# sin(-1230)-cos{(2n+1)π+π/3}=? #

2 Answers
Dean R.
Jun 2, 2018

#sin (-1230^circ ) - cos ( (2n +1) pi + pi/3 ) = 0 #

Explanation:

We'll assume integer #n# and the first argument is degrees; the second radians. That's a lot of assumptions to do this silly problem.

#sin (-1230^circ ) - cos ( (2n +1) pi + pi/3 ) #

#= sin(-1230^circ + 3 \times 360^circ) - cos( 2pi n + {4pi}/3 )#

#= sin(-150^circ) - cos({4pi}/3)#

#= - sin 150^circ - cos 240^circ #

#= - sin 30^circ - cos 120^circ #

#= - sin 30^circ + cos 60^circ #

#= - sin 30^circ + sin 30^circ #

#= 0#

Even when we don't even need to evaluate a trig function, the question writers stick to 30/60/90 or 45/45/90. They just can't help themselves. Students deserve to be taught a subject that handles more than two triangles.

maganbhai P.
Jun 2, 2018

#sin(-1230)-cos{(2n+1)pi+pi/3}=0#

Explanation:

Here,

#sin(-1230)=-sin1230...to[becausesin(-x)=-sinx]#

#color(white)(sin(-1230))=-sin(180xx7-30)#

#color(white)(sin(-1230))=-sin(7pi-pi/6)toII^(nd)Quadrant#

#color(white)(sin(-1230))=-sin(pi/6)#

#sin(-1230)=-1/2...to(1)#

#And #

#{(2n+1)pi+pi/3}toIII^(rd)Quadrant#

#cos{(2n+1)pi+pi/3}=-cos(pi/3)=-1/2to(2)#

#From# #(1) and (2)#we get

#sin(-1230)-cos{(2n+1)pi+pi/3}=(-1/2)-(-1/2)#

#sin(-1230)-cos{(2n+1)pi+pi/3}=0#

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