evaluate sin[2cos-1(-3/5)].
Asked by monami chatterjee | 6th Jun, 2011, 04:16: PM
Expert Answer:
Remember this identity.
sin(2x) = 2sinxcosx
= sin(2arccos(-3/5))
= 2sin(arccos(-3/5))cos(arccos(-3/5))
= 2 * ?(1 - [cos(arccos(-3/5))]^2) * -3/5
= 2 * ?(1 - (-3/5)^2) * -3/5
= -6/5 * ?([25 - 9]/25) *
= -6/5 * ?(16/25) *
= -6/5 * 4/5
= -24/25 (Answer)
sin(2x) = 2sinxcosx
= sin(2arccos(-3/5))
= 2sin(arccos(-3/5))cos(arccos(-3/5))
= 2 * ?(1 - [cos(arccos(-3/5))]^2) * -3/5
= 2 * ?(1 - (-3/5)^2) * -3/5
= -6/5 * ?([25 - 9]/25) *
= -6/5 * ?(16/25) *
= -6/5 * 4/5
= -24/25 (Answer)
Answered by | 7th Jun, 2011, 10:03: AM
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