Free math worksheets

- 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 \(\sin^2(\alpha)+\cos^2(\alpha)=1\).

- Substitute the known trigonometric value into the Pythagorean identity before isolating the missing squared value. - Keep both square-root signs until you use the quadrant. - Use the sign of the x-coordinate in Quadrant I to choose the final value.

1. Substitute \(\sin\theta=\frac{2}{5}\) into \(\sin^2\theta+\cos^2\theta=1\): \(\frac{4}{25}+\cos^2\theta=1\). 2. Then \(\cos^2\theta=\frac{21}{25}\), so \(\cos\theta=\pm\frac{\sqrt{21}}{5}\). 3. Quadrant I has positive cosine, so \(\cos\theta=\frac{\sqrt{21}}{5}\).

a) \(\frac{4}{25}+\cos^2\theta=1\) b) \(\cos\theta=\frac{\sqrt{21}}{5}\)

- Use the Pythagorean identity to determine the magnitude of the missing cosine value. - Do not choose a square-root sign until the magnitude calculation is complete. - Use the sign of the x-coordinate in Quadrant II for the final sign.

1. Substitute \(\sin\theta=\frac{4}{7}\) into \(\sin^2\theta+\cos^2\theta=1\): \(\frac{16}{49}+\cos^2\theta=1\). 2. Then \(\cos^2\theta=\frac{33}{49}\), so \(\cos\theta=\pm\frac{\sqrt{33}}{7}\). 3. Cosine is negative in Quadrant II, so \(\cos\theta=-\frac{\sqrt{33}}{7}\).

a) \(\frac{16}{49}+\cos^2\theta=1\) b) \(\cos\theta=-\frac{\sqrt{33}}{7}\)

- Substitute the known cosine value into the Pythagorean identity before taking a square root. - Separate finding the magnitude from deciding the sign. - Use the sign of the y-coordinate in Quadrant IV to choose the final value.

1. Substitute \(\cos\theta=\frac{5}{8}\) into \(\sin^2\theta+\cos^2\theta=1\): \(\sin^2\theta+\frac{25}{64}=1\). 2. Then \(\sin^2\theta=\frac{39}{64}\), so \(\sin\theta=\pm\frac{\sqrt{39}}{8}\). 3. Sine is negative in Quadrant IV, so \(\sin\theta=-\frac{\sqrt{39}}{8}\).

a) \(\sin^2\theta+\frac{25}{64}=1\) b) \(\sin\theta=-\frac{\sqrt{39}}{8}\)

- 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.

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

- 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, \(\sin^2\theta+\cos^2\theta=1\).

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

- Begin with the equation that defines every point on the unit circle. - Replace the generic coordinates with the trigonometric coordinates of \(P\). - Consider what squaring does to positive and negative coordinate values.

1. Every point \((x,y)\) on the unit circle satisfies \(x^2+y^2=1\). 2. For \(P=(\cos\theta,\sin\theta)\), substitute \(x=\cos\theta\) and \(y=\sin\theta\). 3. This gives \(\cos^2\theta+\sin^2\theta=1\), or equivalently \(\sin^2\theta+\cos^2\theta=1\). 4. The unit-circle equation holds in every quadrant, and squaring the coordinates handles either sign, so no quadrant restriction is needed.

\(\sin^2\theta+\cos^2\theta=1\). The derivation is valid in every quadrant because every unit-circle point satisfies \(x^2+y^2=1\), regardless of the signs of \(x\) and \(y\).

- In part a), use the supplied rounded values exactly as written before rounding the final sum. - In part b), start from the equation satisfied by every point on the unit circle. - Compare the coordinates of \(P\) with the variables in that circle equation.

1. Using the rounded values, \(0.940^2+0.342^2=0.883600+0.116964=1.000564\approx1.001\). 2. Any point \(P=(\cos\theta,\sin\theta)\) on the unit circle satisfies \(x^2+y^2=1\). 3. Substituting \(x=\cos\theta\) and \(y=\sin\theta\) gives \(\cos^2\theta+\sin^2\theta=1\). 4. The value in part a) differs slightly from \(1\) because \(0.940\) and \(0.342\) are rounded approximations rather than exact trigonometric values.

a) \(1.001\) b) Since \(P=(\cos\theta,\sin\theta)\) lies on \(x^2+y^2=1\), \(\sin^2\theta+\cos^2\theta=1\) exactly. The small difference in part a) is caused by rounding the sine and cosine values.

