# How to change the form $A\cos \omega t + B\sin \omega t$ into the form $\sqrt{A^2+B^2}\cos(\omega t-\phi)$ [duplicate]

I'm preparing notes for teaching next week's class in differential equations. My hand calculation solution of $$y''+192y=0, y(0)=1/6, y'(0)=-1$$ is $$y=\frac16\cos 8\sqrt3 t-\frac{\sqrt3}{24}\sin 8\sqrt3 t$$ Now a few more hand calculations provides this equivalent solution: $$y=\frac{\sqrt{19}}{24}\cos(8\sqrt3 t-\tan^{-1}(-\sqrt3/4))$$

Now, when we solve the equation using DSolveValue, we get the following result.

sol = DSolveValue[{u''[t] + 192 u[t] == 0, u == 1/6, u' == -1}, u[t],
t]


Out= 1/24 (4 Cos[8 Sqrt t] - Sqrt Sin[8 Sqrt t])

Which agrees with my hand calculated solution. Moreover, I was able to check that my final form was equivalent to the output provided by DSolveValue.

sol - Sqrt/24*Cos[8 Sqrt t - ArcTan[-Sqrt/4]] // Simplify


Out= 0

Now my question. Is there a simple way in Mathematica to convert the expression

$$\frac16\cos 8\sqrt3 t-\frac{\sqrt3}{24}\sin 8\sqrt3 t$$

to

$$\frac{\sqrt{19}}{24}\cos(8\sqrt3 t-\tan^{-1}(-\sqrt3/4))$$

These are the hand calculations used to make the change. • How did you come up with that little graph to make that simplification? It definitely works but I'm not sure what method you are using there? Dec 12 '16 at 10:45
• @user32882 If you have an expression such as $a\cos\theta+b\sin\theta$, the first step is to factor out $\sqrt{a^2+b^2}$, giving $\sqrt{a^2+b^2}(\frac{a}{\sqrt{a^2+b^2}}\cos\theta+\frac{b}{\sqrt{a^2+b^2}}\sin\theta)$. Then, using my picture, you can replace $\frac{a}{\sqrt{a^2+b^2}}$ with $\cos\phi$ and $\frac{b}{\sqrt{a^2+b^2}}$ with $\sin\phi$. Dec 13 '16 at 17:06

Not sure this is what you need :

A = 1/6; B = -Sqrt/24; ω = 8 Sqrt;
r = Sqrt[A^2 + B^2]
ClearAll[t];
sol = Solve[ A Cos[ω t] + B Sin[ω t] == r Cos[ω t - ϕ], ϕ];
t = 0;
ϕ /. sol[]
(* ConditionalExpression[-ArcCos[4/Sqrt] - 2 π C,  C ∈ Integers] *)


Clear[TrigShrink];
TrigShrink[exp_, trgt_, lag_: ϕ] :=
Module[{sign, xExp, xRes, xTrig, cCos, cSin, tan, xlag, clag},
xExp = exp // TrigExpand // Collect[#, {Sin[trgt], Cos[trgt]}] &;
xRes = xExp /. {Sin[trgt] -> 0, Cos[trgt] -> 0};
xTrig = xExp - xRes;
cCos = xTrig /. {Sin[trgt] -> 0, Cos[trgt] -> 1};
cSin = xTrig /. {Sin[trgt] -> 1, Cos[trgt] -> 0};
sign = If[Sign[cSin] // NumberQ, Sign[cSin], 1];
xlag = ArcTan[cCos/cSin] // Simplify;
clag = sign Sqrt[cSin^2 + cCos^2] // Simplify;
{C[lag] Sin[trgt + lag] + xRes, lag -> xlag, C[lag] -> clag} //
Simplify // Flatten
]


Here are some test cases:

TrigShrink[2 Sin[x + α] + Sin[x + 8] + Cos[t], x, θ]
(*res: {Cos[t] + C[θ] Sin[x + θ], θ ->ArcTan[(Sin + 2 Sin[α])/(Cos + 2 Cos[α])], C[θ] -> Sqrt[5 + 4 Cos[8 - α]]}*)

TrigShrink[ Sin[x] + Cos[x], x, θ]
(*{C[θ] Sin[x + θ], θ -> π/4, C[θ] -> Sqrt}*)


TrigShrink[1/24 (4 Cos[x] - Sqrt Sin[x]), x, θ] /. x -> 8 Sqrt t