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anonymous
 5 years ago
find the exact value of tan x/2, givin that sin x=8/17 and 90degrees <x<180degrees
anonymous
 5 years ago
find the exact value of tan x/2, givin that sin x=8/17 and 90degrees <x<180degrees

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dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0For this problem you need to use your trig halfangle formulas sin(x/2) = sqrt((1cosx)/2) cos(x/2) = sqrt((1+cosx)/2) tanx = sinx/cosx so tan(x/2) = sin(x/2)/cos(x/2) >tan(x/2) = sqrt((1cosx)/2) / sqrt((1+cosx)/2) But now the function is defined in terms of cosx and we are given value of sinx so we use a trig identity sin^2 + cos^2 = 1 > cosx = sqrt(1sinx^2) > cosx = sqrt(1(8/17)^2) = sqrt((17^264)/17^2) = 15/17 sub that into the above equation for cosx with a little simplifying >tan(x/2) = sqrt(1/17) / sqrt(16/17) = (1/sqrt(17))*(sqrt(17)/sqrt(16)) = 1/4

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0actually tan(x/2) = 1/4 due to restrictions on x remember 1/sqrt(16) = +(1/4)
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