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Why is -3sin(2t+30) equal to 3cos(2t+120)?
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Look at the angles, 30 and 120. |dw:1363833373137:dw|
One angle's sin is another angle's cos.
cos(A)=sin(90-A) Simplifying RHS:- 3cos(2t+120) =3sin(90-(2t+120) ) =3sin(-2t-30) =3sin(-(2t+30) ) Now we know that , sin(-A)=-sinA A=2t+30 Hence(proceeding from where we had left the simplification), =-3sin(2t+30) LHS=RHS A lengthier process would be using the formulae: - \[\sin (A+B)= \sin A \cos B+\cos A \sin B\] \[\cos (A+B)=cosAcosB-sinAsinB\] and using the known values of cos30 and sin30
As wio said we can also do that by looking at the quadrants
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