$ \sin(3x) = 3 \sin(x) − 4 \sin^3 (x) $$ $$ \cos(3x) = 4 \cos^3(x) − 3 \cos(x) $$

$$ {\rm tg}(3x) = \frac{ 3 {\rm tg}(x) − {\rm tg}^3(x) }{ 1 − 3 {\rm tg}^2(x) } $$ $$ {\rm ctg}(3x) = \frac{ {\rm ctg}^3(x) − 3 {\rm ctg}(x) }{ 3 {\rm ctg}^2(x) − 1 } $$

$$ \sin(a) \cdot \sin(b) = 0.5 ( \cos(a−b) − \cos(a+b) ) $$ $$ \sin(a) \cdot \cos(b) = 0.5 ( \sin(a−b) + \sin(a+b) ) $$

$$ \cos(a) \cdot \cos(b) = 0.5 ( \cos(a−b) + \cos(a+b) ) $$ $$ \cos(a) \cdot \sin(b) = 0.5 ( \sin(a+b) − \sin(a−b) ) $$

$$ {\rm tg}(a) \cdot {\rm tg}(b) = \frac{ {\rm tg}(a) + {\rm tg}(b) }{ {\rm ctg}(a) + {\rm ctg}(b) } $$ $$ {\rm ctg}(a) \cdot {\rm ctg}(b) = \frac{ {\rm ctg}(a) + {\rm ctg}(b) }{ {\rm tg}(a) + {\rm tg}(b) } $$

$$ \sin(a+b) \cdot \sin(a−b) = \cos^2(b) − \cos^2(a) $$ $$ \cos(a+b) \cdot \cos(a−b) = \cos^2(b) − \sin^2(a) $$

$$ \sin(a) + \sin(b) = 2 \sin[(a+b)/2] \cdot \cos[(a−b)/2] $$ $$ \cos(a) + \cos(b) = 2 \cos[(a+b)/2] \cdot \cos[(a−b)/2] $$

$$ \sin(a) − \sin(b) = 2 \cos[(a+b)/2] \cdot \sin[(a−b)/2] $$ $$ \cos(a) − \cos(b) = − 2 \sin[(a+b)/2] \cdot \sin[(a−b)/2] $$

$$ \cos(a) + \sin(a) = \sqrt{2} \cdot \sin(\pi/4+a) = \sqrt{2} \cdot \cos(\pi/4−a) $$ $$ \cos(a) − \sin(a) = \sqrt{2} \cdot \sin(\pi/4−a) = \sqrt{2} \cdot \cos(\pi/4+a) $$

$$ {\rm tg}(a) \pm {\rm tg}(b) = \frac{ \sin(a \pm b) }{ \cos(a) \cdot \cos(b) } $$ $$ {\rm ctg}(a) \pm {\rm ctg}(b) = \pm \frac{ \sin(a \pm b) }{ \sin(a) \cdot \sin(b) } $$

$$ {\rm tg}(a) + {\rm ctg}(b) = \frac{ \cos(a−b) }{ \cos(a) \cdot \sin(b) } $$ $$ {\rm ctg}(a) − {\rm tg}(b) = \frac{\cos(a+b) }{ \sin(a) \cdot \cos(b) } $$

$$\sin^4(x) - \cos^4(x) = \sin^2(x) - \cos^2(x) = \cos(2x)$$ $$ \left( \sin(x) + \cos(x) \right)^2 = 1 + \sin(2x) $$

$$\sin(x) \pm \cos(x) = \sqrt{2} \pm \sin\left(x \pm \frac{\pi}{4}\right) $$ $$ \cos^6(x) + \sin^6(x) = \frac{5 + 3 \cos(4x)}{8} = \frac{1 + 3 \cos^2(2x)}{4} $$

$$\sin(x)\pm \sqrt{3} \cos(x) = 2 \sin\left(x \pm \frac{\pi}{3}\right) $$ $$ \cos^6(x) - \sin^6(x) = \frac{1}{16} ( 15 \cos(2x) + \cos(6x) ) $$

