Question #bab8a

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Question #bab8a

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Answer 1 · 2016-08-16T07:49:31

If $ϕ$ $phi$ is the angle of 3rd quadrant then

$cos ϕ \to - v e and sin ϕ \to - v e$ $cosphi->-ve and sinphi->-ve$

Now $sin 2 ϕ = 2 sin ϕ cos ϕ$ $sin2phi=2sinphicosphi$

$= - 2 \sqrt{1 - {cos}^{2} ϕ} \cdot cos ϕ$ $=-2sqrt(1-cos^2phi)*cosphi$

$= - 2 \sqrt{1 - {(- \frac{15}{17})}^{2}} \cdot (- \frac{15}{17})$ $=-2sqrt(1-(-15/17)^2)*(-15/17)$

$= + 2 \frac{\sqrt{17^{2} - 15^{2}}}{17} \cdot \times \frac{15}{17}$ $=+2sqrt(17^2-15^2)/17*xx15/17$

$= \frac{2 \cdot 8 \cdot 15}{17^{2}} = \frac{240}{289}$ $=(2*8*15)/17^2=240/289$

$2 ϕ = {sin}^{- 1} (\frac{240}{289})$ $2phi=sin^-1(240/289)$
$\approx {56.14}^{\circ} \to 2 ϕ is in the 1st quadrant$ $~~56.14^@color(red)(->2phi" is in the 1st quadrant")$

Again

$cos ϕ = - \frac{15}{17}$ $cosphi=-15/17$

$ϕ = {cos}^{- 1} (- \frac{15}{17}) = 360 - 151.93 = 208.07$ $phi=cos^-1(-15/17)=360-151.93=208.07$
$(As ϕ is in the 3rd quadrant)$ $color(red)(("As "phi" is in the 3rd quadrant"))$

$:.2phi=416.14=360+56 .14$

So $2phi$ is the angle of 1st quadrant.

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Question #bab8a

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