How do you find the exact value of #Sin(arcsin(3/5)+(arctan-2))#?

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Answer 1 · 2016-01-26T10:05:15

#sin(arcsin(3/5)+(arctan (-2)))=color(red)(-sqrt5/5)# and

#sin(arcsin(3/5)+(arctan (-2)))=color(red)(sqrt5/5)#

There are 2 possible answers:

First solution

#sin (arcsin (3/5)+arctan (-2))#

#sin (A+B)#

Let #A=arcsin (3/5)# and #B=arctan (color(blue)(-2)/1)#

then #sin A=3/5# and

computed using Pythagorean relation #c^2=a^2+b^2#

#cos A=4/5#

also

#sin B=(-2)/sqrt5#

#cos B=1/sqrt5#

compute #sin (A+B)#
#sin (A+B)=sin A cos B + cos A sin B#

#sin (A+B)=3/5* 1/sqrt5 + 4/5 *(-2)/sqrt5=-5/(5sqrt5)#

#sin (A+B)=-1/sqrt5=-sqrt5/5#
#color(green) ("The 4th quadrant angle")=A+B=-63.4349^@#

#sin (arcsin (3/5)+arctan (-2))#

#sin (A+B)#

Let #A=arcsin (3/5)# and #B=arctan (2/color (blue)(-1))#

then #sin A=3/5# and

computed using Pythagorean relation #c^2=a^2+b^2#

#cos A=4/5#

also

#sin B=2/sqrt5#

#cos B=(-1)/sqrt5#

compute #sin (A+B)#
#sin (A+B)=sin A cos B + cos A sin B#

#sin (A+B)=3/5* (-1)/sqrt5 + 4/5 *2/sqrt5=5/(5sqrt5)#

#sin (A+B)=1/sqrt5=sqrt5/5#

#color(green) ("The 2nd quadrant angle")=A+B=116.565^@#
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