$$\cos(x)\pm \sqrt{3} \sin(x) = 2 \sin\left(x \pm \frac{\pi}{6}\right) $$ $$ \cos^8(x) - \sin^8(x) = \frac{1}{4} \cos(2x) \left( 3 + \cos(4x) \right) $$

$$ \sin^2{(x)} + \cos^2{(x)} = 1 $$		$$ {\rm tg}{(x)} \cdot {\rm ctg}{(x)} = 1 $$
$$ {\rm tg}(x) = \frac{\sin (x)}{\cos (x)} $$		$$ {\rm ctg}(x) = \frac{\cos (x)}{\sin (x)} $$
$$ 1 + {\rm tg}^2(x) = \frac{1}{\cos^2 (x)} $$		$$ 1 + {\rm ctg}^2(x) = \frac{1}{\sin^2(x)} $$

$$ 2 \sin^2(x) = 1 − \cos(2x) $$		$$ 2 \cos^2(x) = 1 + \cos(2x) $$
$$ 4 \sin^3(x) = 3 \sin(x) − \sin(3x) $$		$$ 4 \cos^3(x) = 3 \cos(x) + \cos(3x) $$
$$ 8 \sin^4(x) = 3 − 4 \cos(2x) + \cos(4x) $$		$$ 8 \cos^4(x) = 3 + 4 \cos(2x) + \cos(4x) $$

$$ \sin (a+b) = \sin(a) \cdot \cos(b) + \cos(a) \cdot \sin(b) $$		$$ \cos(a+b) = \cos(a) \cdot \cos(b) − \sin(a) \cdot \sin(b) $$
$$ \sin(a−b) = \sin(a) \cdot \cos(b) − \cos(a) \cdot \sin(b) $$		$$ \cos(a−b) = \cos(a) \cdot \cos(b) + \sin(a) \cdot \sin(b) $$
$$ {\rm tg}(a+b) = \frac{ {\rm tg}(a) + {\rm tg}(b) }{ 1 − {\rm tg}(a) \cdot {\rm tg(b)} } $$		$$ {\rm ctg}(a+b) = \frac{ {\rm ctg}(a) \cdot {\rm ctg}(b) − 1 }{ {\rm ctg}(a) + {\rm ctg}(b) } $$
$$ {\rm tg}(a−b) = \frac{{\rm tg}(a) − {\rm tg}(b) }{ 1 + {\rm tg(a)} \cdot {\rm tg}(b) } $$		$$ {\rm ctg}(a-b) = \frac{ {\rm ctg}(a) \cdot {\rm ctg}(b) + 1 }{ {\rm ctg}(a) - {\rm ctg}(b) } $$

$$ \sin(2x) = 2 \sin(x) \cdot \cos(x) = \frac{ 2 {\rm tg}(x) }{ 1 + {\rm tg}^2(x) } $$		$$ \cos(2x) = \cos^2(x) − \sin^2(x) = 2 \cos^2(x) − 1 = 1 − 2 \sin^2(x) = $$ $$ = \frac{ 1 − {\rm tg}^2(x) }{ 1 + {\rm tg}^2(x) } $$
$$ {\rm tg}(2x) = \frac{ 2 {\rm tg}(x) }{ 1 − {\rm tg}^2(x) } = \frac{ 2 }{ {\rm ctg}(x) − {\rm tg}(x) } $$		$$ {\rm ctg}(2x) = \frac{ {\rm ctg}^2(x) − 1 }{ 2 {\rm ctg}(x) } = \frac{ {\rm ctg}(x) − {\rm tg}(x) }{ 2 } $$
$$ \sin(3x) = 3 \sin(x) − 4 \sin^3 (x) $$		$$ \cos(3x) = 4 \cos^3(x) − 3 \cos(x) $$
$$ {\rm tg}(3x) = \frac{ 3 {\rm tg}(x) − {\rm tg}^3(x) }{ 1 − 3 {\rm tg}^2(x) } $$		$$ {\rm ctg}(3x) = \frac{ {\rm ctg}^3(x) − 3 {\rm ctg}(x) }{ 3 {\rm ctg}^2(x) − 1 } $$

