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Mathematics 10 Online
OpenStudy (anonymous):

tan(x/2)=(sinx)/1+cosx for all values of x true or false

OpenStudy (anonymous):

sinx=2sinx/2 cosx/2........1 1+cosx=2(cosx)^2............2 1/2=tan(x/2)..... seems to be right!!!!!

OpenStudy (anonymous):

1+cosx=2*(cosx/2)^2 edit it while doing....

OpenStudy (anonymous):

\[\sin(x)/(1+\cos(x))=\sin(2(x/2))/(1+\cos(2(x/2))=\] \[2\sin(x/2)\cos(x/2)/(1+\cos^2(x/2)-\sin^2(x/2))=\] \[2\sin(x/2)\cos(x/2)/(\sin^2(x/2)+\cos^2(x/2)+\cos^2(x/2)-\sin^2(x/2))=\] \[\frac{2\sin(x/2)\cos(x/2)}{2\cos^2(x/2)}=\frac{\sin(x/2)}{\cos(x/2)}=\tan(x/2)\] Using the identities: \[\sin(2x)=2\sin(x)\cos(x); \cos(2x)=\cos^2(x)-\sin^2(x); \cos^2(x)+\sin^2(x)=1\]

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