Is sin⁡(ωt+π)=−sin⁡ωt always true no matter what ω is?For example, it's true that sin⁡(t+π)=−sin⁡(t). You flip one over the x-axis, and you get the other. But sin⁡(2t+π)=sin⁡(2t). Because sin⁡(2t) has a period of π, so if you translate it by π, you are still gonna get sin⁡(2t)

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Is sin⁡(ωt+π)=−sin⁡ωt always true no matter what ω is?For example, it&#039;s true that sin⁡(t+π)=−sin⁡(t). You flip one over the x-axis, and you get the other. But sin⁡(2t+π)=sin⁡(2t). Because sin⁡(2t) has a period of π, so if you translate it by π, you are still gonna get sin⁡(2t)

Sabrina Campos · Accepted Answer

You are wrong: let u=2t; thensin⁡(2t+π)=sin⁡(u+π)=−sin⁡(u)=−sin⁡(2t)You are right when you assert that sin⁡(2t+π) is periodic with period π, but what you extract from that is thatsin⁡(2(t−π)+π)=2sin⁡(2t+π)in other words,sin⁡(2t−π)=sin⁡(2t+π)

Is \(\displaystyle{\sin{{\left(\omega{t}+\pi\right)}}}=-{\sin{\omega}}{t}\) always true no matter what

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