How do you multiply #\sin ( - \frac { 50\pi } { 3} ) \cdot \cos \frac { 53\pi } { 6}#?

2 Answers
Ratnaker Mehta
Mar 2, 2018

# 3/4#.

Explanation:

#sin(-50/3pi)cos(53/6pi)#,

#=-sin(50/3pi)cos(53/6pi)......[because, sin(-x)=-sinx]#,

#=-sin(((51-1)/3)pi)cos(((54-1)/6)pi)#,

#=-sin((51/3-1/3)pi)cos((54/6-1/6)pi)#,

#=-sin(17pi-1/3pi)cos(9pi-1/3pi)#.

Now, #(17pi-1/3pi)" is in QII, where "sin > 0#,

#:. sin(17pi-1/3pi)=sin(1/3pi)=sqrt3/2#.

Similarly, #cos(9pi-1/6pi)=-cos(1/6pi)=-sqrt3/2#.

#rArr sin(-50/3pi)cos(53/6pi)=(-sqrt3/2)(-sqrt3/2)=3/4#.

Nghi N
Mar 3, 2018

First, simplify the 2 factors -->
#sin ((- 50pi)/3) = sin (- (2pi)/3 - (48pi)/3) = sin ((- 2pi)/3 - 16pi) =#
#= sin ((-2pi)/3) = - sin ((2pi)/3) = - sqrt3/2#
#cos ((53pi)/6) = cos ((5pi)/6 + (48pi)/6) = cos ((5pi)/6 + 8pi) =#
#= cos ((5pi)/6) = - sqrt3/2#
Finally,
#sin ((-50pi)/3).cos ((53pi)/6) = (-sqrt3/2)(-sqrt3/2) = 3/4#

Related questions
Impact of this question
1605 views around the world
You can reuse this answer
Creative Commons License