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How do you find the value?

$cos$ $cos$ $\frac{α}{2}$ $alpha/2$ if $tan α = \frac{40}{9}, (180 < α < 270)$ $tan alpha= 40/9, (180 < alpha<270)$

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Answer 1 · 2018-02-19T18:43:35

$cos (\frac{α}{2}) = - \frac{4}{\sqrt{41}}$ $cos(alpha/2)=-4/sqrt41$

Explanation:

Given:
$(180 < α < 270)$ $(180ltalpha<270)$
indicating that $cos α$ $cosalpha$ is negative

$tan α = \frac{40}{9}$ $tanalpha=40/9$
opposite side is 40
adjacent side is 9
hypotenuse will be 41,
because
$\sqrt{9^{2} + 40^{2}} = \sqrt{81 + 1600} = \sqrt{1681} = 41$ $sqrt(9^2+40^2)=sqrt(81+1600)=sqrt1681=41$
$cos α = (adjacent side)/(hypotenuse)$ $cosalpha="(adjacent side)/(hypotenuse)"$
$cos α = - \frac{9}{41}$ $cosalpha=-9/41$
$cos (\frac{α}{2}) = \pm \sqrt{\frac{1 + cos α}{2}} = \pm \sqrt{\frac{1 - \frac{9}{41}}{2}}$ $cos(alpha/2)=+-sqrt((1+cosalpha)/2)=+-sqrt((1-9/41)/2$
$= \pm \sqrt{\frac{41 - 9}{41 \times 2}} = \pm \sqrt{\frac{32}{2} \times \frac{1}{41}} = \pm \frac{\sqrt{16}}{\sqrt{41}}$ $=+-sqrt((41-9)/(41xx2))=+-sqrt(32/2xx1/41)=+-sqrt(16)/sqrt(41)$

$cos (\frac{α}{2}) = \pm \frac{4}{\sqrt{41}}$ $cos(alpha/2)=+-4/sqrt41$

As mentioned, $180^{\circ} < α < 270^{\circ},$ $180^@ltalphalt270^@,$
it follows that
$\frac{180}{2} < \frac{α}{2} < \frac{270}{2}$ $180/2ltalpha/2<270/2$

$90 < \frac{α}{2} < 135$ $90ltalpha/2<135$ , where $cos (\frac{α}{2})$ $cos(alpha/2)$ is negative

Hence,
$cos (\frac{α}{2}) = - \frac{4}{\sqrt{41}}$ $cos(alpha/2)=-4/sqrt41$

Answer 2 · 2018-02-19T19:04:56

Start with the identity:

${tan}^{2} (α) + 1 = {sec}^{2} (α)$ $tan^2(alpha) +1 = sec^2(alpha)$

Substitute ${sec}^{2} (α) = \frac{1}{{cos}^{2} (α)}$ $sec^2(alpha)= 1/cos^2(alpha)$

${tan}^{2} (α) + 1 = \frac{1}{{cos}^{2} (α)}$ $tan^2(alpha)+ 1= 1/cos^2(alpha)$

Multiply both sides by $\frac{{cos}^{2} (α)}{{tan}^{2} (α) + 1}$ $cos^2(alpha)/(tan^2(alpha) + 1)$ :

${cos}^{2} (α) = \frac{1}{{tan}^{2} (α) + 1}$ $cos^2(alpha) = 1/(tan^2(alpha) + 1)$

Use the square root operation on both sides:

$cos (α) = \pm \sqrt{\frac{1}{{tan}^{2} (α) + 1}}$ $cos(alpha) = +-sqrt(1/(tan^2(alpha) + 1))$

We are told that $180^{\circ} < α < 270^{\circ}$ $180^@ < alpha < 270^@$ , therefore we choose the negative value:

$cos (α) = - \sqrt{\frac{1}{{tan}^{2} (α) + 1}}$ $cos(alpha) = -sqrt(1/(tan^2(alpha) + 1))$

Add 1 to both sides:

$1 + cos (α) = 1 - \sqrt{\frac{1}{{tan}^{2} (α) + 1}}$ $1+ cos(alpha) = 1-sqrt(1/(tan^2(alpha) + 1))$

Multiply both sides by $\frac{1}{2}$ $1/2$ :

$\frac{1 + cos (α)}{2} = \frac{1 - \sqrt{\frac{1}{{tan}^{2} (α) + 1}}}{2}$ $(1+ cos(alpha))/2 = (1-sqrt(1/(tan^2(alpha) + 1)))/2$

Use the square root operation on both sides:

$\pm \sqrt{\frac{1 + cos (α)}{2}} = \pm \sqrt{\frac{1 - \sqrt{\frac{1}{{tan}^{2} (α) + 1}}}{2}}$ $+-sqrt((1+ cos(alpha))/2) = +-sqrt((1-sqrt(1/(tan^2(alpha) + 1)))/2)$

Substitute $\pm \sqrt{\frac{1 + cos (α)}{2}} = cos (\frac{α}{2})$ $+-sqrt((1+ cos(alpha))/2)= cos(alpha/2)$

$cos (\frac{α}{2}) = \pm \sqrt{\frac{1 - \sqrt{\frac{1}{{tan}^{2} (α) + 1}}}{2}}$ $cos(alpha/2) = +-sqrt((1-sqrt(1/(tan^2(alpha) + 1)))/2)$

From $180^{\circ} < α < 270^{\circ}$ $180^@ < alpha < 270^@$ we derive $90^{\circ} < \frac{α}{2} < 135^{\circ}$ $90^@ < alpha/2 < 135^@$ and conclude that the cosine function is negative within the specified domain:

$cos (\frac{α}{2}) = - \sqrt{\frac{1 - \sqrt{\frac{1}{{tan}^{2} (α) + 1}}}{2}}$ $cos(alpha/2) = -sqrt((1-sqrt(1/(tan^2(alpha) + 1)))/2)$

Substitute ${tan}^{2} (α) = {(\frac{40}{9})}^{2}$ $tan^2(alpha) = (40/9)^2$ :

$cos (\frac{α}{2}) = - \sqrt{\frac{1 - \sqrt{\frac{1}{{(\frac{40}{9})}^{2} + 1}}}{2}}$ $cos(alpha/2) = -sqrt((1-sqrt(1/((40/9)^2 + 1)))/2)$

I used WolframAlpha to simplify the above into an exact form:

$cos (\frac{α}{2}) = - \frac{4 \sqrt{41}}{41}$ $cos(alpha/2) = -(4sqrt41)/41$

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How do you find the value?

$cos$ $cos$ $\frac{α}{2}$ $alpha/2$ if $tan α = \frac{40}{9}, (180 < α < 270)$ $tan alpha= 40/9, (180 < alpha<270)$

2 Answers

Explanation:

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$cos$ $cos$ $\frac{α}{2}$ $alpha/2$ if $tan α = \frac{40}{9}, (180 < α < 270)$ $tan alpha= 40/9, (180 < alpha<270)$