How do you find exact value of sin (19pi/12)?
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#sin((19pi)/12) = sin((15pi)/12 + (4pi)/12)#
# = sin((5pi)/4 + pi/3)#
# = sin((5pi)/4)cos(pi/3) + cos((5pi)/4)sin(pi/3)#
# = (-sqrt(2)/2)(1/2) + (-sqrt(2)/2)(sqrt(3)/2)#
# = (-sqrt(2)/4) + (-sqrt(6)/4)#
# = (-sqrt(2) - sqrt(6))/4#
# sin 285^circ= -1/4 (sqrt {2} + sqrt{6}) #
I like to complain that every trig problem uses one of two triangles, 30/60/90 or 45/45/90. This one uses both!
The other answer is fine. Let's do it in degrees here.
# {19 pi }/ 12 times 360^circ/{2pi } = 285^circ #
# sin 285^circ = sin( -360^circ + 285^circ ) = sin( -75^circ) = - sin 75^circ = -sin(30^circ + 45^circ)#
There they are.
# sin 285^circ= -sin(30^circ + 45^circ) #
#= -( sin 30^circ cos 45^circ + cos 30^circ sin 45^circ ) #
#= -((1/2) (sqrt{2}/2) + (sqrt{3}/2)(sqrt{2}/2)) #
#= -1/4 (sqrt {2} + sqrt{6}) #
