Aimathic – Pythagorean identity – Free Math Worksheets

Pythagorean identity

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Difficulty: 1 – 5

51506411

Let \(\alpha = 70^\circ\). a) Use unit-circle reasoning to estimate \(\sin(70^\circ)\) and \(\cos(70^\circ)\). b) Use a calculator to find more accurate values, rounded to the nearest thousandth. c) Use your rounded values from part b) to evaluate \((\sin(70^\circ))^2 + (\cos(70^\circ))^2\). What value should this expression equal theoretically for any angle?

Hints

- Which coordinate of a point on the unit circle represents sine, and which represents cosine? - How does the Pythagorean theorem apply to a right triangle with hypotenuse \(1\)? - Use degree mode on your calculator.

Solution

1. On the unit circle, the point at \(70^\circ\) lies in Quadrant I. Because \(70^\circ\) is closer to \(90^\circ\) than to \(0^\circ\), the y-coordinate is close to \(1\) and the x-coordinate is much smaller. A reasonable estimate is \(\sin(70^\circ) \approx 0.9\) and \(\cos(70^\circ) \approx 0.3\). 2. A calculator gives \(\sin(70^\circ) \approx 0.940\) and \(\cos(70^\circ) \approx 0.342\). 3. Using the rounded values, \(0.940^2 + 0.342^2 = 0.883600 + 0.116964 = 1.000564 \approx 1.001\). 4. Using exact values, the Pythagorean identity gives \(\sin^2(\alpha) + \cos^2(\alpha) = 1\) for every angle. The small difference in part c) is caused by rounding.

Answer

a) A reasonable estimate is \(\sin(70^\circ) \approx 0.9\); \(\cos(70^\circ) \approx 0.3\) b) \(\sin(70^\circ) \approx 0.940\); \(\cos(70^\circ) \approx 0.342\) c) Using the rounded values, the sum is approximately \(1.001\). The theoretical value is exactly \(1\).

51512611

A student claims that \(\sin(\alpha) + \cos(\alpha) = 1\) for every acute angle \(\alpha\). a) Test the claim for \(\alpha = 45^\circ\). b) Use the Pythagorean theorem on the unit circle to state the correct relationship between \(\sin(\alpha)\) and \(\cos(\alpha)\). Explain why the student’s claim is not true in general.

Hints

- Evaluate sine and cosine at \(45^\circ\). - Picture the right triangle formed by a point on the unit circle and the coordinate axes. - What are the leg lengths of that triangle in terms of sine and cosine? - Apply the Pythagorean theorem.

Solution

1. At \(45^\circ\), \(\sin(45^\circ)=\frac{\sqrt{2}}{2}\) and \(\cos(45^\circ)=\frac{\sqrt{2}}{2}\). Their sum is \(\sqrt{2} \approx 1.414\), not \(1\), so the claim is false. 2. A point on the unit circle has coordinates \((\cos(\alpha), \sin(\alpha))\). 3. By the Pythagorean theorem, \((\cos(\alpha))^2 + (\sin(\alpha))^2 = 1^2\). 4. Therefore, the correct identity is \(\sin^2(\alpha)+\cos^2(\alpha)=1\), not \(\sin(\alpha)+\cos(\alpha)=1\).

Answer

a) \(\sin(45^\circ)+\cos(45^\circ)=\sqrt{2} \approx 1.414 \ne 1\), so the claim is false. b) The correct relationship is \(\sin^2(\alpha)+\cos^2(\alpha)=1\).

51516211

A point \(P\) on the unit circle corresponds to an angle \(\alpha\) and has coordinates \(P(-0.8, 0.6)\). a) In which quadrant is \(P\)? b) State \(\cos(\alpha)\) and \(\sin(\alpha)\). c) Evaluate \(\sin^2(\alpha)+\cos^2(\alpha)\). Which identity does this confirm? d) Give the coordinates of the point \(Q\) corresponding to \(180^\circ-\alpha\).

Hints

- Use the signs of \(x\) and \(y\) to identify the quadrant. - On the unit circle, how are cosine and sine related to coordinates? - Apply the Pythagorean theorem to the coordinates. - How does reflection across the y-axis change an ordered pair?

Solution

1. Since the x-coordinate is negative and the y-coordinate is positive, \(P\) lies in Quadrant II. 2. On the unit circle, \(\cos(\alpha)=x=-0.8\) and \(\sin(\alpha)=y=0.6\). 3. Then \(\sin^2(\alpha)+\cos^2(\alpha)=0.6^2+(-0.8)^2=0.36+0.64=1\). This confirms the Pythagorean identity. 4. The angle \(180^\circ-\alpha\) reflects the point across the y-axis, so \(Q=(0.8, 0.6)\).

Answer

a) Quadrant II b) \(\cos(\alpha)=-0.8\); \(\sin(\alpha)=0.6\) c) \(1\); the Pythagorean identity d) \(Q(0.8, 0.6)\)

52864811

The identity \(\sin^2(\alpha)+\cos^2(\alpha)=1\) relates sine and cosine. 1. Solve this identity for \(\cos(\alpha)\), assuming cosine is positive. 2. Find \(\cos(\alpha)\) when \(\sin(\alpha)=0.96\). 3. For \(0^\circ \le \alpha \le 90^\circ\), find \(\sin(\alpha)\) when \(\cos(\alpha)=0.5\). Give an exact radical answer. 4. Given \(\sin(20^\circ)\approx0.342\), approximate \(\cos(20^\circ)\) to the nearest thousandth.

Hints

- Isolate the squared trigonometric expression before taking a square root. - Remember that \(\sin^2(\alpha)\) means \((\sin(\alpha))^2\). - Simplify radicals by writing the radicand as a fraction when useful. - Keep enough decimal places before the final rounding step.

Solution

1. Rearrange: \(\cos^2(\alpha)=1-\sin^2(\alpha)\). Since cosine is positive, \(\cos(\alpha)=\sqrt{1-\sin^2(\alpha)}\). 2. \(\cos(\alpha)=\sqrt{1-0.96^2}=\sqrt{0.0784}=0.28\). 3. \(\sin(\alpha)=\sqrt{1-0.5^2}=\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}\). 4. \(\cos(20^\circ)\approx\sqrt{1-0.342^2}=\sqrt{0.883036}\approx0.940\).

Answer

1. \(\cos(\alpha)=\sqrt{1-\sin^2(\alpha)}\) 2. \(\cos(\alpha)=0.28\) 3. \(\sin(\alpha)=\frac{\sqrt{3}}{2}\) 4. \(\cos(20^\circ)\approx0.940\)

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