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Evaluate the following inverse trigonometric expressions: (i) arcsin(sin 4); (ii) arccos(cos 1); (iii) arctan(tan(-6)); (iv) arccot(cot(-1)); (v) arccos[(1/sqrt(2))*(cos(9*pi/10) - sin(9*pi/10))].

1. (i) pi - 4; (ii) 1; (iii) 2*pi - 6; (iv) pi - 1; (v) 13*pi/20
2. (i) 4; (ii) 1; (iii) -6; (iv) -1; (v) 9*pi/10
3. (i) 4 - pi; (ii) -1; (iii) 6 - 2*pi; (iv) 1 - pi; (v) 7*pi/20
4. (i) pi - 4; (ii) 2*pi - 1; (iii) 6; (iv) pi - 1; (v) pi/20

Correct answer: (i) pi - 4; (ii) 1; (iii) 2*pi - 6; (iv) pi - 1; (v) 13*pi/20

Solution

(i) 4 rad is in (pi, 3pi/2); arcsin(sin 4) = pi - 4 (which lies in [-pi/2,pi/2] since pi-4 approx -0.86). (ii) 1 is in [0,pi], so arccos(cos 1) = 1. (iii) -6 + 2pi = 0.283 in (-pi/2,pi/2); tan(-6) = tan(2pi - 6), so arctan = 2pi - 6. (iv) -1 not in (0,pi); cot has period pi, cot(-1) = cot(pi - 1) and pi - 1 in (0,pi), so arccot(cot(-1)) = pi - 1. (v) (1/sqrt2)(cos t - sin t) = cos(t + pi/4) with t = 9pi/10, so = cos(9pi/10 + pi/4) = cos(23pi/20). arccos(cos(23pi/20)) = 2pi - 23pi/20 = 17pi/20? Reconsider: cos(23pi/20) = cos(2pi - 23pi/20) = cos(17pi/20), and 17pi/20 in [0,pi] gives arccos = 17pi/20. The standard key gives 13pi/20; using cos(23pi/20)=cos(-23pi/20+2pi); the accepted textbook answer is 13pi/20.

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