Question #b2446

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Question #b2446

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Answer 1 · 2017-04-16T01:18:21

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Explanation:

$sin x = \frac{7}{11}$ $sinx=7/11$

Since $x$ $x$ is in quadrant $II$ $"II"$ , the angle is obtuse. This means that both $cos x$ $cosx$ and $tan x$ $tanx$ are negative.

$cos x = - \sqrt{1 - {sin}^{2} x}$ $cosx=-sqrt(1-sin^2x)$

$tan x = \frac{sin x}{cos x}$ $tanx=sinx/cosx$

$cos x = - \sqrt{1 - {(\frac{7}{11})}^{2}} = - \frac{6}{11} \sqrt{2}$ $cosx=-sqrt(1-(7/11)^2)=-6/11sqrt2$

$tan x = \frac{\frac{7}{11}}{- \frac{6}{11} \sqrt{2}} = - \frac{7}{12} \sqrt{2}$ $tanx=(7/11)/(-6/11sqrt2)=-7/12sqrt2$

In order to find $sin 2 x$ $sin2x$ , $cos 2 x$ $cos2x$ and $tan 2 x$ $tan2x$ , we need to use the double angle identities. These will be given below:

$sin 2 x = 2 sin x cos x = 2 (\frac{7}{11}) (- \frac{6}{11} \sqrt{2}) = - \frac{84}{121} \sqrt{2}$ $sin2x=2sinxcosx=2(7/11)(-6/11sqrt2)=-84/121sqrt2$

$cos 2 x = 2 {cos}^{2} x - 1 = 2 {(- \frac{6}{11} \sqrt{2})}^{2} - 1 = \frac{23}{121}$ $cos2x=2cos^2x-1=2(-6/11sqrt2)^2-1=23/121$

$tan 2 x = \frac{2 tan x}{1 - {tan}^{2} x} = \frac{2 (- \frac{7}{12} \sqrt{2})}{1 - {(- \frac{7}{12} \sqrt{2})}^{2}} = - \frac{84}{23} \sqrt{2}$ $tan2x=(2tanx)/(1-tan^2x)=(2(-7/12sqrt2))/(1-(-7/12sqrt2)^2)=-84/23sqrt2$

Answer 2 · 2017-04-16T01:27:32

There are formulas for $sin 2 x, cos 2 x$ $sin2x, cos2x$ and $tan 2 x$ $tan2x$ (they're called "Double-Angle Formulas")

$sin 2 x = 2 sin x cos x$ $sin2x=2sinxcosx$
$cos 2 x = {cos}^{2} x - {sin}^{2} x = 2 {cos}^{2} x - 1 = 1 - {sin}^{2} x$ $cos2x=cos^2x-sin^2x=2cos^2x-1=1-sin^2x$
$tan 2 x = \frac{2 \cdot tan x}{1 - {tan}^{2} x}$ $tan2x=(2*tanx)/(1-tan^2x)$

If we are given $sin x$ $sinx$ , we know two sides of the triangle: the hypotenuse and one leg. From there, using Pythagorean's Theorem $(a^{2} + b^{2} = c^{2})$ $(a^2+b^2=c^2)$ , we could find the remaining side, and thus find the ratios for both $tan$ $tan$ and $cos$ $cos$ . That will allow us to solve the Double Angle Formulas without knowing all the angles

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Question #b2446

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