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

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

- Recover the missing cosine value from the Pythagorean identity before forming the tangent ratio. - Use the quadrant to decide the sign of cosine after finding its magnitude. - Keep the tangent value exact when simplifying the ratio of sine to cosine.

1. Substitute into the identity: \(\frac{4}{49}+\cos^2\theta=1\). 2. Then \(\cos^2\theta=\frac{45}{49}\), so \(\cos\theta=\pm\frac{3\sqrt{5}}{7}\). 3. Cosine is positive in Quadrant IV, so \(\cos\theta=\frac{3\sqrt{5}}{7}\). 4. Therefore, \(\tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{-2/7}{3\sqrt{5}/7}=-\frac{2}{3\sqrt{5}}=-\frac{2\sqrt{5}}{15}\).

a) Identity equation: \(\frac{4}{49}+\cos^2\theta=1\); \(\cos\theta=\frac{3\sqrt{5}}{7}\) b) \(\tan\theta=-\frac{2\sqrt{5}}{15}\)

- 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

- A tangent value fixes the ratio of sine to cosine, not either value separately. - Use the quadrant before assigning signs to the common-scale representation. - Substitute the scaled sine and cosine expressions into the Pythagorean identity to determine their common scale.

1. In Quadrant IV, sine is negative and cosine is positive. The tangent ratio gives \(\sin\theta=-2k\) and \(\cos\theta=3k\), where \(k>0\). 2. Substitute into \(\sin^2\theta+\cos^2\theta=1\): \(4k^2+9k^2=1\). 3. Thus \(13k^2=1\), so \(k=\frac{1}{\sqrt{13}}\). 4. Therefore, \(\sin\theta=-\frac{2}{\sqrt{13}}=-\frac{2\sqrt{13}}{13}\) and \(\cos\theta=\frac{3}{\sqrt{13}}=\frac{3\sqrt{13}}{13}\).

a) \(\sin\theta=-2k\), \(\cos\theta=3k\), \(k>0\); identity equation: \(4k^2+9k^2=1\) b) \(\sin\theta=-\frac{2\sqrt{13}}{13}\), \(\cos\theta=\frac{3\sqrt{13}}{13}\)

- Positive tangent means sine and cosine have the same sign, not necessarily that both are positive. - Represent the two values with one common scale factor before using the identity. - Keep both possible signs of the scale factor and connect them with unit-circle quadrants.

1. A positive tangent means sine and cosine have the same sign, so the angle can lie in Quadrant I or Quadrant III; it does not force both values to be positive. 2. Since \(\tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{2}{5}\), write \(\sin\theta=2k\) and \(\cos\theta=5k\) for a nonzero real scale factor \(k\). 3. The identity gives \(4k^2+25k^2=1\), so \(29k^2=1\) and \(k=\pm\frac{1}{\sqrt{29}}\). 4. If \(k=\frac{1}{\sqrt{29}}\), then \((\cos\theta,\sin\theta)=\left(\frac{5\sqrt{29}}{29},\frac{2\sqrt{29}}{29}\right)\), which is in Quadrant I. 5. If \(k=-\frac{1}{\sqrt{29}}\), then \((\cos\theta,\sin\theta)=\left(-\frac{5\sqrt{29}}{29},-\frac{2\sqrt{29}}{29}\right)\), which is in Quadrant III.

a) A positive tangent allows Quadrant I or Quadrant III, so both sine and cosine need not be positive. b) \(\sin\theta=2k\), \(\cos\theta=5k\); identity equation: \(4k^2+25k^2=1\), so \(k=\pm\frac{1}{\sqrt{29}}\) c) \(\left(\frac{5\sqrt{29}}{29},\frac{2\sqrt{29}}{29}\right)\) in Quadrant I or \(\left(-\frac{5\sqrt{29}}{29},-\frac{2\sqrt{29}}{29}\right)\) in Quadrant III

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

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