Find $sin θ$ $sintheta$ , $cos θ$ $costheta$ , and $tan θ$ $tantheta$ under the given conditions: $sin 2 θ = \frac{1}{5}, \frac{π}{2} \leq 2 θ < π$ $sin2theta=1/5, pi/2 <= 2theta < pi$ ?

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Find $sin θ$ $sintheta$ , $cos θ$ $costheta$ , and $tan θ$ $tantheta$ under the given conditions: $sin 2 θ = \frac{1}{5}, \frac{π}{2} \leq 2 θ < π$ $sin2theta=1/5, pi/2 <= 2theta < pi$ ?

I know the answers are:
$sin θ = \sqrt{\frac{1}{2} + \frac{\sqrt{6}}{5}}$ $sintheta=sqrt(1/2+sqrt6/5)$
$cos θ = \sqrt{\frac{1}{2} - \frac{\sqrt{6}}{5}}$ $costheta = sqrt(1/2 - sqrt6/5)$
$tan θ = \sqrt{\frac{5 + 2 \sqrt{6}}{5 - 2 \sqrt{6}}}$ $tantheta = sqrt((5+2sqrt6)/(5-2sqrt6)$

Thanks in advance.

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Answer 1 · 2018-04-07T20:19:24

See below.

Explanation:

$\frac{π}{2} \leq 2 θ \leq π \to \frac{π}{4} \leq θ \leq \frac{π}{2}$ $pi/2<=2theta<=pi -> pi/4<=theta<=pi/2$

Thus, all double-angle cosines must be negative due to the double angle being in the second quadrant, and all single-angle sines, cosines, and tangents must be positive due to the single angle being in a portion of the first quadrant.

Keeping that in mind, recall that

${sin}^{2} θ + {cos}^{2} θ = 1$ $sin^2theta+cos^2theta=1$

We may optimize this to

${sin}^{2} (2 θ) + {cos}^{2} (2 θ) = 1$ $sin^2(2theta)+cos^2(2theta)=1$

${sin}^{2} (2 θ) = {(\frac{1}{5})}^{2} = \frac{1}{25}$ $sin^2(2theta)=(1/5)^2=1/25$

${cos}^{2} (2 θ) = \frac{25}{25} - \frac{1}{25}$ $cos^2(2theta)=25/25-1/25$

$cos (2 θ) = \pm \frac{\sqrt{24}}{5}$ $cos(2theta)=+-sqrt(24)/5$

We want the negative solution.

$cos (2 θ) = - 2 \frac{\sqrt{6}}{5}$ $cos(2theta)=-2sqrt(6)/5$

Now, we want to relate these double angles to single angle trig functions.

The power-reduction identities are the way to go.

Recall

${cos}^{2} θ = \frac{1}{2} (1 + cos 2 θ)$ $cos^2theta=1/2(1+cos2theta)$

Thus,

${cos}^{2} θ = \frac{1}{2} (1 - 2 \frac{\sqrt{6}}{5}) = \frac{1}{2} - \frac{\sqrt{6}}{5}$ $cos^2theta=1/2(1-2sqrt6/5)=1/2-sqrt6/5$

$cos θ = \pm \sqrt{\frac{1}{2} - \frac{\sqrt{6}}{5}}$ $costheta=+-sqrt(1/2-sqrt6/5)$

We want the positive answer.

$cos θ = \sqrt{\frac{1}{2} - \frac{\sqrt{6}}{5}}$ $costheta=sqrt(1/2-sqrt6/5)$

Furthermore, recall

${sin}^{2} θ = \frac{1}{2} (1 - cos 2 θ)$ $sin^2theta=1/2(1-cos2theta)$

Then,

${sin}^{2} θ = \frac{1}{2} (1 - (- 2 \frac{\sqrt{6}}{5}))$ $sin^2theta=1/2(1-(-2sqrt6/5))$

${sin}^{2} θ = \frac{1}{2} + \frac{\sqrt{6}}{5}$ $sin^2theta=1/2+sqrt6/5$

$sin θ = \pm \sqrt{\frac{1}{2} + \frac{\sqrt{6}}{5}}$ $sintheta=+-sqrt(1/2+sqrt6/5)$

We want the positive answer.

$sin θ = \sqrt{\frac{1}{2} + \frac{\sqrt{6}}{5}}$ $sintheta=sqrt(1/2+sqrt6/5)$

$tan θ = \frac{sin θ}{cos θ},$ $tantheta=sintheta/costheta,$ so

$tantheta=sqrt(1/2+sqrt6/5)/sqrt(1/2-sqrt6/5)=sqrt((1/2+sqrt6/5)/(1/2-sqrt6/5))=sqrt(((5+2sqrt6)/cancel(10))/((5-2sqrt6)/cancel(10)))=sqrt((5+2sqrt6)/(5-2sqrt6)$

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Find $sin θ$ $sintheta$ , $cos θ$ $costheta$ , and $tan θ$ $tantheta$ under the given conditions: $sin 2 θ = \frac{1}{5}, \frac{π}{2} \leq 2 θ < π$ $sin2theta=1/5, pi/2 <= 2theta < pi$ ?

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Explanation:

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Find sinthetasinθsintheta, costhetacosθcostheta, and tanthetatanθtantheta under the given conditions: sin2theta=1/5, pi/2 <= 2theta < pisin2θ=15,π2≤2θ<πsin2theta=1/5, pi/2 <= 2theta < pi?

I know the answers are: sintheta=sqrt(1/2+sqrt6/5)sinθ=√12+√65sintheta=sqrt(1/2+sqrt6/5) costheta = sqrt(1/2 - sqrt6/5)cosθ=√12−√65costheta = sqrt(1/2 - sqrt6/5) tantheta = sqrt((5+2sqrt6)/(5-2sqrt6)tanθ=√5+2√65−2√6tantheta = sqrt((5+2sqrt6)/(5-2sqrt6) Thanks in advance.

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Explanation:

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Find $sin θ$ $sintheta$ , $cos θ$ $costheta$ , and $tan θ$ $tantheta$ under the given conditions: $sin 2 θ = \frac{1}{5}, \frac{π}{2} \leq 2 θ < π$ $sin2theta=1/5, pi/2 <= 2theta < pi$ ?