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i f N(-7,-1) is a point on the terminal side of ∅ in standard form, find the exact values of the trigonometric functions of ∅.

Sagot :

MrRoyal

Answer:

[tex]sin\ \varnothing = \frac{-1}{10}\sqrt 2[/tex]

[tex]cos\ \varnothing = \frac{-7}{10}\sqrt 2[/tex]

[tex]tan\ \varnothing = \frac{1}{7}[/tex]

[tex]cot\ \varnothing = 7[/tex]

[tex]sec\ \varnothing = \frac{-5}{7}\sqrt 2[/tex]

[tex]csc\ \varnothing = -5\sqrt 2[/tex]

Step-by-step explanation:

Given

[tex]N = (-7,-1)[/tex] --- terminal side of [tex]\varnothing[/tex]

Required

Determine the values of trigonometric functions of [tex]\varnothing[/tex].

For [tex]\varnothing[/tex], the trigonometry ratios are:

[tex]sin\ \varnothing = \frac{y}{r}[/tex]       [tex]cos\ \varnothing = \frac{x}{r}[/tex]       [tex]tan\ \varnothing = \frac{y}{x}[/tex]

[tex]cot\ \varnothing = \frac{x}{y}[/tex]       [tex]sec\ \varnothing = \frac{r}{x}[/tex]       [tex]csc\ \varnothing = \frac{r}{y}[/tex]

Where:

[tex]r^2 = x^2 + y^2[/tex]

[tex]r = \sqrt{x^2 + y^2[/tex]

In [tex]N = (-7,-1)[/tex]

[tex]x = -7[/tex] and [tex]y = -1[/tex]

So:

[tex]r = \sqrt{(-7)^2 + (-1)^2[/tex]

[tex]r = \sqrt{50[/tex]

[tex]r = \sqrt{25 * 2[/tex]

[tex]r = \sqrt{25} * \sqrt 2[/tex]

[tex]r = 5 * \sqrt 2[/tex]

[tex]r = 5 \sqrt 2[/tex]

Solving the trigonometry functions

[tex]sin\ \varnothing = \frac{y}{r}[/tex]

[tex]sin\ \varnothing = \frac{-1}{5\sqrt 2}[/tex]

Rationalize:

[tex]sin\ \varnothing = \frac{-1}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}[/tex]

[tex]sin\ \varnothing = \frac{-\sqrt 2}{5*2}[/tex]

[tex]sin\ \varnothing = \frac{-\sqrt 2}{10}[/tex]

[tex]sin\ \varnothing = \frac{-1}{10}\sqrt 2[/tex]

[tex]cos\ \varnothing = \frac{x}{r}[/tex]

[tex]cos\ \varnothing = \frac{-7}{5\sqrt 2}[/tex]

Rationalize

[tex]cos\ \varnothing = \frac{-7}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}[/tex]

[tex]cos\ \varnothing = \frac{-7*\sqrt 2}{5*2}[/tex]

[tex]cos\ \varnothing = \frac{-7\sqrt 2}{10}[/tex]

[tex]cos\ \varnothing = \frac{-7}{10}\sqrt 2[/tex]

[tex]tan\ \varnothing = \frac{y}{x}[/tex]

[tex]tan\ \varnothing = \frac{-1}{-7}[/tex]

[tex]tan\ \varnothing = \frac{1}{7}[/tex]

[tex]cot\ \varnothing = \frac{x}{y}[/tex]

[tex]cot\ \varnothing = \frac{-7}{-1}[/tex]

[tex]cot\ \varnothing = 7[/tex]

[tex]sec\ \varnothing = \frac{r}{x}[/tex]

[tex]sec\ \varnothing = \frac{5\sqrt 2}{-7}[/tex]

[tex]sec\ \varnothing = \frac{-5}{7}\sqrt 2[/tex]

[tex]csc\ \varnothing = \frac{r}{y}[/tex]

[tex]csc\ \varnothing = \frac{5\sqrt 2}{-1}[/tex]

[tex]csc\ \varnothing = -5\sqrt 2[/tex]

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