Free Math Worksheets

- Rewrite the expressions using common trigonometric functions. - In part a, use a rearrangement of the Pythagorean identity. - In part b, rewrite tangent as a quotient. - Look for \(\sin^2\alpha+\cos^2\alpha\).

1. For a), use \(1-\sin^2\alpha=\cos^2\alpha\). Then \(\frac{\cos^2\alpha}{1-\sin^2\alpha}=\frac{\cos^2\alpha}{\cos^2\alpha}=1\). 2. For b), replace \(\tan\alpha\) with \(\frac{\sin\alpha}{\cos\alpha}\): \(\frac{\sin\alpha}{\cos\alpha}\sin\alpha\cos\alpha+\cos^2\alpha\). 3. Simplify to \(\sin^2\alpha+\cos^2\alpha=1\).

a) \(1\) b) \(1\)

- In part a, rewrite tangent as a quotient. - In part b, expand the square before combining terms. - Identify terms that cancel. - Use the Pythagorean identity.

1. For a), write \(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}\). Then \(\frac{\sin\alpha}{\cos\alpha}\cdot\frac{1}{\sin\alpha}=\frac{1}{\cos\alpha}\). 2. For b), expand the square: \(\sin^2\alpha+2\sin\alpha\cos\alpha+\cos^2\alpha\). 3. Subtract \(2\sin\alpha\cos\alpha\), leaving \(\sin^2\alpha+\cos^2\alpha=1\).

a) \(\frac{1}{\cos\alpha}\) b) \(1\)

- In part a, rewrite the denominator as a quotient. - Cancel only complete factors. - In part b, recognize a difference-of-squares pattern. - Use a rearrangement of the Pythagorean identity.

1. For a), replace \(\tan\alpha\) with \(\frac{\sin\alpha}{\cos\alpha}\): \(\frac{\sin\alpha\cos\alpha}{\sin\alpha/\cos\alpha}=\sin\alpha\cos\alpha\cdot\frac{\cos\alpha}{\sin\alpha}=\cos^2\alpha\). 2. For b), use the difference of squares: \((1-\cos\alpha)(1+\cos\alpha)=1-\cos^2\alpha\). 3. By the Pythagorean identity, \(1-\cos^2\alpha=\sin^2\alpha\).

a) \(\cos^2\alpha\) b) \(\sin^2\alpha\)

- In part a, rewrite the numerator using the Pythagorean identity. - Cancel only common factors. - In part b, rewrite tangent as a quotient. - Check whether all factors cancel.

1. For a), use \(1-\cos^2\alpha=\sin^2\alpha\). Then \(\frac{\sin^2\alpha}{\sin\alpha}=\sin\alpha\). 2. For b), replace \(\tan\alpha\) with \(\frac{\sin\alpha}{\cos\alpha}\). Then \(\frac{\sin\alpha}{\cos\alpha}\cdot\frac{\cos\alpha}{\sin\alpha}=1\).

a) \(\sin\alpha\) b) \(1\)

- Rewrite tangent in terms of sine and cosine. - When dividing by a fraction, multiply by its reciprocal. - In part b, distribute first. - Look for the Pythagorean identity.

1. For a), replace \(\tan\alpha\) with \(\frac{\sin\alpha}{\cos\alpha}\): \(\frac{\sin\alpha}{\sin\alpha/\cos\alpha}=\cos\alpha\). 2. For b), distribute: \(\cos^2\alpha+\cos^2\alpha\tan^2\alpha\). 3. Replace \(\tan^2\alpha\) with \(\frac{\sin^2\alpha}{\cos^2\alpha}\). The expression becomes \(\cos^2\alpha+\sin^2\alpha=1\).

a) \(\cos\alpha\) b) \(1\)

- Use the Pythagorean identity. - Express tangent in terms of sine and cosine. - Because \(\alpha\) is acute, determine which square-root sign applies.

