How do you find the five remaining trigonometric function satisfying $tan θ = \frac{5}{4}$ $tantheta=5/4$ , $cos θ < 0$ $costheta<0$ ?

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How do you find the five remaining trigonometric function satisfying $tan θ = \frac{5}{4}$ $tantheta=5/4$ , $cos θ < 0$ $costheta<0$ ?

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Answer 1 · 2017-01-16T04:06:00

Find values of trig functions

Explanation:

Use trig identity:
$1 + {tan}^{2} x = \frac{1}{{cos}^{2} x}$ $1 + tan^2 x = 1/(cos^2 x)$
${cos}^{2} x = \frac{1}{1 + {tan}^{2} x}$ $cos^2 x = 1/(1 + tan^2 x)$
${sin}^{2} x = \frac{1}{1 + {cot}^{2} x}$ $sin^2 x = 1/(1 + cot^2 x)$
In this case:
${cos}^{2} x = \frac{1}{1 + \frac{25}{16}} = \frac{1}{\frac{41}{16}} = \frac{16}{41}$ $cos^2 x = 1/(1 + 25/16) = 1/(41/16) = 16/41$
$cos x = - \frac{4}{\sqrt{41}} = - \frac{4 \sqrt{41}}{41}$ $cos x = - 4/sqrt41 = - (4sqrt41)/41$
Find sin x by the same way:
${sin}^{2} x = \frac{1}{1 + \frac{16}{25}} = \frac{1}{\frac{41}{25}} = \frac{25}{41}$ $sin^2 x = 1/(1 + 16/25) = 1/(41/25) = 25/41$
$sin x = - \frac{5}{\sqrt{41}}$ $sin x = - 5/(sqrt41)$ (because $tan x = \frac{5}{4}$ $tan x = 5/4$ > 0)
$tan x = \frac{sin}{cos} = \frac{5}{4}$ $tan x = sin/(cos) = 5/4$
$cot x = \frac{1}{tan} = \frac{4}{5}$ $cot x = 1/(tan) = 4/5$
$sec x = \frac{1}{cos} = - \frac{\sqrt{41}}{4}$ $sec x = 1/(cos) = - sqrt41/4$
$csc x = \frac{1}{sin} = - \frac{\sqrt{41}}{5}$ $csc x = 1/(sin) = - sqrt41/5$

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How do you find the five remaining trigonometric function satisfying $tan θ = \frac{5}{4}$ $tantheta=5/4$ , $cos θ < 0$ $costheta<0$ ?

1 Answer

Explanation:

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How do you find the five remaining trigonometric function satisfying $tan θ = \frac{5}{4}$ $tantheta=5/4$ , $cos θ < 0$ $costheta<0$ ?