Differential-Equations · Unit 8 · Video 2 · Interactive Practice

The Sinusoidal Identity and Its Right Triangle

IKey Formulas

Formula	Name	What you need
$a\cos\theta + b\sin\theta = A\cos(\theta - \phi)$	Sinusoidal identity	The same angle $\theta$ (same frequency) in both terms
$A = \sqrt{a^2 + b^2}$	Amplitude = hypotenuse	Legs $a$ (horizontal) and $b$ (vertical)
$\tan\phi = \dfrac{b}{a}, \quad \phi = \tan^{-1}\dfrac{b}{a}$	Phase angle, opposite $b$	$a, b > 0$ , so $0 < \phi < \tfrac{\pi}{2}$ ; if $a < 0$ , add $\pi$ (Problem 3)
$\cos\omega t + \frac{\omega}{k}\sin\omega t = \sqrt{1+(\omega/k)^2}\,\cos(\omega t - \phi)$	Applied to the Cartesian answer	$a = 1$ , $b = \omega/k$ , $\theta = \omega t$

Key Insight: Memorize the triangle, not two formulas: legs $a$ and $b$ , hypotenuse $A$ , and the angle $\phi$ at the corner opposite $b$ , so $b$ goes on top in $\tan\phi = b/a$ .

IIThe Triangle Behind the Identity

The point $(a, b)$ is the corner of a right triangle whose hypotenuse is $A$ and whose angle is $\phi$ .

a 1 b 1

IIISame Frequency Required

A cosine and a sine combine into one cosine only when they share the same frequency.

Second term

IVConverting the Cartesian Answer

With $a = 1$ and $b = \omega/k$ , the triangle turns the Cartesian answer into amplitude $1/\sqrt{1+(\omega/k)^2}$ and phase lag $\phi$ .

ω/k 1

VQuiz Questions

Problem 1 · Read the Triangle

Given: $3\cos\theta + 4\sin\theta = A\cos(\theta - \phi)$ — find the amplitude $A$ and the phase angle $\phi$ .

What is the amplitude A?

What is the phase angle φ?

✅ Correct! The triangle has legs

3

and

4

, so it is the 3-4-5 triangle:

3\cos\theta + 4\sin\theta = 5\cos(\theta - 53.1\degree)

❌ Not quite. The hypotenuse is not the sum of the legs; Pythagoras gives

A = \sqrt{3^2 + 4^2}

❌ Not quite.

3^2 + 4^2 = 25

A^2

; the amplitude is its square root,

A = 5

❌ Not quite.

\sqrt{3 + 4}

skips squaring the legs; Pythagoras gives

A = \sqrt{3^2 + 4^2}

❌ Check the amplitude.

A

is the hypotenuse of the triangle with legs

a = 3

and

b = 4

: square, add, then take the square root.

❌ Close, but the ratio is upside down.

b = 4

is the side opposite

\phi

, so it goes on top:

\tan\phi = 4/3

❌ Not quite.

\tfrac{4}{5} = b/A

\sin\phi

, not

\tan\phi

; the tangent is opposite over adjacent,

b/a = 4/3

❌ Not quite.

\tfrac{4}{3}

\tan\phi

, not the angle itself; take the inverse tangent:

\phi = \tan^{-1}\tfrac{4}{3} \approx 0.927

rad.

❌ Check the angle.

\tan\phi

is opposite over adjacent,

b/a

, and

\phi

is the inverse tangent of that ratio.

Show solution

Here $a = 3$ and $b = 4$ . Draw the triangle with horizontal leg $3$ and vertical leg $4$ .

A = \sqrt{a^2 + b^2} = \sqrt{9 + 16} = \sqrt{25} = 5

The angle $\phi$ sits at the corner where the hypotenuse meets the horizontal leg, opposite $b$ :

\tan\phi = \frac{b}{a} = \frac{4}{3}, \qquad \phi = \tan^{-1}\frac{4}{3} \approx 53.1\degree \approx 0.927 \text{ rad}

So $3\cos\theta + 4\sin\theta = 5\cos(\theta - 53.1\degree)$ . Since $a, b > 0$ , $\phi$ lies between $0$ and $90\degree$ , as it should.

Problem 2 · Which Side Is on Top?

Given: $\cos 2t + \sqrt{3}\sin 2t$ — write it as a single cosine $A\cos(2t - \phi)$ .

✅ Correct! Legs

1

and

\sqrt{3}

give hypotenuse

2

and

\tan\phi = \sqrt{3}

, so

\phi = \pi/3

; the frequency

2

is untouched.

❌ Close, but the ratio is inverted.

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

puts

a

on top. The side opposite

\phi

b = \sqrt{3}

, so

\tan\phi = \sqrt{3}/1

❌ Not quite.

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

is the square of the amplitude; the amplitude itself is

\sqrt{4} = 2

❌ Not quite. The identity keeps the angle: here

\theta = 2t

, so the single cosine has the same frequency

2

, not

1

❌ Not quite. Read

a

b

and

\theta

off the expression, then use the triangle.

Show solution

Match $a\cos\theta + b\sin\theta$ : $a = 1$ , $b = \sqrt{3}$ , $\theta = 2t$ .

A = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{4} = 2, \qquad \tan\phi = \frac{b}{a} = \sqrt{3} \;\Rightarrow\; \phi = \frac{\pi}{3}

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

Check: $2\cos(2t - \tfrac{\pi}{3}) = 2\left(\cos 2t \cdot \tfrac{1}{2} + \sin 2t \cdot \tfrac{\sqrt{3}}{2}\right) = \cos 2t + \sqrt{3}\sin 2t$ ✓

Problem 3 · A Negative Coefficient

Given: $-\cos t + \sin t$ — write it as $A\cos(t - \phi)$ with $A > 0$ . Here $a < 0$ , so the corner $(a, b)$ is not in the first quadrant.

✅ Correct! The corner

(-1, 1)

points up and to the left, at angle

135\degree = 3\pi/4

❌ Close, but that is the calculator's angle.

\tan^{-1}(1/(-1)) = -\pi/4

points to

(1, -1)

, and

\sqrt{2}\cos(t + \pi/4) = \cos t - \sin t

, the negative of what you want. With

a < 0

, add

\pi

\phi = -\pi/4 + \pi = 3\pi/4

❌ Not quite.

\sqrt{2}\cos(t - \pi/4) = \cos t + \sin t

: the minus sign on the cosine was dropped.

❌ Not quite. The angle is right, but

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

, not

(-1)^2 + 1^2 = 2

Show solution

Here $a = -1$ , $b = 1$ . The amplitude is still the hypotenuse: $A = \sqrt{(-1)^2 + 1^2} = \sqrt{2}$ .

The angle must satisfy $\cos\phi = a/A = -\tfrac{1}{\sqrt{2}}$ and $\sin\phi = b/A = \tfrac{1}{\sqrt{2}}$ , so $\phi$ is in the second quadrant: $\phi = \tfrac{3\pi}{4}$ . The plain inverse tangent gives $\tan^{-1}(-1) = -\tfrac{\pi}{4}$ , which is off by $\pi$ .

-\cos t + \sin t = \sqrt{2}\cos\!\left(t - \tfrac{3\pi}{4}\right)

Check: $\sqrt{2}\left(\cos t \cos\tfrac{3\pi}{4} + \sin t \sin\tfrac{3\pi}{4}\right) = \sqrt{2}\left(-\tfrac{1}{\sqrt{2}}\cos t + \tfrac{1}{\sqrt{2}}\sin t\right) = -\cos t + \sin t$ ✓

Problem 4 · Convert a Multiplied-Out Answer

Given: with $k = 1$ and $\omega = \sqrt{3}$ , a response left in the form $a\cos + b\sin$ reads $y = \tfrac{1}{4}\cos\sqrt{3}\,t + \tfrac{\sqrt{3}}{4}\sin\sqrt{3}\,t$ — find its amplitude and phase lag.

What is the amplitude?

What is the phase lag φ?

✅ Correct!

y = \tfrac{1}{2}\cos\!\left(\sqrt{3}\,t - \tfrac{\pi}{3}\right)

, matching the polar formulas

1/\sqrt{1+(\omega/k)^2} = 1/2

and

\tan^{-1}(\omega/k) = \pi/3

❌ Not quite.

2 = \sqrt{1 + (\omega/k)^2}

is the hypotenuse of the triangle with legs

1

and

\omega/k

; the front factor

\tfrac{1}{1+(\omega/k)^2} = \tfrac{1}{4}

still multiplies it.

❌ Check the amplitude. The amplitude is

\sqrt{a^2 + b^2}

with

a = \tfrac14

and

b = \tfrac{\sqrt3}{4}

, not a sum of the coefficients or the sum of their squares.

❌ Close, but the ratio is inverted. The sine coefficient is opposite

\phi

\tan\phi = b/a = \sqrt{3}

, not

1/\sqrt{3}

❌ Not quite.

\phi = \pi/4

needs equal coefficients (

\tan\phi = 1

); here

b/a = \sqrt{3}

❌ Check the lag.

\tan\phi = b/a

uses both coefficients; the common factor

\tfrac14

cancels.

Show solution

Read off $a = \tfrac{1}{4}$ , $b = \tfrac{\sqrt{3}}{4}$ , $\theta = \sqrt{3}\,t$ . (These are $\tfrac{k^2}{k^2+\omega^2}$ and $\tfrac{k\omega}{k^2+\omega^2}$ with $k = 1$ , $\omega = \sqrt{3}$ .)

A = \sqrt{\tfrac{1}{16} + \tfrac{3}{16}} = \sqrt{\tfrac{4}{16}} = \tfrac{1}{2}, \qquad \tan\phi = \frac{\sqrt{3}/4}{1/4} = \sqrt{3} \;\Rightarrow\; \phi = \frac{\pi}{3}

So $y = \tfrac{1}{2}\cos\!\left(\sqrt{3}\,t - \tfrac{\pi}{3}\right)$ .

Same answer by the video's route: $\omega/k = \sqrt{3}$ , so the amplitude is $\dfrac{1}{\sqrt{1 + 3}} = \dfrac{1}{2}$ and the lag is $\tan^{-1}\sqrt{3} = \dfrac{\pi}{3}$ ✓

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