Hi,

Can someone help me express " i(t) = 0.5cos(25πt) + 0.25sin(25πt) - 1.2) " as a sine function please?

Printable View

- Dec 13th 2012, 05:09 AMaritechSine function
Hi,

Can someone help me express " i(t) = 0.5cos(25πt) + 0.25sin(25πt) - 1.2) " as a sine function please?

- Dec 13th 2012, 05:45 AMHallsofIvyRe: Sine function
sin(A+ B)= sin(A)cos(B)+ cos(A)sin(B).

Set $\displaystyle B= 0.25\pi t$. Of course, there is no "A" such that sin(A)= .5 and cos(A)= 0.25 because [itex]sin^2(A)+ cos^2(A)= 1\ne 0.5^2+ 0.25^2[/tex] but you can both multiply and divide by [itex]\sqrt{0.5^2+ 0.25^2}= \sqrt{5/16}= \sqrt{5}/4[/tex] to get $\displaystyle \frac{\sqrt{5}}{4}(\frac{4}{\sqrt{5}}cos(25\pi t)+ \frac{4}{\sqrt{5}}sin(25\pi t))$.

Now one problem is that you have and extra ")" with no corresponding "(". Did you mean $\displaystyle sin(25\pi t- 1.2)$? If so then use the fact that $\displaystyle sin(25\pi t- 1.2)= sin(25\pi)cos(1.2)- cos(25\pi t)sin(1.2)$ first.