Question

cos24×cos48×cos96×cos192=​

Answer 1

Step-by-step explanation:

OK, you're given:cos(24°)×cos(48°)×cos(96°)×cos(192°)

={2sin(24°)×cos(24°)×cos(48°)×cos(96°)×cos(192°)}/2sin(24°)

={2sin(48°)×cos(48°)×cos(96°)×cos(192°)}/4sin(24°)

={2sin(96°)×cos(96°)×cos(192°)}/8sin(24°)

={2sin(192°)×cos(192°)}/16sin(24°)

=sin(384°)/16sin(24°)

=sin(24°)/16sin(24°)

=1/16…and you get it.

The above is a fine application of the formula 2sinAcosA=sin2A.

Sin(384°)=Sin(4×90°+24°)=Sin(24°).