Trig Identities

[igl1$]

Hi could someone please lead me through the problem below,
3sinωt + 4cosωt = 5sin(ωt+0.09395) Verify the resultant using the double angle formula sin(A+B).

I understand how the resultant 5sin(ωt+0.09395) was formed but I am struggling with the verification. The LHS must be proved to equal the RHS.

 

[soroban]

Hello, igl1$ !

\(\displaystyle 3\sin\omega t + 4\cos\omega t \:=\:5\sin(\omega + 0.09395)\) . This is not true!
Verify the result using the double-angle formula \(\displaystyle \sin(A+B)\)

I understand how the result \(\displaystyle 5\sin(\omega t + 0.09395)\) was formed
but I am struggling with the verification.
The LHS must be proved to equal the RHS.

The right side should be: .\(\displaystyle 5\sin(\omega t + \color{red}{0.9273})\)

 

[SAMUELK]

Hi could someone please lead me through the problem below,
3sinωt + 4cosωt = 5sin(ωt+0.09395) Verify the resultant using the double angle formula sin(A+B).

I understand how the resultant 5sin(ωt+0.09395)was formed but I am struggling with the verification. The LHS must be proved to equal the RHS.

I think the electrical engineers have a name for it, which I cannot recall. But:
Set
A sin wt + B cos wt = C sin (wt + phi)

= C [sin wt cos phi + cos wt sin phi]​

= C cos phi sin wt + C sin phi cos wt​

Now set:

C sin phi = B​

C cos phi = A​

And tan phi = B/A


Take it from there.

 

[igl1$]

I think the electrical engineers have a name for it, which I cannot recall. But:
Set
A sin wt + B cos wt = C sin (wt + phi)

= C [sin wt cos phi + cos wt sin phi]​

= C cos phi sin wt + C sin phi cos wt​

Now set:

C sin phi = B​

C cos phi = A​

And tan phi = B/A


Take it from there.

Hi, The method shown is how I derived the resultant, using the 'compound angle' formula. I need to know how I can verify the resultant using the double angle formula sin(A+B). In the compound angle formula if we let B = A then this would take the form sin2A = 2sinAcosA but how is this applied to give verification of the resulatant?

Sorry for the error Soroban, the phase shift is indeed 0.9273 radians as you corrected, thanks.

 

[Deleted member 4993]

Hi could someone please lead me through the problem below,
3sinωt + 4cosωt = 5sin(ωt+0.09273) Verify the resultant using the double angle formula sin(A+B).

I understand how the resultant 5sin(ωt+0.09273)was formed but I am struggling with the verification. The LHS must be proved to equal the RHS.

Verification would mean, you need to show:

5sin(ωt+0.09273) → 3sinωt + 4cosωt

5sin(ωt+0.09273)

= 5 * [sin(ωt) * cos(0.09273) + cos(ωt) * sin(0.09273)] ← using sin(A+B) = sin(A)*cos(B) + cos(A)*sin(B)

= 5 * [sin(ωt) * 0.599996 + cos(ωt) * 0.800003]

Now continue.....

 

[igl1$]

Verification would mean, you need to show:

5sin(ωt+0.09273) → 3sinωt + 4cosωt

5sin(ωt+0.09273)

= 5 * [sin(ωt) * cos(0.09273) + cos(ωt) * sin(0.09273)] ← using sin(A+B) = sin(A)*cos(B) + cos(A)*sin(B)

= 5 * [sin(ωt) * 0.599996 + cos(ωt) * 0.800003]

Now continue.....

Many thanks Subhotosh, I think I was trying to overcomplicate the verification method. You have made this very simple and I thankyou.