How do you find the exact value of $cos [2 arcsin (- \frac{3}{5}) - arctan (\frac{5}{12})]$ $cos[2 arcsin (-3/5) - arctan (5/12)]$ ?

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How do you find the exact value of $cos [2 arcsin (- \frac{3}{5}) - arctan (\frac{5}{12})]$ $cos[2 arcsin (-3/5) - arctan (5/12)]$ ?

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Answer 1 · 2016-03-28T06:20:38

Possible values are $\pm 0.1108$ $+-0.1108$ or $\pm 0.6277$ $+-0.6277$

Explanation:

$arcsin (- \frac{3}{5}) = x$ $arcsin(−3/5)=x$ means $sin x = (- \frac{3}{5}) = - 0.6$ $sinx=(-3/5)=-0.6$ .

As $sin x = 0.6$ $sinx=0.6$ for $x = {36.87}^{o}$ $x=36.87^o$ and sine is negative in third and fourth quadrant, $x = 180^{\oplus} {36.87}^{o}$ $x=180^o+36.87^o$ or ${216.87}^{o}$ $216.87^o$ and $x = 360^{o} - {36.87}^{o} = {323.13}^{o}$ $x=360^o-36.87^o=323.13^o$ .

$arctan (\frac{5}{12}) = x$ $arctan(5/12)=x$ means $tan x = (\frac{5}{12})$ $tanx=(5/12)$ .

As $tan x = \frac{5}{12}$ $tanx=5/12$ for $x = {22.62}^{o}$ $x=22.62^o$ and tan is positive in first and third quadrant, $x = {22.62}^{o}$ $x=22.62^o$ or $x = 180^{o} + {22.62}^{o}$ $x=180^o +22.62^o$ .or ${202.62}^{o}$ $202.62^o$ .

Hence $cos {2 arcsin (- \frac{3}{5}) - arctan (\frac{5}{12})] = cos [2 \times {216.87}^{o} - {22.62}^{o}] = {cos 411.12}^{o} = {cos 51.12}^{o} = 0.6277$ $cos{2arcsin(−3/5)-arctan(5/12)]=cos[2xx216.87^o-22.62^o]=cos411.12^o=cos51.12^o=0.6277$ or

$cos {2 arcsin (- \frac{3}{5}) - arctan (\frac{5}{12})] = cos [2 \times {216.87}^{o} - {202.62}^{o}] = {cos 411.12}^{o} = {cos 231.12}^{o} = - 0.6277$ $cos{2arcsin(−3/5)-arctan(5/12)]=cos[2xx216.87^o-202.62^o]=cos411.12^o=cos231.12^o=-0.6277$ or

$cos {2 arcsin (- \frac{3}{5}) - arctan (\frac{5}{12})] = cos [2 \times {323.13}^{o} - {22.62}^{o}] = {cos 623.64}^{o} = - 0.1108$ $cos{2arcsin(−3/5)-arctan(5/12)]=cos[2xx323.13^o-22.62^o]=cos623.64^o=-0.1108$ or

$cos {2 arcsin (- \frac{3}{5}) - arctan (\frac{5}{12})] = cos [2 \times {323.13}^{o} - {202.62}^{o}] = {cos 443.64}^{o} = 0.1108$ $cos{2arcsin(−3/5)-arctan(5/12)]=cos[2xx323.13^o-202.62^o]=cos443.64^o=0.1108$

Answer 2 · 2016-03-28T07:39:13

Values in exactitude are $\pm \frac{12}{125}$ $+-12/125$ and $\pm \frac{68}{125}$ $+-68/125$ .

Explanation:

Let $A = a r c sin (- \frac{3}{5})$ $A = arc sin (-3/5)$ and $B = a r c tan (\frac{5}{12})$ $B=arc tan (5/12)$ .
Then, $sin A = - \frac{3}{5}, cos A = \pm \frac{4}{5}$ $sin A = -3/5, cos A = +-4/5$ .

$cos 2 A = 1 - 2 {sin}^{2} A = \frac{7}{25}$ $cos 2A = 1-2 sin^2A=7/25$

$sin 2 A = 2 sin A cos A = - \frac{24}{25} and \frac{24}{25}$ $sin 2A=2 sin A cos A=-24/25 and 24/25$ , respectively..

Also, $tan B = \frac{5}{12}, (sin B = \frac{5}{13}, cos B = \frac{12}{13}) and (sin B = - \frac{5}{13}, cos B = - \frac{12}{13})$ $tan B=5/12, (sin B =5/13, cos B=12/13) and (sin B =-5/13, cos B=-12/13)$

The given expression is

cos(2A-B) = cos 2A cos B+sin 2A sin B

= $\pm \frac{84 \pm 120}{375}$ $+-(84+-120)/375$ .

Note that both sin B and cos B have the same sign + or $-$ $-$ .

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