Prove that sin120°.sin140°.sin160°=root 3/8
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Prove that sin120° .sin140° .sin160° =3/8
Answer=
Note
that, sin120sin140sin160=sin20sin40sin60=sin602(cos20−cos60)sin120sin140sin160=sin20sin40sin60=sin602(cos20−cos60)... (1)
sin60,cos60sin60,cos60 are well known values. We will try to get cos20cos20.
Now, cos3x=4cos3x−3cosxcos3x=4cos3x−3cosx. Substituting x=20x=20, we get,
4cos320−3cos20=124cos320−3cos20=12
That is,
8cos320−6cos20−1=08cos320−6cos20−1=0
Therefore, cos20cos20 is the root of the equation 8x3−6x−1=08x3−6x−1=0. One can solve this cubic equation to find the value of cos20cos20. Substitute it back in (1) to obtain the desired value.
Answer=
Note
that, sin120sin140sin160=sin20sin40sin60=sin602(cos20−cos60)sin120sin140sin160=sin20sin40sin60=sin602(cos20−cos60)... (1)
sin60,cos60sin60,cos60 are well known values. We will try to get cos20cos20.
Now, cos3x=4cos3x−3cosxcos3x=4cos3x−3cosx. Substituting x=20x=20, we get,
4cos320−3cos20=124cos320−3cos20=12
That is,
8cos320−6cos20−1=08cos320−6cos20−1=0
Therefore, cos20cos20 is the root of the equation 8x3−6x−1=08x3−6x−1=0. One can solve this cubic equation to find the value of cos20cos20. Substitute it back in (1) to obtain the desired value.
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