1. From \(\sin^2(\alpha)+\cos^2(\alpha)=1\), \(\sin(\alpha)=\sqrt{1-\cos^2(\alpha)}\). The positive square root is used because \(\alpha\) is acute. 2. Since \(\tan(\alpha)=\frac{\sin(\alpha)}{\cos(\alpha)}\), \(\tan(\alpha)=\frac{\sqrt{1-\cos^2(\alpha)}}{\cos(\alpha)}\). 3. When \(\cos(\alpha)=0.6=\frac{3}{5}\), \(\sin(\alpha)=\sqrt{1-\frac{9}{25}}=\frac{4}{5}\). 4. Therefore, \(\tan(\alpha)=\frac{4/5}{3/5}=\frac{4}{3}\).

a) \(\sin(\alpha)=\sqrt{1-\cos^2(\alpha)}\) and \(\tan(\alpha)=\frac{\sqrt{1-\cos^2(\alpha)}}{\cos(\alpha)}\) b) \(\sin(\alpha)=\frac{4}{5}\) and \(\tan(\alpha)=\frac{4}{3}\)

- Use the Pythagorean identity to find cosine. - Express tangent as a quotient of sine and cosine. - Substitute both exact values into the identity.

1. The Pythagorean identity gives \(\cos^2(\alpha)=1-\left(\frac{4}{5}\right)^2=\frac{9}{25}\). 2. Since \(\alpha\) is acute, \(\cos(\alpha)=\frac{3}{5}\). 3. Then \(\tan(\alpha)=\frac{\sin(\alpha)}{\cos(\alpha)}=\frac{4/5}{3/5}=\frac{4}{3}\). 4. Substitution gives \(\left(\frac{4}{5}\right)^2+\left(\frac{3}{5}\right)^2=\frac{16}{25}+\frac{9}{25}=1\).

\(\cos(\alpha)=\frac{3}{5}\) \(\tan(\alpha)=\frac{4}{3}\) The identity is verified because \(\frac{16}{25}+\frac{9}{25}=1\).

- Use the Pythagorean identity in part a. - Rewrite tangent as a quotient of sine and cosine before simplifying part b.

1. For part a, use \(1-\sin^2(\alpha)=\cos^2(\alpha)\). Then \(\frac{\cos^2(\alpha)}{\cos(\alpha)}=\cos(\alpha)\). 2. For part b, \(\frac{\sin(\alpha)}{\tan(\alpha)}=\cos(\alpha)\). The expression becomes \(\cos^2(\alpha)+\sin^2(\alpha)=1\).

a) \(\cos(\alpha)\) b) \(1\)

- Rewrite tangent in terms of sine and cosine. - In part b, simplify the product involving tangent before combining terms.

1. For part a, rewrite \(\tan^2(\alpha)\) as \(\frac{\sin^2(\alpha)}{\cos^2(\alpha)}\). Then the expression is \(\frac{1-\sin^2(\alpha)}{\cos^2(\alpha)}=\frac{\cos^2(\alpha)}{\cos^2(\alpha)}=1\). 2. For part b, \(\tan(\alpha)\cos(\alpha)=\sin(\alpha)\). The numerator becomes \(2\sin(\alpha)\), so the expression equals \(2\).

a) \(1\) b) \(2\)

- How do you divide by a fraction? - Check the change from Leon’s first line to his second. - After multiplying by the reciprocal, simplify and look for the Pythagorean identity.

1. Leon correctly replaces \(\tan(\alpha)\) with \(\frac{\sin(\alpha)}{\cos(\alpha)}\). 2. His second step is incorrect. Dividing by a fraction requires multiplying by its reciprocal. 3. Correctly, \(T=\cos(\alpha)\cdot\frac{\cos(\alpha)}{\sin(\alpha)}\cdot\sin(\alpha)+\sin^2(\alpha)\). 4. Canceling \(\sin(\alpha)\) gives \(T=\cos^2(\alpha)+\sin^2(\alpha)=1\).

Leon is incorrect. He mishandles division by \(\frac{\sin(\alpha)}{\cos(\alpha)}\). The correct result is \(T=1\).

