Pythagorean Identity: Proof and Applications

On the unit circle, a point at angle $\theta$ has coordinates $(x, y)$.
Which of the following correctly identifies $\sin\theta$ and $\cos\theta$?

In which quadrant are both $\sin\theta$ and $\cos\theta$ negative?

The unit circle is the set of all points $(x, y)$ satisfying a specific
equation. A point on the unit circle satisfies $x^{2} + y^{2} = $   ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲   .
Enter your answer.

Given $\cos\theta = \dfrac{4}{5}$ and $\theta$ is in Quadrant I, use
the Pythagorean identity to find $\sin\theta$. Express your answer as
a fraction in simplest form.

Given $\sin\theta = -\dfrac{5}{13}$ and $\theta$ is in Quadrant III,
use the Pythagorean identity to find $\cos\theta$. Express your answer
as a fraction in simplest form.

Given $\cos\theta = -\dfrac{2}{3}$ and $\theta$ is in Quadrant II,
use the Pythagorean identity to find $\sin\theta$. Express your answer
in simplest radical form (e.g., write sqrt(5)/3 for $\dfrac{\sqrt{5}}{3}$).

Given $\cos\theta = \dfrac{3}{5}$ and $\theta$ is in Quadrant I, find
$\tan\theta$. Express your answer as a fraction in simplest form.
(Hint: first find $\sin\theta$, then use $\tan\theta = \sin\theta / \cos\theta$.)

Given $\sin\theta = \dfrac{5}{13}$ and $\theta$ is in Quadrant II, find
$\tan\theta$. Express your answer as a fraction in simplest form.

A student says the Pythagorean identity $\sin^{2}\theta + \cos^{2}\theta = 1$
is only valid for special angles like $\dfrac{\pi}{6}$, $\dfrac{\pi}{4}$,
and $\dfrac{\pi}{3}$ because those are the only angles with exact values.
Which response best refutes this claim?

The expression $\sin^{2}\theta$ means $(\sin\theta)^{2}$, NOT
$\sin(\theta^{2})$.

For $\theta = \dfrac{\pi}{6}$:
$\sin\dfrac{\pi}{6} = \dfrac{1}{2}$, so
$\sin^{2}\dfrac{\pi}{6} = \left(\dfrac{1}{2}\right)^{2} =$   ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲   .

Then $\cos^{2}\dfrac{\pi}{6} = 1 - \sin^{2}\dfrac{\pi}{6} = 1 - \dfrac{1}{4} =$   ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲   .

If $\sin\theta = \dfrac{1}{4}$ and $\theta$ is in Quadrant II, which
of the following is the correct value of $\cos\theta$?

The Pythagorean identity can be verified for any angle. Use it to
verify the identity at $\theta = \dfrac{3\pi}{4}$.

Recall: $\sin\dfrac{3\pi}{4} = \dfrac{\sqrt{2}}{2}$ and
$\cos\dfrac{3\pi}{4} = -\dfrac{\sqrt{2}}{2}$.

Compute $\sin^{2}\dfrac{3\pi}{4} + \cos^{2}\dfrac{3\pi}{4}$.
Enter your answer.

Given $\tan\theta = -\dfrac{3}{4}$ and $\theta$ is in Quadrant II,
which of the following is the value of $\sec\theta$?
(Use the derived identity $\tan^{2}\theta + 1 = \sec^{2}\theta$.)

An angle $\theta$ is in Quadrant IV. A physics student knows that the
horizontal component of a unit vector at angle $\theta$ is
$\cos\theta = \dfrac{7}{25}$.

What is the vertical component $\sin\theta$?
(The vertical component is negative in QIV.)
Express your answer as a fraction in simplest form.

An angle $\alpha$ is in Quadrant III. Which combination of signs is
possible for $(\sin\alpha,\, \cos\alpha,\, \tan\alpha)$?

A student solved the following problem and made an error.
Identify the mistake and select the correct final answer.

A student made an algebra error when working with the Pythagorean
identity. Identify the error and choose the correction.

Given $\tan\theta = 2$ and $\theta$ in Quadrant III, find all six
trigonometric values and then state $\sec\theta$.

Use the derived identity $\sec^{2}\theta = 1 + \tan^{2}\theta$.
In Quadrant III, $\cos\theta < 0$ and therefore $\sec\theta < 0$.

Express $\sec\theta$ in simplest radical form
(e.g., write -sqrt(5) for $-\sqrt{5}$).

Write a complete proof of $\sin^{2}\theta + \cos^{2}\theta = 1$
starting from the unit circle equation $x^{2} + y^{2} = 1$.
Your proof should:

1. State what the unit circle is.
2. Identify the coordinates of a general point on the unit circle.
3. Substitute those coordinates into the circle equation.
4. Explain why the result holds for every value of $\theta$, not just
   special angles.

Write 3–5 complete sentences.