Question

Ctg300° + tg210° + sin(-225°) - cos 405°​

Answer 1

$- \frac{ \sqrt{3} }{3} + \frac{ \sqrt{3} }{3} + \frac{ \sqrt{2} }{2} - \frac{ \sqrt{2} }{2} = 0$

Ответ 0

Так как сокращаются

Answer 2

To evaluate the given trigonometric expressions, we can first find the exact values of trigonometric functions for the given angles and then perform the calculations.

1. $\cot(300°)$

The cotangent function is the reciprocal of the tangent function: $\cot(x) = \frac{1}{\tan(x)}$

To find $\cot(300°)$, first, we find $\tan(300°)$: $\tan(300°) = \tan(300° - 1 \times 360°) = \tan(-60°)$

Using the periodicity of the tangent function, $\tan(-60°) = \tan(60°)$.

Since $\tan(60°) = \sqrt{3}$,

$\cot(300°) = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$

2. $\tan(210°)$

To find $\tan(210°)$, we use the fact that $\tan(210°) = \tan(210° - 1 \times 180°) = \tan(-30°)$.

Since $\tan(-30°) = -\frac{\sqrt{3}}{3}$,

$\tan(210°) = -\frac{\sqrt{3}}{3}$

3. $\sin(-225°)$

Using the fact that the sine function is an odd function ($\sin(-x) = -\sin(x)$),

$\sin(-225°) = -\sin(225°)$

Using the periodicity of the sine function, $\sin(225°) = \sin(225° - 1 \times 360°) = \sin(-135°)$.

Since $\sin(-135°) = -\frac{\sqrt{2}}{2}$,

$\sin(-225°) = -\left(-\frac{\sqrt{2}}{2}\right) = \frac{\sqrt{2}}{2}$

4. $\cos(405°)$

Using the periodicity of the cosine function, $\cos(405°) = \cos(405° - 1 \times 360°) = \cos(45°)$.

Since $\cos(45°) = \frac{\sqrt{2}}{2}$,

$\cos(405°) = \frac{\sqrt{2}}{2}$

Now, the given expression becomes:

$\frac{\sqrt{3}}{3} - \frac{\sqrt{3}}{3} + \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}$

The terms with $\sqrt{3}$ and $\sqrt{2}$ cancel out, and the result is $0$.