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Trigonometry 17 Online
OpenStudy (anonymous):

sin(2x)/cos(x)-cos^2(x)=sin^2(x) Thank you in advance

OpenStudy (phi):

Have you tried anything ?

OpenStudy (mathmale):

Hello, Emily! Welcome to OpenStudy!! Why not break this problem down into smaller parts to make it more comprehensible? Look at the first term:\[\sin(2x)/\cos(x)=\frac{ \sin 2x }{ \cos x }\] Please look at your table of trig identities (find one on the Internet if need be), and rewrite sin 2x in a format that includes both sin x and cos x.

OpenStudy (anonymous):

Finally I have: 2sin=1; p/6+2pn But I am not sure

OpenStudy (mathmale):

It'd likely be very helpful for both of us if you were to show all of your work. There is an identity for sin 2x; have you looked for it yet, Emily? If not, please do that. You could look up "double angle formula" or just "trig identities." Have you a textbook? If not, have you an online reference? If not, are you willing to try an Internet search?

OpenStudy (anonymous):

sinx=1/2

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