How do you use the product to sum formulas to write #4cos(pi/3)sin((5pi)/6)# as a sum or difference?

2 Answers
Ratnaker Mehta
Jul 8, 2017

# 1.#

Explanation:

#4cos(pi/3)sin(5pi/6)=4cos(pi/3)sin(pi-pi/6),#

#=4cos(pi/3)sin(pi/6),#

#=2{2cos(pi/3)sin(pi/6)},#

#=2{sin(pi/3+pi/6)-sin(pi/3-pi/6)},#

#=2{sin(pi/2)-sin(pi/6)},#

#=2(1-1/2),#

#=1.#

Nghi N
Jul 9, 2017

#2[cos ((2pi)/3) - cos pi]#

Explanation:

#y = 4cos (pi/3)sin ((5pi)/6)#
Because,
#cos (pi/3) = sin (pi/2 - pi/3) = sin (pi/6)#
Therefor,
#y = 4sin (pi/6).sin ((5pi)/6)#
Apply the trig identity:
#sin a.sin b = (1/2)[cos (a - b) - cos (a + b)]#
In this case:
#y = 2[cos ((2pi)/3) - cos pi] #
That is the answer. However, we can go further, knowing
#cos ((2pi)/3) = - 1/2# and #cos pi = - 1#.
#y = 2(- 1/2 + 1) = 2/2 = 1#

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