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If $v(t)=10\cos(10t+30°)$, and $i(t)=-5\sin(10t-70°)$, my classmate said the phase of $v(t),\theta_v,$ is leading the phase of $i(t),\theta_i $ for $10°$.

Here is his proof

$i(t)=-5\sin(10t-70°)=5\sin(5t-70°+180)=5\cos(10t+110°-90°)=5\cos(10t+20°)$$v(t)=10\cos(10t+30°)$

So $\theta_i=20°$, and $\theta_v=30°$, so $\theta_v$ is leading $\theta_i $ for $10°$.

However, here is my calculation

$v(t)=10\cos(10t+30°)=10\cos(10t+(90°-60°))=10\sin(10t+60°)$$i(t)=-5\sin(10t-70°)=5\sin(10t-70°+180°)=5\sin(10t+110°)$

So $\theta_i=110°$, and $\theta_v=60°$, so $\theta_v$ is lagging $\theta_i $ for $50°$.

The answer shows me my classmate is right,but i don't know where am i wrong in my calculation,can anyone tell me?

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1 Answer

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$$10\cos\left(10t+(90^\circ-60^\circ)\right) \\=-10\sin(10t-60^\circ) \hspace{1 cm}\color{orange}{ \because[\cos(90^\circ + x)=-\sin x]} \\=10\sin(60^\circ-10t) \\ = 10\sin\left(180^\circ-(60^\circ-10t)\right) \\ =10\sin(10t+120^\circ) \ne 10\sin(10t+60^\circ)$$

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