- Rewrite tangent using sine and cosine. - Combine the terms over a common denominator. - Apply the Pythagorean identity.

1. Rewrite tangent: \(\frac{1}{\cos(\alpha)}-\frac{\sin^2(\alpha)}{\cos(\alpha)}\). 2. Combine the terms: \(\frac{1-\sin^2(\alpha)}{\cos(\alpha)}\). 3. Since \(1-\sin^2(\alpha)=\cos^2(\alpha)\), the expression simplifies to \(\cos(\alpha)\).

\(\cos(\alpha)\)

- Express tangent as a quotient of sine and cosine. - Substitute into the Pythagorean identity so only one trigonometric function remains. - Use the fact that an acute angle has positive sine and cosine.

1. Since \(\tan(\beta)=\frac{\sin(\beta)}{\cos(\beta)}=\sqrt{3}\), \(\sin(\beta)=\sqrt{3}\cos(\beta)\). 2. Substitute into \(\sin^2(\beta)+\cos^2(\beta)=1\): \(3\cos^2(\beta)+\cos^2(\beta)=1\). 3. Thus, \(4\cos^2(\beta)=1\), so \(\cos(\beta)=\frac{1}{2}\) because \(\beta\) is acute. 4. Then \(\sin(\beta)=\sqrt{3}\left(\frac{1}{2}\right)=\frac{\sqrt{3}}{2}\).

\(\sin(\beta)=\frac{\sqrt{3}}{2}\) and \(\cos(\beta)=\frac{1}{2}\)

- Use the Pythagorean identity to find cosine. - Express tangent as a quotient of sine and cosine. - Use the cofunction identity for complementary angles.

1. From the Pythagorean identity, \(\cos^2(\beta)=1-\left(\frac{2}{3}\right)^2=\frac{5}{9}\). Because \(\beta\) is acute, \(\cos(\beta)=\frac{\sqrt{5}}{3}\). 2. Then \(\tan(\beta)=\frac{2/3}{\sqrt{5}/3}=\frac{2}{\sqrt{5}}=\frac{2\sqrt{5}}{5}\). 3. By the cofunction identity, \(\sin(90^\circ-\beta)=\cos(\beta)=\frac{\sqrt{5}}{3}\).

a) \(\cos(\beta)=\frac{\sqrt{5}}{3}\) b) \(\tan(\beta)=\frac{2\sqrt{5}}{5}\) c) \(\sin(90^\circ-\beta)=\frac{\sqrt{5}}{3}\)

- Replace sine and cosine with their right-triangle side ratios. - Combine the two fractions. - Apply the Pythagorean theorem to the numerator.

1. In the triangle, \(\sin(\alpha)=\frac{a}{c}\) and \(\cos(\alpha)=\frac{b}{c}\). 2. Substitute: \(\sin^2(\alpha)+\cos^2(\alpha)=\frac{a^2}{c^2}+\frac{b^2}{c^2}=\frac{a^2+b^2}{c^2}\). 3. By the Pythagorean theorem, \(a^2+b^2=c^2\), so the expression becomes \(\frac{c^2}{c^2}=1\).

Substitution gives \(\sin^2(\alpha)+\cos^2(\alpha)=\frac{a^2+b^2}{c^2}\). Since \(a^2+b^2=c^2\), the expression equals \(1\).

- Rewrite tangent using sine and cosine. - Put the terms over a common denominator. - Use the Pythagorean identity in the numerator. - In part b, use the positive square root because the angle is acute.

1. Replace tangent with sine over cosine: \(1+\tan^2(\alpha)=1+\frac{\sin^2(\alpha)}{\cos^2(\alpha)}\). 2. Write \(1\) with the common denominator: \(\frac{\cos^2(\alpha)+\sin^2(\alpha)}{\cos^2(\alpha)}=\frac{1}{\cos^2(\alpha)}\). 3. For \(\tan(\alpha)=0.75\), \(1+0.75^2=1.5625=\frac{1}{\cos^2(\alpha)}\). 4. Thus, \(\cos^2(\alpha)=\frac{1}{1.5625}=0.64\). Because \(\alpha\) is acute, \(\cos(\alpha)=0.8\).

a) \(1+\tan^2(\alpha)=\frac{\cos^2(\alpha)+\sin^2(\alpha)}{\cos^2(\alpha)}=\frac{1}{\cos^2(\alpha)}\) b) \(\cos(\alpha)=0.8\)

- Expand the product on the left. - Rewrite tangent using sine and cosine. - Simplify the resulting product.

