The terminal side of $θ$ $theta$ in standard position contains (-6,8), how do you find the exact values of the six trigonometric functions of $θ$ $theta$ ?

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The terminal side of $θ$ $theta$ in standard position contains (-6,8), how do you find the exact values of the six trigonometric functions of $θ$ $theta$ ?

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Answer 1 · 2018-02-28T23:26:33

See explanation.

Explanation:

If the point in the angle's terminal side is $P = (x, y)$ $P=(x,y)$ then the trigonometric functions can be calculated as:

$sin α = \frac{y}{r}$ $sin alpha=y/r$

$cos α = \frac{x}{r}$ $cos alpha=x/r$

$tan α = \frac{y}{x}$ $tan alpha=y/x$

$cot α = \frac{x}{y}$ $cot alpha=x/y$

$sec α = \frac{r}{x}$ $sec alpha=r/x$

$csc α = \frac{r}{y}$ $csc alpha=r/y$

where $r = \sqrt{x^{2} + y^{2}}$ $r=sqrt(x^2+y^2)$

For the given point we have:

$r = \sqrt{{(- 6)}^{2} + 8^{2}} = \sqrt{36 + 64} = 10$ $r=sqrt((-6)^2+8^2)=sqrt(36+64)=10$

So the functions are:

$sin α = \frac{8}{10} = \frac{4}{5}$ $sin alpha=8/10=4/5$

$cos α = \frac{- 6}{10} = - \frac{3}{5}$ $cos alpha=(-6)/10=-3/5$

$tan α = \frac{8}{- 6} = - \frac{4}{3}$ $tan alpha=8/(-6)=-4/3$

$cot α = \frac{- 6}{8} = - \frac{3}{4}$ $cot alpha=(-6)/8=-3/4$

$sec α = \frac{10}{- 6} = - \frac{5}{3} = - 1 \frac{2}{3}$ $sec alpha=10/(-6)=-5/3=-1 2/3$

$csc α = \frac{10}{8} = \frac{5}{4} = 1 \frac{1}{4}$ $csc alpha=10/8=5/4=1 1/4$

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The terminal side of $θ$ $theta$ in standard position contains (-6,8), how do you find the exact values of the six trigonometric functions of $θ$ $theta$ ?

1 Answer

Explanation:

$sin α = \frac{y}{r}$ $sin alpha=y/r$

$cos α = \frac{x}{r}$ $cos alpha=x/r$

$tan α = \frac{y}{x}$ $tan alpha=y/x$

$cot α = \frac{x}{y}$ $cot alpha=x/y$

$sec α = \frac{r}{x}$ $sec alpha=r/x$

$csc α = \frac{r}{y}$ $csc alpha=r/y$

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The terminal side of θtheta in standard position contains (-6,8), how do you find the exact values of the six trigonometric functions of θtheta?

1 Answer

Explanation:

sinα=yrsin alpha=y/r

cosα=xrcos alpha=x/r

tanα=yxtan alpha=y/x

cotα=xycot alpha=x/y

secα=rxsec alpha=r/x

cscα=rycsc alpha=r/y

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Impact of this question

The terminal side of $θ$ $theta$ in standard position contains (-6,8), how do you find the exact values of the six trigonometric functions of $θ$ $theta$ ?

$sin α = \frac{y}{r}$ $sin alpha=y/r$

$cos α = \frac{x}{r}$ $cos alpha=x/r$

$tan α = \frac{y}{x}$ $tan alpha=y/x$

$cot α = \frac{x}{y}$ $cot alpha=x/y$

$sec α = \frac{r}{x}$ $sec alpha=r/x$

$csc α = \frac{r}{y}$ $csc alpha=r/y$