Chebyshev’s Theorem.. please help.

Mathematics

1 answer:

Alex2 years ago

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Answer:

I don't know nothing bout this what grade

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Find the exact values of the six trigonometric functions of the given angle. If any are not defined, say "not defined." Do not u

Solnce55 [7]

Answer:

Step-by-step explanation:

Let's take a look at the given angle 135°

The sketch of the angle which corresponds to $-\dfrac{3\pi}{4}$ unit circle and can be seen in the attached image below;

The trigonometric ratios are as follows for an angle θ on the unit circle:

Trigonometric ratio related ratio on coordinate axes

sin θ $\dfrac{y}{1}$

cos θ $\dfrac{x}{1}$

tan θ $\dfrac{y}{x}$

csc θ $\dfrac{1}{y}$

sec θ $\dfrac{1}{x}$

cot θ $\dfrac{x}{y}$

From the sketch of the image attached below;

The six trigonometric ratio for 135° can be expressed as follows:

$sin (-\dfrac{3\pi}{4})= \dfrac{y}{1}$

$sin (-\dfrac{3\pi}{4})=- \dfrac{\sqrt{2}}{2}$

$cos (-\dfrac{3\pi}{4})= \dfrac{x}{1}$

$cos (-\dfrac{3\pi}{4})= -\dfrac{\sqrt{2}}{2}$

$tan (-\dfrac{3\pi}{4})= \dfrac{y}{x}$

$tan (-\dfrac{3\pi}{4})= \dfrac{-\dfrac{\sqrt{2}}{2}}{-\dfrac{\sqrt{2}}{2}}$

$tan (-\dfrac{3\pi}{4})= -\dfrac{\sqrt{2}}{2}} \times {-\dfrac{2}{\sqrt{2}}$

$tan (-\dfrac{3\pi}{4})= 1$

$csc (-\dfrac{3\pi}{4})= \dfrac{1}{y} \\ \\ csc (-\dfrac{3\pi}{4})=\dfrac{1}{-\dfrac{\sqrt{2}}{2}} \\ \\ csc=1 \times -\dfrac{2}{\sqrt{2}} \\ \\csc =-\sqrt{2}$

$sec (-\dfrac{3 \pi}{4})=\dfrac{1}{x} \\ \\ sec = \dfrac{1}{(-\dfrac{\sqrt{2}}{2})} \\ \\ sec = 1 \times -\dfrac{2}{\sqrt{2}} \\ \\ sec = - \sqrt{2}$

$cot(-\dfrac{3 \pi}{4}) = \dfrac{x}{y} \\ \\ cot(-\dfrac{3 \pi}{4}) = \dfrac{-\dfrac{\sqrt{2}}{2} }{-\dfrac{\sqrt{2}}{2}} \\ \\ cot(-\dfrac{3 \pi}{4})= -\dfrac{\sqrt{2}}{2} } \times {-\dfrac{2}{\sqrt{2}}} \\ \\ cot (-\dfrac{3 \pi}{4}) = 1$