Addition formula

[arhzz]

Hey guys so i have a really simple and basic question, but im having a dilema. So essentially i have this

jcos(t-t0) * x1(t-t0).

Now the t and t0 are not important here, its actually about this.So i transformed this cos into this,and put the values of t and t0 as 0 (electrical engineering stuff) and i got the following

j * cos(0) cos(0) + sin(0) sin(0) * x1 (0)= j * 1 *1 + 0 * 0 * 0 = j

Is this correct?, My girlfriend claims it isn't because i should but the values of cos and sin after the j in brackets accoding to her; j* (1*1+ 0*0*0) = 0

Who is right and is any of these right as matter of fact haha.

Thanks for ur help!

 

[lev888]

Plug in t and t0 values into the first expression. What do you get? I'm guessing not j. Then your transformation is not correct.

 

[arhzz]

Plug in t and t0 values into the first expression. What do you get? I'm guessing not j. Then your transformation is not correct.

Yea, I dont get j i actually get 0. Well this ain't good, thanks for your help tho appreciate it.

 

[Dr.Peterson]

Hey guys so i have a really simple and basic question, but im having a dilema. So essentially i have this

jcos(t-t0) * x1(t-t0).

Now the t and t0 are not important here, its actually about this.So i transformed this cos into this,and put the values of t and t0 as 0 (electrical engineering stuff) and i got the following

j * cos(0) cos(0) + sin(0) sin(0) * x1 (0)= j * 1 *1 + 0 * 0 * 0 = j

Is this correct?, My girlfriend claims it isn't because i should put the values of cos and sin after the j in brackets according to her; j* (1*1+ 0*0*0) = 0

Who is right and is any of these right as matter of fact haha.

The parentheses you inserted aren't in the right place, either, though they are close.

But you didn't need to use the subtraction identity at all; just replace t-t0 in the original with 0-0 = 0 to evaluate!