$$ \sin(a) \cdot \sin(b) = 0.5 ( \cos(a−b) − \cos(a+b) ) $$		$$ \sin(a) \cdot \cos(b) = 0.5 ( \sin(a−b) + \sin(a+b) ) $$
$$ \cos(a) \cdot \cos(b) = 0.5 ( \cos(a−b) + \cos(a+b) ) $$		$$ \cos(a) \cdot \sin(b) = 0.5 ( \sin(a+b) − \sin(a−b) ) $$
$$ {\rm tg}(a) \cdot {\rm tg}(b) = \frac{ {\rm tg}(a) + {\rm tg}(b) }{ {\rm ctg}(a) + {\rm ctg}(b) } $$		$$ {\rm ctg}(a) \cdot {\rm ctg}(b) = \frac{ {\rm ctg}(a) + {\rm ctg}(b) }{ {\rm tg}(a) + {\rm tg}(b) } $$
$$ \sin(a+b) \cdot \sin(a−b) = \cos^2(b) − \cos^2(a) $$		$$ \cos(a+b) \cdot \cos(a−b) = \cos^2(b) − \sin^2(a) $$

$$ \sin(a) + \sin(b) = 2 \sin[(a+b)/2] \cdot \cos[(a−b)/2] $$		$$ \cos(a) + \cos(b) = 2 \cos[(a+b)/2] \cdot \cos[(a−b)/2] $$
$$ \sin(a) − \sin(b) = 2 \cos[(a+b)/2] \cdot \sin[(a−b)/2] $$		$$ \cos(a) − \cos(b) = − 2 \sin[(a+b)/2] \cdot \sin[(a−b)/2] $$
$$ \cos(a) + \sin(a) = \sqrt{2} \cdot \sin(\pi/4+a) = \sqrt{2} \cdot \cos(\pi/4−a) $$		$$ \cos(a) − \sin(a) = \sqrt{2} \cdot \sin(\pi/4−a) = \sqrt{2} \cdot \cos(\pi/4+a) $$
$$ {\rm tg}(a) \pm {\rm tg}(b) = \frac{ \sin(a \pm b) }{ \cos(a) \cdot \cos(b) } $$		$$ {\rm ctg}(a) \pm {\rm ctg}(b) = \pm \frac{ \sin(a \pm b) }{ \sin(a) \cdot \sin(b) } $$
$$ {\rm tg}(a) + {\rm ctg}(b) = \frac{ \cos(a−b) }{ \cos(a) \cdot \sin(b) } $$		$$ {\rm ctg}(a) − {\rm tg}(b) = \frac{\cos(a+b) }{ \sin(a) \cdot \cos(b) } $$

$$\sin^4(x) - \cos^4(x) = \sin^2(x) - \cos^2(x) = \cos(2x)$$		$$ \left( \sin(x) + \cos(x) \right)^2 = 1 + \sin(2x) $$
$$\sin(x) \pm \cos(x) = \sqrt{2} \pm \sin\left(x \pm \frac{\pi}{4}\right) $$		$$ \cos^6(x) + \sin^6(x) = \frac{5 + 3 \cos(4x)}{8} = \frac{1 + 3 \cos^2(2x)}{4} $$
$$\sin(x)\pm \sqrt{3} \cos(x) = 2 \sin\left(x \pm \frac{\pi}{3}\right) $$		$$ \cos^6(x) - \sin^6(x) = \frac{1}{16} ( 15 \cos(2x) + \cos(6x) ) $$
$$\cos(x)\pm \sqrt{3} \sin(x) = 2 \sin\left(x \pm \frac{\pi}{6}\right) $$		$$ \cos^8(x) - \sin^8(x) = \frac{1}{4} \cos(2x) \left( 3 + \cos(4x) \right) $$