1. Expand the left side: \(\cos^2(\alpha)-\cos^2(\alpha)\tan^2(\alpha)\). 2. Substitute \(\tan^2(\alpha)=\frac{\sin^2(\alpha)}{\cos^2(\alpha)}\). 3. The second term simplifies to \(\sin^2(\alpha)\), leaving \(\cos^2(\alpha)-\sin^2(\alpha)\), which is the right side.

The left side simplifies as \(\cos^2(\alpha)-\cos^2(\alpha)\frac{\sin^2(\alpha)}{\cos^2(\alpha)}=\cos^2(\alpha)-\sin^2(\alpha)\), so the identity is true.

- Rewrite the reciprocal of tangent squared using sine and cosine. - Combine the fractions once they have the same denominator. - Use the Pythagorean identity to simplify the numerator.

1. Since \(\tan^2(\alpha)=\frac{\sin^2(\alpha)}{\cos^2(\alpha)}\), \(\frac{1}{\tan^2(\alpha)}=\frac{\cos^2(\alpha)}{\sin^2(\alpha)}\). 2. The left side becomes \(\frac{1-\cos^2(\alpha)}{\sin^2(\alpha)}\). 3. Use \(1-\cos^2(\alpha)=\sin^2(\alpha)\) to obtain \(\frac{\sin^2(\alpha)}{\sin^2(\alpha)}=1\).

After replacing \(\frac{1}{\tan^2(\alpha)}\) with \(\frac{\cos^2(\alpha)}{\sin^2(\alpha)}\), the left side becomes \(\frac{1-\cos^2(\alpha)}{\sin^2(\alpha)}=1\).

- Use the Pythagorean identity to rewrite the numerator. - Rewrite tangent using sine and cosine. - Check the claim by substituting exact values into the original expression.

1. Use \(1-\sin^2(\alpha)=\cos^2(\alpha)\) and \(\cos(\alpha)\tan(\alpha)=\sin(\alpha)\). 2. Thus, \(A=\frac{\cos^2(\alpha)}{\sin(\alpha)}=\frac{1-\sin^2(\alpha)}{\sin(\alpha)}=\frac{1}{\sin(\alpha)}-\sin(\alpha)\). 3. For \(\alpha=45^\circ\), \(A=\frac{1/2}{\sqrt{2}/2}=\frac{\sqrt{2}}{2}\approx 0.707\). 4. Therefore, the claim is false.

a) \(A=\frac{1}{\sin(\alpha)}-\sin(\alpha)\) b) The claim is false. For \(\alpha=45^\circ\), \(A=\frac{\sqrt{2}}{2}\approx 0.707\), not \(1\).

- On the unit circle, \(x=\cos(\alpha)\) and \(y=\sin(\alpha)\). - Use \(x^2+y^2=1\). - Determine the signs of both coordinates in Quadrant III.

1. Since the x-coordinate on the unit circle equals cosine, \(\cos(\alpha)=-0.6\). 2. Use \(x^2+y^2=1\): \((-0.6)^2+y^2=1\), so \(0.36+y^2=1\). 3. Therefore, \(y^2=0.64\), so \(y=\pm0.8\). 4. In Quadrant III, the y-coordinate is negative, so \(y=-0.8\). 5. Since the y-coordinate equals sine, \(\sin(\alpha)=-0.8\).

a) \(\cos(\alpha)=-0.6\) b) \(y=-0.8\) c) \(\sin(\alpha)=-0.8\)

- Use the Pythagorean identity. - Remember both signs when taking a square root. - Determine where sine is positive. - Use the quadrant to choose the cosine sign.

