Deriving the Fundamental Identity

Prove that $\sin^2\theta + \cos^2\theta = 1$ using the unit circle.

Start with a point on the unit circle: Any point on the unit circle has coordinates $(\cos\theta, \sin\theta)$ = $(x, y) = (\cos\theta, \sin\theta)$

Apply the unit circle equation: The unit circle is defined by $x^2 + y^2 = 1$ = $x^2 + y^2 = 1$

Substitute the coordinates: Replace $x$ with $\cos\theta$ and $y$ with $\sin\theta$ = $(\cos\theta)^2 + (\sin\theta)^2 = 1$

Write in standard form: Use exponent notation: $\sin^2\theta + \cos^2\theta$ = $\sin^2\theta + \cos^2\theta = 1$ ✓

Answer: The identity $\sin^2\theta + \cos^2\theta = 1$ is proven by the definition of the unit circle!

Finding a Trig Value Using an Identity

If $\sin\theta = \frac{3}{5}$ and $\theta$ is in Quadrant I, find $\cos\theta$.

Write the fundamental identity: $\sin^2\theta + \cos^2\theta = 1$ = Identity ready to use

Substitute the known value: $\left(\frac{3}{5}\right)^2 + \cos^2\theta = 1$ = $\frac{9}{25} + \cos^2\theta = 1$

Solve for $\cos^2\theta$: $\cos^2\theta = 1 - \frac{9}{25} = \frac{25}{25} - \frac{9}{25}$ = $\cos^2\theta = \frac{16}{25}$

Take the square root: $\cos\theta = \pm\frac{4}{5}$ = $\cos\theta = \pm\frac{4}{5}$

Choose the correct sign: In Quadrant I, cosine is positive = $\cos\theta = \frac{4}{5}$

Answer: $\cos\theta = \frac{4}{5}$

Deriving the Second Pythagorean Identity

Derive $1 + \tan^2\theta = \sec^2\theta$ from the fundamental identity.

Start with the fundamental identity: $\sin^2\theta + \cos^2\theta = 1$ = Starting point

Divide both sides by $\cos^2\theta$: $\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\cos^2\theta} = \frac{1}{\cos^2\theta}$ = Each term divided by $\cos^2\theta$

Simplify using trig definitions: $\tan\theta = \frac{\sin\theta}{\cos\theta}$, so $\frac{\sin^2\theta}{\cos^2\theta} = \tan^2\theta$ = $\tan^2\theta + 1 = \frac{1}{\cos^2\theta}$

Apply the secant definition: $\sec\theta = \frac{1}{\cos\theta}$, so $\frac{1}{\cos^2\theta} = \sec^2\theta$ = $\tan^2\theta + 1 = \sec^2\theta$

Rearrange: Write in standard form = $1 + \tan^2\theta = \sec^2\theta$ ✓

Answer: By dividing the fundamental identity by $\cos^2\theta$, we obtain $1 + \tan^2\theta = \sec^2\theta$.

Simplifying a Trigonometric Expression

Simplify: $\sec^2\theta - \tan^2\theta$

Recognize the identity pattern: This matches the second Pythagorean identity: $1 + \tan^2\theta = \sec^2\theta$ = Identity identified

Rearrange the identity: $\sec^2\theta - \tan^2\theta = 1$ = Subtract $\tan^2\theta$ from both sides

Apply to our expression: Our expression is exactly $\sec^2\theta - \tan^2\theta$ = $\sec^2\theta - \tan^2\theta = 1$

Answer: $\sec^2\theta - \tan^2\theta = 1$

Mistake: Forgetting to consider the quadrant when taking square roots

Why: The equation $\cos^2\theta = \frac{16}{25}$ gives $\cos\theta = \pm\frac{4}{5}$. Students often forget to determine which sign applies.

Correct: Always check which quadrant the angle is in to determine the sign of the trig function.

Mistake: Writing $\sin^2\theta$ as $\sin\theta^2$

Why: $\sin^2\theta$ means $(\sin\theta)^2$, not $\sin(\theta^2)$. The notation is shorthand for squaring the result of the sine function.

Correct: $\sin^2\theta = (\sin\theta)^2$, meaning "find sine of theta, then square the result."

Mistake: Applying an identity where a function is undefined

Why: The identity $1 + \tan^2\theta = \sec^2\theta$ is undefined when $\cos\theta = 0$ (at $\theta = 90°, 270°$, etc.).

Correct: Check that the functions in the identity are defined for the given angle.

Signal Processing

Engineers use Pythagorean identities when analyzing radio and audio signals that are modeled by sine and cosine waves.

When combining two signals $A\sin\theta$ and $A\cos\theta$, the total power is $A^2(\sin^2\theta + \cos^2\theta) = A^2$, which is constant.

Physics: Simple Harmonic Motion

The position and velocity of an oscillating object are related through sine and cosine. Their energy relationship uses the Pythagorean identity.

If position is $x = A\cos(\omega t)$ and velocity is $v = -A\omega\sin(\omega t)$, then $\frac{x^2}{A^2} + \frac{v^2}{A^2\omega^2} = \cos^2(\omega t) + \sin^2(\omega t) = 1$.

Q: Why are they called "Pythagorean" identities?

A: They derive from the Pythagorean theorem $a^2 + b^2 = c^2$. On the unit circle, the legs are $\cos\theta$ and $\sin\theta$, and the hypotenuse is 1.

Q: Do these identities work for any angle?

A: The fundamental identity $\sin^2\theta + \cos^2\theta = 1$ works for all angles. The other two identities work except where tan, cot, sec, or csc are undefined (division by zero).

Q: How do I remember all three identities?

A: Memorize only $\sin^2\theta + \cos^2\theta = 1$. Derive the others by dividing by $\cos^2\theta$ or $\sin^2\theta$ as needed.