Question

Sin120.sin140.sin160

Answer 1

Note that, sin120sin140sin160=sin20sin40sin60=sin602(cos20−cos60)sin⁡120sin⁡140sin⁡160=sin⁡20sin⁡40sin⁡60=sin⁡602(cos⁡20−cos⁡60)... (1)

sin60,cos60sin⁡60,cos⁡60 are well known values. We will try to get cos20cos⁡20.

Now, cos3x=4cos3x−3cosxcos⁡3x=4cos3⁡x−3cos⁡x. Substituting x=20x=20, we get,

4cos320−3cos20=124cos3⁡20−3cos⁡20=12

That is,

8cos320−6cos20−1=08cos3⁡20−6cos⁡20−1=0

Therefore, cos20cos⁡20 is the root of the equation 8x3−6x−1=08x3−6x−1=0. One can solve this cubic equation to find the value of cos20cos⁡20. Substitute it back in (1) to obtain the desired value.

Answer 2

Sin 120= Sin (180-120)
=Sin 60
Sin 140 = Sin 40
Sin160=Sin 20

Now Sin 60= √3/2
Sin 40=0.64 (approx)
Sin 20= 0.34 (approx)
Now Sin20xsin40 x sin 60 = √3/2 x 64/100 x 34/100 = 1088√3/10000