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Mathematics 16 Online
OpenStudy (ammarah):

Solve: sinx + cosx = 1

OpenStudy (anonymous):

\[\sin x+\cos x=1\] squaring both sides \[\sin ^2 x+\cos ^2x+2\sin x \cos x=1\] \[1+\sin 2x=1,\sin 2x=0=\sin n \pi,2x=n \pi,x=\frac{ n \pi }{ 2 }\] where n is an integer.

OpenStudy (ammarah):

wait why square both sides

OpenStudy (anonymous):

\[\sin ^2x+\cos ^2x=1\]

OpenStudy (anonymous):

\[\sin 2x=2\sin x \cos x\]

OpenStudy (ammarah):

so sin x+cosx =sin 2x

OpenStudy (ammarah):

could u show me step by step process, i will memorize the identities.

OpenStudy (ammarah):

like how did u get from sinx + cosx to sin 2xcos2x??

OpenStudy (anonymous):

i know the identity \[\sin ^2x+\cos ^2x=1\] sin 2x=2sinx cosx the idea is to change in one trigonomety ratio only.

OpenStudy (ammarah):

ok

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