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

Question

3219 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-28T06:20:38

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

$arcsin(−3/5)=x$ means $sinx=(-3/5)=-0.6$ .

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

$arctan(5/12)=x$ means $tanx=(5/12)$ .

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

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

$cos{2arcsin(−3/5)-arctan(5/12)]=cos[2xx216.87^o-202.62^o]=cos411.12^o=cos231.12^o=-0.6277$ or

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

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

Answer 2 · 2016-03-28T07:39:13

Values in exactitude are $+-12/125$ and $+-68/125$ .

Let $A = arc sin (-3/5)$ and $B=arc tan (5/12)$ .
Then, $sin A = -3/5, cos A = +-4/5$ .

$cos 2A = 1-2 sin^2A=7/25$

$sin 2A=2 sin A cos A=-24/25 and 24/25$ , respectively..

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

The given expression is

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

= $+-(84+-120)/375$ .

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

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