Free math worksheets

- Substitute the known sine value into the Pythagorean identity. - Solving for a squared trigonometric value produces two possible signs. - Use the quadrant only after finding the possible magnitudes.

1. Use \(\sin^2\theta+\cos^2\theta=1\). 2. Substitute \(\sin\theta=\frac{3}{5}\): \(\frac{9}{25}+\cos^2\theta=1\). 3. Then \(\cos^2\theta=\frac{16}{25}\), so \(\cos\theta=\pm\frac{4}{5}\). 4. Quadrant I has positive cosine, so \(\cos\theta=\frac{4}{5}\).

\(\cos\theta=\frac{4}{5}\)

- Use the identity to find the magnitude of the missing cosine value. - Do not choose a square-root sign until you use the quadrant. - Recall the sign of the x-coordinate in Quadrant II.

1. Substitute into \(\sin^2\theta+\cos^2\theta=1\): \(\frac{25}{169}+\cos^2\theta=1\). 2. Then \(\cos^2\theta=\frac{144}{169}\), so \(\cos\theta=\pm\frac{12}{13}\). 3. Cosine is negative in Quadrant II, so \(\cos\theta=-\frac{12}{13}\).

\(\cos\theta=-\frac{12}{13}\)

- Substitute the known cosine value into the identity. - Separate the magnitude calculation from the sign decision. - Use the sign of the y-coordinate in Quadrant IV.

1. Substitute into \(\sin^2\theta+\cos^2\theta=1\): \(\sin^2\theta+\frac{144}{169}=1\). 2. Then \(\sin^2\theta=\frac{25}{169}\), so \(\sin\theta=\pm\frac{5}{13}\). 3. Sine is negative in Quadrant IV, so \(\sin\theta=-\frac{5}{13}\).

\(\sin\theta=-\frac{5}{13}\)

- A unit-circle point must satisfy the equation \(x^2+y^2=1\). - Substitute the two coordinates and simplify the sum. - Then connect the coordinate order with cosine and sine.

1. Compute \(\left(\frac{5}{13}\right)^2+\left(\frac{12}{13}\right)^2=\frac{25}{169}+\frac{144}{169}=1\). 2. Since the coordinate squares sum to \(1\), the point lies on the unit circle. 3. On the unit circle, the x-coordinate is cosine and the y-coordinate is sine.

a) Yes, because \(\frac{25}{169}+\frac{144}{169}=1\). b) \(\cos\theta=\frac{5}{13}\) and \(\sin\theta=\frac{12}{13}\)

- Identify the two legs and the hypotenuse in the shown right triangle. - Write the Pythagorean theorem using the labels carried by those three segments. - Simplify the square of the unit-circle radius.

1. Triangle \(OHP\) is right, with legs \(OH=\cos\theta\) and \(HP=\sin\theta\), and hypotenuse \(OP=1\). 2. By the Pythagorean theorem, \((\cos\theta)^2+(\sin\theta)^2=1^2\). 3. Therefore, \(\sin^2\theta+\cos^2\theta=1\).

\(\sin^2\theta+\cos^2\theta=1\)

- 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.

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.

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 \(1\).

- 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.

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 \(\sin(\alpha)+\cos(\alpha)=1\).

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

- 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?

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 \(Q=(0.8, 0.6)\).

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

- 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.

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. \(\cos(20^\circ)\approx\sqrt{1-0.342^2}=\sqrt{0.883036}\approx0.940\).

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

- Square the given cosine value carefully before substituting it. - The identity determines the magnitude of the missing sine value first. - Use the quadrant to choose between the two square-root signs.

1. Substitute into \(\sin^2\theta+\cos^2\theta=1\): \(\sin^2\theta+\frac{7}{16}=1\). 2. Then \(\sin^2\theta=\frac{9}{16}\), so \(\sin\theta=\pm\frac{3}{4}\). 3. Sine is negative in Quadrant III, so \(\sin\theta=-\frac{3}{4}\).

\(\sin\theta=-\frac{3}{4}\)

- Use the identity to recover the missing coordinate first. - Choose its sign from the quadrant before using it in another ratio. - After finding sine and cosine, connect tangent to those two values.

1. Substitute into the identity: \(\frac{49}{625}+\cos^2\theta=1\). 2. Then \(\cos^2\theta=\frac{576}{625}\), so \(\cos\theta=\pm\frac{24}{25}\). 3. Cosine is positive in Quadrant IV, so \(\cos\theta=\frac{24}{25}\). 4. Therefore, \(\tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{-7/25}{24/25}=-\frac{7}{24}\).

a) \(\cos\theta=\frac{24}{25}\) b) \(\tan\theta=-\frac{7}{24}\)

- Notice that the identity determines \(\cos^2\theta\), not the sign of \(\cos\theta\). - Ask which quadrants are compatible with a positive sine value. - Check the sign of cosine in each compatible quadrant.

1. The identity gives \(\frac{64}{289}+\cos^2\theta=1\), so \(\cos^2\theta=\frac{225}{289}\). 2. Taking square roots gives \(\cos\theta=\pm\frac{15}{17}\). 3. The positive sine value places \(\theta\) in Quadrant I or II, where cosine can be positive or negative. 4. Without a quadrant or other location information, both cosine values are possible.

The student ignored the negative square-root possibility. The possible values are \(\cos\theta=\frac{15}{17}\) and \(\cos\theta=-\frac{15}{17}\).

- Use the identity to recover the other squared trigonometric value. - Apply the stated sign condition to sine, but do not invent a sign condition for cosine. - Translate the possible sign pairs into quadrants.

1. From \(\sin^2\theta+\cos^2\theta=1\), \(\sin^2\theta=1-\frac{7}{16}=\frac{9}{16}\). 2. Since \(\sin\theta>0\), \(\sin\theta=\frac{3}{4}\). 3. From \(\cos^2\theta=\frac{7}{16}\), \(\cos\theta=\pm\frac{\sqrt{7}}{4}\). 4. Positive sine places \(\theta\) in Quadrant I or II, matching the two possible cosine signs.

a) \(\sin\theta=\frac{3}{4}\) b) \(\cos\theta=\pm\frac{\sqrt{7}}{4}\) c) Quadrant I or Quadrant II

- The target is a product, while the given information is a sum. - Consider an algebraic operation on the sum that creates a product term. - Look for where the Pythagorean identity can replace two squared terms with a single value.

1. Square the given equation: \((\sin\theta+\cos\theta)^2=\frac{1}{4}\). 2. Expand: \(\sin^2\theta+2\sin\theta\cos\theta+\cos^2\theta=\frac{1}{4}\). 3. Use \(\sin^2\theta+\cos^2\theta=1\): \(1+2\sin\theta\cos\theta=\frac{1}{4}\). 4. Therefore, \(2\sin\theta\cos\theta=-\frac{3}{4}\), so \(\sin\theta\cos\theta=-\frac{3}{8}\).

\(\sin\theta\cos\theta=-\frac{3}{8}\)