How do you show that sin(x+60) + sin(x+120) =(√3)cosx ?

2 Answers
Feb 28, 2018

See the answer below...

Explanation:

#sin(x+60)+sin(x+120)#

#=2 cdot sin{((x+60)+(x+120))/2} cdot cos{((x+60)-(x+120))/2}#

#=2 cdot sin{(2x+180)/2} cdot cos{(-60)/2}#

#=2 cdot sin{90+x} cdot cos{30}#

#=2 cdot sin{90-(-x)} cdot cos30#

#=2 cdot cosx cdot sqrt3/2#

#=sqrt3 cdot cosx#

Feb 28, 2018

see below

Explanation:

We have,#sin(x+60) +sin(x+120)# on L.H.S
According to formula, #sin(a+b)=sinacosb+cosasinb#
Applying it,we get,
#sinxcos(60) +cosxsin(60) +sinxcos(120) + cosxsin(120)#
or, #sinx//2 +cosx xx (sqrt3)//2 -sinx//2 +cosx xx(sqrt3)//2#
{#color(blue)(as cos "120 " is -1//2 and sin "120 " is (sqrt3)//2#}

Thus,the expression comes out to be,#2xx (sqrt3)//2 xxcosx# which is #color(red)(sqrt3 cosx)#, the R.H.S