cosx=-4/5 x属于(π,3π/2)求1+tanx/2除以1-tanx/2
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cosx=-4/5 x属于(π,3π/2)求1+tanx/2除以1-tanx/2
cosx=-4/5
x∈(π,3π/2)
sinx=-√[1-(-4/5)^2]=-3/5
所以[1+tan(x/2)]/[1-tan(x/2)]
=[cos(x/2)+sin(x/2)]/[cos(x/2)-sin(x/2)](分子分母同时乘以cos(x/2))
=[cos(x/2)+sin(x/2)]^2/[cos(x/2)-sin(x/2)][cos(x/2)+sin(x/2)]
=(1+sinx)/cosx
=(1-3/5)/(-4/5)
=-1/2
如果不懂,请Hi我,祝学习愉快!
x∈(π,3π/2)
sinx=-√[1-(-4/5)^2]=-3/5
所以[1+tan(x/2)]/[1-tan(x/2)]
=[cos(x/2)+sin(x/2)]/[cos(x/2)-sin(x/2)](分子分母同时乘以cos(x/2))
=[cos(x/2)+sin(x/2)]^2/[cos(x/2)-sin(x/2)][cos(x/2)+sin(x/2)]
=(1+sinx)/cosx
=(1-3/5)/(-4/5)
=-1/2
如果不懂,请Hi我,祝学习愉快!
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