Question

Sin40/sin80 + sin80/sin20 - sin20/sin40

A 1

B 2

C 3

D 4

Answer 1

there is this standard identity:

f(x)=4sinx*sin(60-x)*sin(60+x)=sin3x

putting x=20,

we have f(20)=root3/2.....(1)


so taking lcm, we get

(sin^240sin20+sin^280sin40-sin^220sin80)/sin20sin40sin80



from 1 this becomes

8(sin^240sin20+sin^280sin40-sin^220sin80)/root3.....(2)


now we focus on 8(sin^240sin20+sin^280sin40-sin^220sin80)

using identity
cos2y=1 – sin^2y and sin(180-z)=sinz

and 2sinAcosB=sin(A+B)+sin(A-B),

we have
2(3sin20+3sin40 – 3sin80+3sin60)
now,
sin80-sin20=2sin30cos50

=sin40
so 2(3sin20+3sin40 – 3sin80+3sin60)

= 6sin60=3root3

from (2), we have

sin40/sin80 + sin80/sin20 - sin20/sin40=

3root3/root3
= 3

So option (c) is the answer