Given that sin^-1 (sin 3π/4) =2π/k, then k = (a) 3 (b) 8 (c) 6 (d) 9
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sin^-1(sin 3π/4) = 2π/k
sin^-1[sin(π-π/4)] = 2π/k
[ now, sin(180-theta)=sin theta]
sin^-1(sinπ/4) = 2π/k
π/4 =2π/k
kπ = 8π
(b) k = 8
sin^-1[sin(π-π/4)] = 2π/k
[ now, sin(180-theta)=sin theta]
sin^-1(sinπ/4) = 2π/k
π/4 =2π/k
kπ = 8π
(b) k = 8
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