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Find the principal values of each of the following:(i) cosec-1 (-√2)(ii) cosec-1 (-2)(iii) cosec-1 (2/√3)(iv) cosec-1 (2 cos(2π/3)) |
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Answer» (i) Given as cosec-1 (-√2) Let y = cosec-1 (-√2) cosec y = -√2 – cosec y = √2 – cosec (π/4) = √2 – cosec (π/4) = cosec (-π/4) [since – cosec θ = cosec (-θ)] So, the range of principal value of cosec-1 [-π/2, π/2] – {0} and cosec (-π/4) = – √2 cosec (-π/4) = – √2 So, the principal value of cosec-1 (-√2) is – π/4 (ii) Given as cosec-1 (-2) Let y = cosec-1 (-2) cosec y = -2 – cosec y = 2 – cosec (π/6) = 2 – cosec (π/6) = cosec (-π/6) [since – cosec θ = cosec (-θ)] So, the range of principal value of cosec-1 [-π/2, π/2] – {0} and cosec (-π/6) = – 2 cosec (-π/6) = – 2 So, the principal value of cosec-1 (-2) is – π/6 (iii) Given as cosec-1 (2/√3) Let y = cosec-1 (2/√3) cosec y = (2/√3) cosec (π/3) = (2/√3) So, the range of principal value of cosec-1 is [-π/2, π/2] – {0} and cosec (π/3) = (2/√3) Hence, the principal value of cosec-1 (2/√3) is π/3 (iv) Given as cosec-1 (2 cos(2π/3)) As we know that cos (2π/3) = – ½ So, 2 cos (2π/3) = 2 × – ½ 2 cos (2π/3) = -1 Substitute these values in cosec-1 (2 cos(2π/3)) we get, cosec-1 (-1) Let y = cosec-1 (-1) – cosec y = 1 – cosec (π/2) = cosec (-π/2) [since –cosec θ = cosec (-θ)] So, the range of principal value of cosec-1 [-π/2, π/2] – {0} and cosec (-π/2) = – 1 cosec (-π/2) = – 1 So, the principal value of cosec-1 (2 cos(2π/3)) is – π/2 |
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