1. From the Pythagorean identity, \(\cos^2(\alpha)=1-(0.6)^2=0.64\), so \(\cos(\alpha)=0.8\) or \(\cos(\alpha)=-0.8\). 2. The first angle is \(\alpha_1=\sin^{-1}(0.6)\approx36.9^\circ\). 3. Sine is also positive in Quadrant II, so \(\alpha_2=180^\circ-36.9^\circ\approx143.1^\circ\). 4. In Quadrant I, cosine is positive, so \(\cos(\alpha_1)=0.8\). In Quadrant II, cosine is negative, so \(\cos(\alpha_2)=-0.8\).

a) \(\cos(\alpha)=0.8\) or \(\cos(\alpha)=-0.8\) b) \(\alpha\approx36.9^\circ\) or \(\alpha\approx143.1^\circ\) c) For \(\alpha\approx36.9^\circ\), \(\cos(\alpha)=0.8\); for \(\alpha\approx143.1^\circ\), \(\cos(\alpha)=-0.8\)

- Use the quadrant to choose the correct angle from the inverse-sine result. - Use the Pythagorean identity to find cosine and choose its sign from the quadrant. - Equal cosine values correspond to unit-circle points reflected across the x-axis.

1. Since \(\alpha\) is in Quadrant II, \(\alpha=180^\circ-\sin^{-1}(0.75)\approx131.41^\circ\). 2. Using the Pythagorean identity and the Quadrant II sign, \(\cos(\alpha)=-\sqrt{1-(0.75)^2}\approx-0.6614\). 3. An angle with the same cosine is the reflection across the x-axis: \(\beta=360^\circ-131.41^\circ\approx228.59^\circ\). 4. Since \(180^\circ < 228.59^\circ < 270^\circ\), \(\beta\) lies in Quadrant III.

a) \(\alpha\approx131.41^\circ\) b) \(\cos(\alpha)\approx-0.6614\) c) \(\beta\approx228.59^\circ\), in Quadrant III

- Which identity relates sine and cosine of the same angle? - How do unit-circle coordinates change under reflection across the y-axis or x-axis? - In which quadrant does \(360^\circ-\alpha\) lie when \(\alpha\) is acute?

1. Use \(\sin^2(\alpha)+\cos^2(\alpha)=1\): \(0.28^2+\cos^2(\alpha)=1\). 2. Then \(\cos^2(\alpha)=1-0.0784=0.9216\). Because \(\alpha\) is acute, cosine is positive, so \(\cos(\alpha)=\sqrt{0.9216}=0.96\). 3. Supplementary angles have equal sine values, so \(\sin(180^\circ-\alpha)=\sin(\alpha)=0.28\). 4. The angles \(\alpha\) and \(360^\circ-\alpha\) have equal cosine values, so \(\cos(360^\circ-\alpha)=\cos(\alpha)=0.96\).

a) \(\cos(\alpha)=0.96\) b) \(\sin(180^\circ-\alpha)=0.28\) and \(\cos(360^\circ-\alpha)=0.96\)

- Use exact sine and cosine values at \(30^\circ\) to test the first claim. - For part b, substitute the right-triangle side ratios for sine and cosine. - Apply the Pythagorean theorem.

1. For part a, \(\sin(30^\circ)=\frac{1}{2}\) and \(\cos(30^\circ)=\frac{\sqrt{3}}{2}\). 2. Their sum is \(\frac{1+\sqrt{3}}{2}\ne 1\), so the first claim is false. 3. For part b, in a right triangle \(\sin(\alpha)=\frac{a}{c}\) and \(\cos(\alpha)=\frac{b}{c}\). 4. Therefore, \(\sin^2(\alpha)+\cos^2(\alpha)=\frac{a^2+b^2}{c^2}=\frac{c^2}{c^2}=1\) by the Pythagorean theorem.

a) False. At \(30^\circ\), \(\sin(30^\circ)+\cos(30^\circ)=\frac{1+\sqrt{3}}{2}\ne 1\). b) True. The identity follows from \(a^2+b^2=c^2\).

