Question

1)sin(-135°) ×cos390°×tg405°×ctg(-330°) 2)sin(-225°) ×cos(-480°) ×ctg(-420°) ×tg300°​

Answer 1

Ответ:

1) $\boxed{\sin(-135^{\circ}) \cdot \cos(390^{\circ}) \cdot \rm tg \ (405^{\circ}) \cdot ctg \ (-330^{\circ}) = -\dfrac{3\sqrt{2} }{4}}$

2) $\boxed{\sin(-225^{\circ}) \cdot \cos(-480^{\circ}) \cdot \rm ctg \ (-420^{\circ}) \cdot tg \ (300^{\circ}) = - \dfrac{\sqrt{2} }{4}}$

Примечание:

Косинус четная тригонометрическая функция, а все остальные тригонометрические функции нечетные, то есть:

$\cos (-\alpha ) = \cos \alpha$

$\sin (-\alpha ) = - \sin \alpha$

$\rm tg \ (- \alpha) = - tg \ \alpha$

$\rm ctg \ (- \alpha) = - ctg \ \alpha$

По формуле приведения:

$\sin \alpha = \sin (180^{\circ} - \alpha)$

$\sin 135^{\circ} = \sin (180^{\circ} - 135^{\circ}) = \sin 45^{\circ} = \dfrac{\sqrt{2} }{2}$

Объяснение:

Период функций любой тригонометрической функции может быть равен: T = 360°.

1) $\sin(-135^{\circ}) \cdot \cos(390^{\circ}) \cdot \rm tg \ (405^{\circ}) \cdot ctg \ (-330^{\circ}) =$

$-1 \cdot (-1) \cdot \sin(180^{\circ} -135^{\circ}) \cdot \cos(390^{\circ} - 360^{\circ}) \cdot \rm tg \ (405^{\circ} - 360^{\circ}) \cdot ctg \ (330^{\circ} - 360^{\circ}) =$

$= - \sin 45^{\circ} \cdot \cos(30^{\circ}) \cdot \rm tg \ (45^{\circ}) \cdot ctg \ ( 30^{\circ}) = - \dfrac{\sqrt{2} }{2} \cdot \dfrac{\sqrt{3} }{2 } \cdot \dfrac{1}{1} \cdot \dfrac{\sqrt{3} }{1} = -\dfrac{3\sqrt{2} }{4}$

2) $\sin(-225^{\circ}) \cdot \cos(-480^{\circ}) \cdot \rm ctg \ (-420^{\circ}) \cdot tg \ (300^{\circ}) =$

$= -1 \cdot (-1)\cdot \sin(225^{\circ}) \cdot \cos(480^{\circ}) \cdot \rm ctg \ (420^{\circ}) \cdot tg \ (300^{\circ}) =$

$= \sin(225^{\circ} - 360^{\circ}) \cdot \cos(480^{\circ}-360^{\circ}) \cdot \rm ctg \ (420^{\circ} - 360^{\circ}) \cdot tg \ (300^{\circ} - 360^{\circ}) =$

$=\sin(-135^{\circ}) \cdot \cos(120^{\circ}) \cdot \rm ctg \ (60^{\circ}) \cdot tg \ (- 60^{\circ}) =$

$= -1 \cdot (-1) \cdot (-1) \cdot \sin(180^{\circ}-135^{\circ}) \cdot \cos(180^{\circ}-120^{\circ}) \cdot \rm ctg \ (60^{\circ}) \cdot tg \ (60^{\circ}) =$

$= - \sin (45^{\circ}) \cdot \cos 60^{\circ} = -\dfrac{\sqrt{2} }{2} \cdot \dfrac{1}{2} = - \dfrac{\sqrt{2} }{4}$

Answer 2

To calculate these trigonometric expressions, we'll use the following trigonometric identities:

1. sin(-x) = -sin(x)
2. cos(-x) = cos(x)
3. tan(-x) = -tan(x)
4. cot(-x) = -cot(x)

Let's calculate each expression step by step:

1. sin(-135°) × cos(390°) × tan(405°) × cot(-330°)

Using the identities above:

sin(-135°) = -sin(135°) cos(390°) = cos(30°) because cos(390°) = cos(360° + 30°) tan(405°) = tan(45°) because tan(405°) = tan(360° + 45°) cot(-330°) = -cot(30°) because cot(-330°) = -cot(360° - 30°)

Now, let's calculate each part:

-sin(135°) = -(-√2/2) = √2/2 cos(30°) = √3/2 tan(45°) = 1 -cot(30°) = -√3

Now, let's multiply these values together:

(√2/2) * (√3/2) * 1 * (-√3) = -√6/4

So, the value of the expression is -√6/4.

2. sin(-225°) × cos(-480°) × cot(-420°) × tan(300°)

Using the identities above:

sin(-225°) = -sin(225°) cos(-480°) = cos(480°) cot(-420°) = -cot(420°) tan(300°) = tan(300°)

Now, let's calculate each part:

-sin(225°) = -(√2/2) = -√2/2 cos(480°) = cos(480° - 360°) = cos(120°) = -√3/2 -cot(420°) = -cot(420° - 360°) = -cot(60°) = -√3 tan(300°) = tan(300° - 180°) = tan(120°) = -√3

Now, let's multiply these values together:

(-√2/2) * (-√3/2) * (-√3) * (-√3) = 9/4

So, the value of the expression is 9/4.