Thank you for the opportunity to help you with your question!
you have to show f(x)-sinx=0 for all pi/2<x<pi. Note that tanx is defined here and tanx=sinx/cosx Then f(x)-sinx=sin^3 x + cos^3 x tan x-sinx=sinx(sin^2 x+cos^2 x-1) But sin^2 x+cos^2 x=1 for any value of x, so we have =sinx(1-1)=sinx(0)=0.
sinx=2/3; note that since sine is continuous function we have that x is a value in the range pi/2<x<pi (there are other ranges, but this is enough). Now sin^2 x=4/9. Then 1-cos^2 x=4/9; rearranging we get cos^2 x = 5/9 cosx=sqrt(5)/3 or -sqrt(5)/3
Please let me know if you need any clarification. I'm always happy to answer your questions.
Jun 29th, 2015
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