Question #f6d94

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Question #f6d94

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Answer 1 · 2017-04-08T00:24:21

$cos (b) = \sqrt{\frac{289}{338}} \approx 0.924680985 (radians)$ $cos(b)=sqrt(289/338) ~~0.924680985 ("radians")$

Explanation:

Here are 2 ways you could do this.

Method 1: ...almost feels like cheating ;)
If $tan (2 b) = \frac{119}{120}$ $tan(2b)=119/120$
use your calculator (or a spreadsheet) to evaluate:
$XXX 2 b = arctan (\frac{119}{120}) \approx 0.7812140874$ $color(white)("XXX")2b=arctan(119/120)~~0.7812140874$
which implies (after dividing by 2)
$XXX b = 0.3906070437$ $color(white)("XXX")b=0.3906070437$
Then
use your calculator again to find
$XXX cos (b) = cos (0.3906070437) \approx 0.9246780985$ $color(white)("XXX")cos(b)=cos(0.3906070437)~~0.9246780985$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Method 2: possibly more virtuous (???)
If $tan (2 b) = \frac{119}{120}$ $tan(2b)=119/120$
then we can think of angle $(2 b)$ $(2b)$ as being the angle of a right triangle in standard position with the horizontal ( $x$ $x$ ) component equal to $120$ $120$ and the vertical ( $y$ $y$ ) component equal to $119$ $119$ .
The hypotenuse would be equal to $\sqrt{120^{2} + 119^{2}} = 169$ $sqrt(120^2+119^2)=169$
(and, yes, I did use a calculator to discover this).

$cos (2 b) = \frac{horizontal}{hypotenuse} = \frac{120}{169}$ $cos(2b)=("horizontal")/("hypotenuse")=120/169$

Then, if we remember the double angle formula:
$XXX cos (2 b) = 2 {cos}^{2} (b) - 1$ $color(white)("XXX")cos(2b)=2cos^2(b)-1$
we can re-arrange the terms to get
$XXX cos (b) = \sqrt{\frac{cos (2 b) + 1}{2}}$ $color(white)("XXX")cos(b)=sqrt((cos(2b)+1)/2)$

$XXXXXX = \sqrt{\frac{\frac{120}{169} + 1}{2}}$ $color(white)("XXXXXX")=sqrt((120/169+1)/2)$

$XXXXXX = \sqrt{\frac{\frac{289}{169}}{2}}$ $color(white)("XXXXXX")=sqrt((289/169)/2)$

$XXXXXX = \sqrt{\frac{289}{338}}$ $color(white)("XXXXXX")=sqrt(289/338)$

and with the aid of my calculator (again)
$XXXXXX \approx 0.924680985$ $color(white)("XXXXXX")~~0.924680985$

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Question #f6d94

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