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Mathematics 7 Online
OpenStudy (anonymous):

How would you solve cos(5x)=sin(10x) ?

OpenStudy (asnaseer):

Simplify using:\[\sin(10x)=2\sin(5x)\cos(5x)\]

OpenStudy (anonymous):

how do you know that?

OpenStudy (asnaseer):

using the rule:\[\sin(2x)=\sin(x+x)=\sin(x)\cos(x)+\cos(x)\sin(x)=2\sin(x)\cos(x)\]

OpenStudy (anonymous):

right, but how come you can separate the 10x to 5x?

OpenStudy (asnaseer):

\[\sin(10x)=\sin(5x+5x)=...\]

OpenStudy (anonymous):

oh! i see, thank you!

OpenStudy (asnaseer):

yw

OpenStudy (anonymous):

does that mean you do \[\cos(5x)= 2\sin(5x)+2\cos(5x)\] ?

OpenStudy (anonymous):

and then \[0= 2\sin(5x)-\cos(5x) ?\]

OpenStudy (asnaseer):

no, it should be:\[\cos(5x)=\sin(10x)=2\sin(5x)\cos(5x)\] then cancel the \(\cos(5x)\) terms from left and right hand sides to leave:\[1=2\sin(5x)\]

OpenStudy (anonymous):

oh dear, I don't know why I added an addition since in there. That makes more sense

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