Back to Tutor Intake Assessment: Prove and use Pythagorean identity

HSF.TF.C.8 Tutor Intake -- Pythagorean Identity and Trig Values

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Grade 9·10 problems·~13 min·Common Core Math - HS Functions·group·hsf-tf-c-8

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Concepts

The Pythagorean identity $\sin^2(\theta) + \cos^2(\theta) = 1$ can be proved by
starting with which equation?

The Pythagorean theorem $a^2 + b^2 = c^2$ applied to a right triangle with hypotenuse 1

The unit circle equation $x^2 + y^2 = 1$ , substituting $x = \cos(\theta)$ and $y = \sin(\theta)$

The definition $\tan(\theta) = \sin(\theta)/\cos(\theta)$

The numerical observation that the identity holds at $\theta = \pi/6$ , $\pi/4$ , and $\pi/3$

The Pythagorean identity $\sin^2(\theta) + \cos^2(\theta) = 1$ holds for
̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ [scope] ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ values of $\theta$ .

scope:

A student applies the Pythagorean identity and finds $\sin^2(\theta) = \dfrac{9}{25}$ .
The student writes $\sin(\theta) = \dfrac{3}{5}$ as the final answer. What is wrong
with this reasoning?

The student should have used $\cos^2(\theta) = 1 - \sin^2(\theta)$ instead

Taking a square root gives $\pm\dfrac{3}{5}$ ; without quadrant information,
both values are possible and the student must use the quadrant to choose the sign

The correct answer is $\sin(\theta) = \dfrac{\sqrt{9}}{\sqrt{25}} = \dfrac{3}{5}$ , so the student is right

The student confused $\sin^2(\theta)$ with $\sin(\theta^2)$

Procedures

Given $\cos(\theta) = \dfrac{3}{5}$ and $\theta$ is in Quadrant I, use the
Pythagorean identity to find $\sin(\theta)$ . Enter your answer as a fraction
(e.g., 4/5).

Given $\sin(\theta) = -\dfrac{5}{13}$ and $\theta$ is in Quadrant III, use the
Pythagorean identity to find $\cos(\theta)$ . Enter your answer as a fraction
(e.g., -12/13).

If $\cos(\theta) = -\dfrac{2}{3}$ and $\theta$ is in Quadrant II, what is
$\sin(\theta)$ ?

$\dfrac{\sqrt{5}}{3}$

$-\dfrac{\sqrt{5}}{3}$

$\dfrac{1}{3}$

$-\dfrac{1}{3}$

Given $\sin(\theta) = \dfrac{5}{13}$ and $\theta$ is in Quadrant II, find
$\tan(\theta)$ . Enter your answer as a fraction (e.g., -5/12).

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