How do you use the double-angle identities to find cot(2x) if cos x= -15/17 and csc x is less than 0?

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How do you use the double-angle identities to find cot(2x) if cos x= -15/17 and csc x is less than 0?

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Answer 1 · 2016-06-18T15:15:35

$\frac{161}{240}$ $161/240$

Explanation:

Given
$cos x = - \frac{15}{17} and csc x < 0 \to sin x < 0$ $cosx =-15/17 and cscx<0->sinx<0$
$both cos x and sin x being < 0, x is in 3rd quadrant$ $"both "cosx and sinx " being " <0 , color(blue)(" x is in 3rd quadrant")$

So $tan x > 0$ $tanx>0$

$sin x = - \sqrt{1 - {cos}^{2} x} = - \sqrt{1 - {(- \frac{15}{17})}^{2}} = - \frac{8}{17}$ $sinx =-sqrt(1-cos^2x)=-sqrt(1-(-15/17)^2)=-8/17$

$tan x = \frac{sin x}{cos x} = \frac{- \frac{8}{17}}{- \frac{15}{17}} = \frac{8}{15}$ $tanx = sinx/cosx=(-8/17)/(-15/17)=8/15$

Now

$cot 2 x = \frac{1}{tan (2 x)} = \frac{1 - {tan}^{2} x}{2 tan x} = \frac{1 - {(\frac{8}{15})}^{2}}{2 \cdot \frac{8}{15}}$ $cot2x=1/tan(2x)=(1-tan^2x)/(2tanx)=(1-(8/15)^2)/(2*8/15)$
$= \frac{161}{15^{2}} \times \frac{15}{16} = \frac{161}{240}$ $=161/15^2xx15/16=161/240$

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How do you use the double-angle identities to find cot(2x) if cos x= -15/17 and csc x is less than 0?

1 Answer

Explanation:

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