can some hlep me solve this trig problem
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skeeter
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draw a reference triangle in quad II ...
tanx = -24/7 = opp/adj ... therefore, hyp = 25
so cosx = adj/hyp = -7/25
sin(x/2) = +/- sqrt[(1 - cosx)/2]
x is in quad II, x/2 will be in quad I ... sin(x/2) > 0
sin(x/2) = sqrt[(1 + 7/25)/2] = sqrt(16/25) = 4/5
tanx = -24/7 = opp/adj ... therefore, hyp = 25
so cosx = adj/hyp = -7/25
sin(x/2) = +/- sqrt[(1 - cosx)/2]
x is in quad II, x/2 will be in quad I ... sin(x/2) > 0
sin(x/2) = sqrt[(1 + 7/25)/2] = sqrt(16/25) = 4/5