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Mathematics 14 Online
OpenStudy (anonymous):

Solve for theta, given sin theta + cos theta = 0 and 0<= theta <=pi

OpenStudy (anonymous):

subtract the cos: sin(theta)=-cos(theta) Divide by cos(theta) -tan(theta)=1 tan is one at pi/4. Its negative in quadrant 2 so 3pi/4

OpenStudy (anonymous):

so you can;t square it?

OpenStudy (anonymous):

You mean try to do sin^2(theta)+cos^2(theta)=1 or square it to get rid of the negative 1?

OpenStudy (anonymous):

yes first one

OpenStudy (anonymous):

Well, if you square it you get (sin(x)+cos(x))^2=sin^2(x)+cos^2(x)+2sin(x)cos(x) So that doesn't quite help. You get 1+sin(2x)=0 (using the double angle formula sin(2x)=2sin(x)cos(x) Then you could subtract) 1=-sin(2x) Either way, you get the same answer :P

OpenStudy (anonymous):

that makes sense - thank you!

OpenStudy (anonymous):

No problem :D

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