Differential-Equations · Unit 8 · Video 3 · Interactive Practice

Three Proofs of the Sinusoidal Identity

IKey Formulas

Formula	Name	What it says
$a\cos\theta + b\sin\theta = A\cos(\theta - \phi)$ , $\ A = \sqrt{a^2 + b^2}$	Sinusoidal identity	$(a, b) = (A\cos\phi, A\sin\phi)$ : $\phi$ is the polar angle of the point $(a, b)$ , equal to $\tan^{-1}(b/a)$ only when $a > 0$
$A\cos(\theta - \phi) = A\cos\phi\,\cos\theta + A\sin\phi\,\sin\theta$	Proof 1: trig expansion	Matching coefficients gives $a = A\cos\phi$ , $b = A\sin\phi$ ; it runs backwards, from the answer to the check
$\langle a, b\rangle \cdot \mathbf{u}_\theta = a\cos\theta + b\sin\theta = A \cdot 1 \cdot \cos(\theta - \phi)$	Proof 2: dot product	$\mathbf{u}_\theta = \langle\cos\theta, \sin\theta\rangle$ ; one dot product, by components and by lengths and angle
$\operatorname{Re}\big[(a - bi)\,e^{i\theta}\big] = a\cos\theta + b\sin\theta$ , $\ (a - bi)\,e^{i\theta} = A\,e^{i(\theta - \phi)}$	Proof 3: complex numbers	One product, in Cartesian and in polar form; the conjugate $a - bi = A\,e^{-i\phi}$ is what gives $+\,b\sin\theta$

Key Insight: The last two proofs run forwards from $a\cos\theta + b\sin\theta$ and work the same way: compute one quantity two ways, then set the two results equal.

IIProof 1 — Guess, Then Match Coefficients

The expansion proof needs a guess for $A$ and $\phi$ first; matching coefficients only checks it.

Target A 2.00 φ 0°

IIIProof 2 — One Dot Product, Two Ways

The dot product $\langle a, b\rangle \cdot \mathbf{u}_\theta$ computed from components equals the one computed from lengths and the angle between.

💡 Challenge: place $\mathbf{u}_\theta$ so the dot product is as large as possible. That largest value is the amplitude $A$ , reached at $\theta = \phi$ .

IVProof 3 — One Product, Cartesian and Polar

As $\theta$ turns, the real part of $(a - bi)\,e^{i\theta} = A\,e^{i(\theta - \phi)}$ traces exactly $a\cos\theta + b\sin\theta$ .

θ 70° First factor

VQuiz Questions

Problem 1 · Amplitude and Phase Lag

Given: $-\cos\theta + \sqrt{3}\,\sin\theta$ — write it in the form $A\cos(\theta - \phi)$ .

✅ Correct!

(a, b) = (-1, \sqrt{3})

is the point

2\big(\cos\tfrac{2\pi}{3}, \sin\tfrac{2\pi}{3}\big)

in the second quadrant, so

A = 2

and

\phi = \tfrac{2\pi}{3}

❌ Close, but…

\tan^{-1}(\sqrt{3}/(-1)) = -\tfrac{\pi}{3}

points to the fourth quadrant, but

(-1, \sqrt{3})

is in the second. Expanding

2\cos(\theta + \tfrac{\pi}{3})

gives

\cos\theta - \sqrt{3}\sin\theta

: both signs are wrong.

❌ Not quite.

2\cos(\theta - \tfrac{\pi}{3}) = \cos\theta + \sqrt{3}\sin\theta

, so the coefficient of

\cos\theta

has the wrong sign. With

a < 0

the angle

\phi

is in the second quadrant.

❌ Close, but…

a^2 + b^2 = 1 + 3 = 4

A^2

; the amplitude is its square root,

A = 2

Show solution

Here $a = -1$ and $b = \sqrt{3}$ , so

A = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{4} = 2.

We need $\cos\phi = a/A = -\tfrac{1}{2}$ and $\sin\phi = b/A = \tfrac{\sqrt{3}}{2}$ . That is the second-quadrant angle $\phi = \tfrac{2\pi}{3}$ (not $\tan^{-1}(-\sqrt{3}) = -\tfrac{\pi}{3}$ , which has the right tangent but sits in the fourth quadrant).

Check by expanding (Proof 1):

2\cos\!\left(\theta - \tfrac{2\pi}{3}\right) = 2\cos\tfrac{2\pi}{3}\cos\theta + 2\sin\tfrac{2\pi}{3}\sin\theta = -\cos\theta + \sqrt{3}\sin\theta.

Problem 2 · Which Factor Is the Conjugate?

Given: $e^{i\theta} = \cos\theta + i\sin\theta$ — which expression equals $3\cos\theta + 2\sin\theta$ ?

✅ Correct! The real terms are

3\cos\theta

and

(-2i)(i\sin\theta) = -2i^2\sin\theta = 2\sin\theta

❌ Close, but…

(2i)(i\sin\theta) = 2i^2\sin\theta = -2\sin\theta

, so this real part is

3\cos\theta - 2\sin\theta

: the wrong sign on the sine term.

❌ Not quite. The imaginary part collects the cross terms:

3\sin\theta + 2\cos\theta

, which swaps the coefficients.

❌ Not quite. The imaginary part of

(3 - 2i)(\cos\theta + i\sin\theta)

3\sin\theta - 2\cos\theta

, not the target.

