How do you find the values of the other five trigonometric functions of the acute angle A with $csc A = 3$ $cscA=3$ ?

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How do you find the values of the other five trigonometric functions of the acute angle A with $csc A = 3$ $cscA=3$ ?

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Answer 1 · 2018-03-21T16:05:05

$sin A = \frac{1}{3}$ $sinA=1/3$

$cos A = \frac{2 \sqrt{2}}{3}$ $cosA=(2sqrt2)/3$

$tan A = \frac{1}{2 \sqrt{2}}$ $tanA=1/(2sqrt2)$

$csc A = \frac{3}{1}$ $cscA=3/1$

$sec A = \frac{3}{2 \sqrt{2}}$ $secA=3/(2sqrt2)$

$cot A = 2 \frac{\sqrt{2}}{1}$ $cotA=2sqrt2/1$

Explanation:

$csc A = 3$ $cscA=3$
$csc A = \frac{hypotenuse}{opposite side} = \frac{3}{1}$ $cscA="hypotenuse"/"opposite side"=3/1$

By pythagoras theorem

$adjacent side = \sqrt{3^{2} - 1^{2}} = 2 \sqrt{2}$ $"adjacent side"=sqrt(3^2-1^2)=2sqrt2$

We have

$opposite side = 1$ $"opposite side"=1$
$adjacent side = 2 \sqrt{2}$ $"adjacent side"=2sqrt2$
$hypotenuse = 3$ $"hypotenuse"=3$

$sin A = \frac{opposite side}{hypotenuse} = \frac{1}{3}$ $sinA="opposite side"/"hypotenuse"=1/3$
$cos A = \frac{adjacent side}{hypotenuse} = \frac{2 \sqrt{2}}{3}$ $cosA="adjacent side"/"hypotenuse"=(2sqrt2)/3$
$tan A = \frac{opposite side}{adjacent side} = \frac{1}{2 \sqrt{2}}$ $tanA="opposite side"/"adjacent side"=1/(2sqrt2)$
$csc A = \frac{hypotenuse}{opposite side} = \frac{3}{1}$ $cscA="hypotenuse"/"opposite side"=3/1$
$sec A = \frac{hypotenuse}{adjacent side} = \frac{3}{2 \sqrt{2}}$ $secA="hypotenuse"/"adjacent side"=3/(2sqrt2)$
$cot A = \frac{adjacent side}{opposite side} = 2 \frac{\sqrt{2}}{1}$ $cotA="adjacent side"/"opposite side"=2sqrt2/1$

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How do you find the values of the other five trigonometric functions of the acute angle A with $csc A = 3$ $cscA=3$ ?

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How do you find the values of the other five trigonometric functions of the acute angle A with cscA=3cscA=3?

1 Answer

Explanation:

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How do you find the values of the other five trigonometric functions of the acute angle A with $csc A = 3$ $cscA=3$ ?