# [SOLVED] Trig Help required

• Mar 5th 2008, 12:37 AM
stuckbigtime
[SOLVED] Trig Help required
Question is v=240sin(100 pi t-pi/6) i=30cos(100 pi t+pi/3)
power = V x i
find max power and time it first occurs

I have expanded v to give me (207.84sin 100 pi t-120 cos 100 pi t)
and expanded i to give me (15 cos 100 pi t - 25.98 sin 100 pi t)

multiply v x i gives me
3117.6 sin 100 pi t cos 100 pi t - 5399.68sinē 100 pi t - 1800 cosē 100 pi t + 3117.6 cos 100 pi t sin 100 pi t

not too sure if I have my + and - in correct places after I have multiplied and i'm stuck after that about how to get my max power and time answers ??
any help would be greatful its quite a big question
• Mar 5th 2008, 02:58 AM
mr fantastic
Quote:

Originally Posted by stuckbigtime
Question is v=240sin(100 pi t-pi/6) i=30cos(100 pi t+pi/3)
power = V x i
find max power and time it first occurs

I have expanded v to give me (207.84sin 100 pi t-120 cos 100 pi t)
and expanded i to give me (15 cos 100 pi t - 25.98 sin 100 pi t)

multiply v x i gives me
3117.6 sin 100 pi t cos 100 pi t - 5399.68sinē 100 pi t - 1800 cosē 100 pi t + 3117.6 cos 100 pi t sin 100 pi t

not too sure if I have my + and - in correct places after I have multiplied and i'm stuck after that about how to get my max power and time answers ??
any help would be greatful its quite a big question

$\displaystyle P = 240 \sin \left( 100 \pi t - \frac{\pi}{6} \right) \times 30 \cos \left( 100 \pi t + \frac{\pi}{3} \right)$

$\displaystyle = 7200 \sin \left( 100 \pi t - \frac{\pi}{6} \right) \cos \left( 100 \pi t + \frac{\pi}{3} \right)$.

Now use the product rule to get dP/dt. Then solve dP/dt = 0 for t. Test the nature - keep the maximum. Substitute this value of t into P to get maximum P.