- Factor a common term from the last two terms. - Look for the Pythagorean identity inside the factored expression. - Apply the same identity again at the end.

1. Factor \(\sin^2(\beta)\) from the last two terms: \(\cos^2(\beta)+\sin^2(\beta)\left(\cos^2(\beta)+\sin^2(\beta)\right)\). 2. Apply the Pythagorean identity inside the parentheses to get \(\cos^2(\beta)+\sin^2(\beta)\). 3. Apply the identity again. The result is \(1\).

\(1\)

- Which identity relates sine and cosine of the same angle? - Solving for a squared value gives two possible signs. - What is the sign of cosine in Quadrant II?

1. Use \(\sin^2(\alpha)+\cos^2(\alpha)=1\). 2. Substitute \(\sin(\alpha)=0.6\): \(0.6^2+\cos^2(\alpha)=1\). 3. Then \(\cos^2(\alpha)=1-0.36=0.64\), so \(\cos(\alpha)=\pm0.8\). 4. An obtuse angle lies in Quadrant II, where cosine is negative. Therefore, \(\cos(\alpha)=-0.8\).

\(\cos(\alpha)=-0.8\)

- Use the unit-circle equation relating the two coordinates. - How many points on the unit circle can have the given x-coordinate? - When solving \(u^2=a\), include both square roots.

1. Use \(\sin^2(\beta)+\cos^2(\beta)=1\). 2. Substitute: \(\sin^2(\beta)+\left(-\frac{12}{13}\right)^2=1\). 3. Thus, \(\sin^2(\beta)=1-\frac{144}{169}=\frac{25}{169}\). 4. Therefore, \(\sin(\beta)=\pm\frac{5}{13}\). 5. A negative cosine places the point in Quadrant II or III, so both signs of sine are possible.

\(\sin(\beta)=\frac{5}{13}\) or \(\sin(\beta)=-\frac{5}{13}\)

- Use the Pythagorean identity to relate sine and cosine. - Because the angle is acute, choose the positive square root. - Sine and cosine interchange for complementary angles.

1. Substitute \(\sin(\alpha)=0.8\) into the Pythagorean identity: \((0.8)^2+\cos^2(\alpha)=1\). 2. Then \(\cos^2(\alpha)=1-0.64=0.36\). 3. Because \(\alpha\) is acute, cosine is positive, so \(\cos(\alpha)=\sqrt{0.36}=0.6\). 4. Since \(\cos(\beta)=0.8\), \(\beta=\cos^{-1}(0.8)\approx36.87^\circ\). 5. Also, \(\sin(\alpha)=\cos(90^\circ-\alpha)\), so \(\beta=90^\circ-\alpha\). The two angles are complementary.

a) \(\cos(\alpha)=0.6\) b) \(\beta\approx36.87^\circ\), and \(\alpha+\beta=90^\circ\).

- Relate a unit-circle point to a right triangle. - What is the length of the hypotenuse? - What happens to a negative sign when the value is squared? - Side lengths are represented by absolute values of coordinates.

1. Every point \(P(x, y)\) on the unit circle is a distance \(1\) from the origin. 2. The lengths of the legs of the corresponding right triangle are \(|x|\) and \(|y|\), and the hypotenuse has length \(1\). 3. The Pythagorean theorem gives \(|x|^2+|y|^2=1^2\). 4. Since \(|x|^2=x^2\) and \(|y|^2=y^2\), substituting \(x=\cos(\alpha)\) and \(y=\sin(\alpha)\) gives \(\cos^2(\alpha)+\sin^2(\alpha)=1\). 5. The negative sign of cosine does not affect its square.

The coordinate lengths satisfy \(|\cos(\alpha)|^2+|\sin(\alpha)|^2=1\). Because squaring a value and squaring its absolute value give the same result, \(\cos^2(\alpha)+\sin^2(\alpha)=1\) in Quadrant II as well.