Show solution

Only the real part is needed, so keep the two real products:

(3 - 2i)(\cos\theta + i\sin\theta) = 3\cos\theta + (-2i)(i\sin\theta) + i(\cdots) = 3\cos\theta + 2\sin\theta + i(\cdots).

The other two products, $3i\sin\theta$ and $-2i\cos\theta$ , are imaginary. With $3 + 2i$ instead, the product of imaginary terms is $2i^2\sin\theta = -2\sin\theta$ , which is why the first factor must be the conjugate $a - bi$ .

Problem 3 · The Complex Proof, Third Quadrant

Given: $-\sqrt{3}\cos\theta - \sin\theta = \operatorname{Re}\big[(a - bi)\,e^{i\theta}\big]$ with $a = -\sqrt{3}$ , $b = -1$ — find the polar form of $a - bi$ , then the single cosine.

What is $a - bi$ in polar form?

So $-\sqrt{3}\cos\theta - \sin\theta$ equals

✅ Correct!

a - bi = 2e^{5\pi i/6} = 2e^{-i\phi}

with

\phi = -\tfrac{5\pi}{6}

, so the product is

2e^{i(\theta + 5\pi/6)}

and its real part is

2\cos(\theta + \tfrac{5\pi}{6})

❌ Close, but…

2e^{-5\pi i/6} = -\sqrt{3} - i

a + bi

. Its conjugate

a - bi = -\sqrt{3} + i

lies in the second quadrant.

❌ Not quite.

2e^{\pi i/6} = \sqrt{3} + i

comes from

\tan^{-1}(b/a) = \tan^{-1}(1/\sqrt{3}) = \tfrac{\pi}{6}

, which ignores the signs;

-\sqrt{3} + i

has negative real part, so it lies in the second quadrant.

❌ Close, but… the modulus is

\sqrt{3 + 1} = 2

, not

3 + 1

❌ Check the polar form. Plot

a - bi = -\sqrt{3} + i

first: which quadrant is it in?

❌ Close, but…

\tan^{-1}(b/a) = \tan^{-1}(1/\sqrt{3}) = \tfrac{\pi}{6}

ignores that

a < 0

2\cos(\theta - \tfrac{\pi}{6}) = \sqrt{3}\cos\theta + \sin\theta

, the negative of the target.

❌ Check the exponent. Multiplying

2e^{5\pi i/6}

e^{i\theta}

adds the exponents, giving

2e^{i(\theta + 5\pi/6)}

Show solution

$a - bi = -\sqrt{3} - (-1)i = -\sqrt{3} + i$ . Its modulus is $\sqrt{3 + 1} = 2$ , and it lies in the second quadrant at angle $\tfrac{5\pi}{6}$ (cosine $-\tfrac{\sqrt{3}}{2}$ , sine $\tfrac{1}{2}$ ):

a - bi = 2e^{5\pi i/6}.

Multiplying adds the exponents:

(a - bi)\,e^{i\theta} = 2e^{i(\theta + 5\pi/6)}, \qquad \operatorname{Re} = 2\cos\!\left(\theta + \tfrac{5\pi}{6}\right).

In the identity's notation $a - bi = Ae^{-i\phi}$ , so $\phi = -\tfrac{5\pi}{6}$ : the point $(a, b) = (-\sqrt{3}, -1)$ is in the third quadrant. Check: $2\cos(\theta + \tfrac{5\pi}{6}) = 2\cos\tfrac{5\pi}{6}\cos\theta - 2\sin\tfrac{5\pi}{6}\sin\theta = -\sqrt{3}\cos\theta - \sin\theta$ .

Problem 4 · An Oscillation in Time

Given: a solution $y(t) = 6\cos 2t + 8\sin 2t$ of an oscillator equation — find its amplitude and the first time $t > 0$ at which it reaches its maximum.

What is the amplitude?

When is the first maximum?

✅ Correct!

y = 10\cos(2t - \phi)

with

\phi = \tan^{-1}\tfrac{4}{3}

, and the cosine first peaks when

2t = \phi

❌ Check the amplitude.

A = \sqrt{a^2 + b^2} = \sqrt{36 + 64}

; it is not

a + b = 14

, not

\sqrt{a + b} = \sqrt{14}

and not

a^2 + b^2 = 100

❌ Close, but… the phase lag

\phi

is an angle in the variable

2t

. Solving

2t = \phi

gives

t = \phi/2

❌ Close, but…

\tan\phi = b/a = 8/6 = 4/3

, not

a/b

❌ Check the time. The maximum of

\cos(2t - \phi)

is at

2t - \phi = 0

Show solution

Apply the identity with the angle $2t$ in place of $\theta$ : $a = 6$ , $b = 8$ , both positive, so

A = \sqrt{6^2 + 8^2} = 10, \qquad \phi = \tan^{-1}\tfrac{8}{6} = \tan^{-1}\tfrac{4}{3} \approx 0.927.

y(t) = 10\cos(2t - \phi).

The cosine first reaches $1$ (for $t > 0$ ) when $2t - \phi = 0$ , so $t = \phi/2 \approx 0.464$ , where $y = 10$ . Check: $y(0.464) = 6\cos 0.927 + 8\sin 0.927 = 6(0.6) + 8(0.8) = 10$ .

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