Question

y1=-3sin8t-4cos8t and y2=Asin(Bt+C) transform y1 into y2

Answer 1

as in the previous problem, we expand \[A\sin( Bt + C)\] to \[A\sin(Bt) \cos C + A \sin C \cos (Bt)\]

Answer 2

and compare it to \[-3\sin(8t)-4\cos(8t)\]

Answer 3

we conclude that the coefficients of t must be the same for both expressions

Answer 4

right

Answer 5

so \(B = 8\). agreed?

Answer 6

yes

Answer 7

i wouldn't have to add them together would I be it's 2 differnet equations though would I? I just want to make sure.

Answer 8

the coefficient of \(\sin (Bt)\) in the first expression is \(A \cos C\) and the coefficient of \(\sin 8t\) is \(-3\), so we have our first relation \(-3=A\cos C\)

Answer 9

i just want to make sure before we combined two to make one to satisfy both but since in this one there equal it doesn't matter right?

Answer 10

yes, we compare "similar terms"

Answer 11

okay

Answer 12

doing the same with \(\cos Bt\) and \(\cos 8t\), we have our 2nd relation \[-4=A\sin C\]

Answer 13

square both sides of the equation and adding them up, we get the value of \(A\)
dividing the 2nd equation by the first equation, we get an equation in \(\tan C\)

Answer 14

what did you get for \(A\)?

Answer 15

I'm working it out still sorry I write stuff out step by step when I'm working a problem so I don't mess up.

Answer 16

take your time. haste makes waste.

Answer 17

but I would get 25= A^2*cos^2(C)+A^2*sin^2(C)

Answer 18

is this right?

Answer 19

on the right track. factor out the \(A^2\), then use a trigonometric identity.

Answer 20

25=A^2(cos^2(C)+sin^2(C))

Answer 21

correct. \[\cos^2 C + \sin^2 C=1\]

Answer 22

so i need to take the square root right of both sides?

Answer 23

yes, you must, to find the value of \(A\).

Answer 24

A will =5

Answer 25

right?

Answer 26

right. it is 5.

Answer 27

I can use either equation then right to find C?

Answer 28

before you proceed with angle C
have you observed that in the two problems, \[A^2 = (-4)^2+(3)^2 \quad \text {previous}\]\[A^2 = (-4)^2+(-3)^2 \quad \text {current}\]

Answer 29

these are the constants in the pair of equations \[\begin{cases} -4=A \cos C \\3 = A\sin C\end{cases}\]and \[\begin{cases} -3=A \cos C \\-4 = A\sin C\end{cases}\]

Answer 30

I haven't noticed the previous one. I thought it was the current one from the beginning until you pointed it out.

Answer 31

they appear everytime in this class of trigonometric problems, so it should be steps shorter if you remember them.

Answer 32

okay

Answer 33

when we're done with this problem, i'll give you an example on finding \(A\).

Answer 34

okay

Answer 35

finding \(C\) is the tricky part. you have to account for the quadrant when choosing the value of C. your calculator may give you an angle but it may not be the right value for C.

Answer 36

okay so c will be in the first or fourth quadrant right?

Answer 37

\(-3=5 \cos C,\quad C\) in quadrant II or III

Answer 38

so i'm looking were cos would be negative then not positive right?

Answer 39

that's why it's the 2 and 3?

Answer 40

\(-4=5\sin C\quad C\) in quadrant III or IV

Answer 41

yes, Q II or Q III for cosine, Q III or Q IV for sine. i have provided a sign check above.

Answer 42

okay.