Free math worksheets

- Recall the identity that relates \(\sin^2(x)\) and \(\cos^2(x)\). - Isolate the term that is not being subtracted from \(1\).

1. Use the Pythagorean identity \(\sin^2(x)+\cos^2(x)=1\). 2. Subtract \(\sin^2(x)\) from both sides to get \(1-\sin^2(x)=\cos^2(x)\).

\(\cos^2(x)\)

- 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 \(\cos(\alpha)\).

\(\cos(\alpha)\)

- An identity must be true for every value in its domain. - Test the second statement at a familiar special angle. - A single counterexample is enough to show that a statement is not an identity.

1. The statement \(\sin^2(x)+\cos^2(x)=1\) is the Pythagorean identity, so it is true for every real \(x\). 2. The statement \(\sin(x)+\cos(x)=1\) is not true for every real \(x\). 3. For example, when \(x=45^\circ\), \(\sin(x)+\cos(x)=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}=\sqrt{2}\ne1\).

\(\sin^2(x)+\cos^2(x)=1\) is an identity. \(\sin(x)+\cos(x)=1\) is an equation that holds only for some values of \(x\). One counterexample is \(x=45^\circ\), for which the left side is \(\sqrt{2}\), not \(1\).

- The radius \(OP\) is the hypotenuse of the right triangle. - Identify which leg is adjacent to \(\alpha\) and which is opposite. - Apply the Pythagorean theorem after writing the two legs as trigonometric functions.

1. The radius \(OP\) has length \(1\), so \(OP\) is the hypotenuse of right triangle \(OHP\). 2. The side \(OH\) is adjacent to \(\alpha\), so \(x=OH=\cos(\alpha)\). 3. The side \(HP\) is opposite \(\alpha\), so \(y=HP=\sin(\alpha)\). 4. By the Pythagorean theorem, \(OH^2+HP^2=OP^2\). 5. Substituting gives \(\cos^2(\alpha)+\sin^2(\alpha)=1^2=1\).

a) \(x=\cos(\alpha)\) and \(y=\sin(\alpha)\) b) \(\sin^2(\alpha)+\cos^2(\alpha)=1\)

- 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, check what the acute-angle condition tells you about the sign of cosine.

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.

- First remove the denominator without changing the identity on its domain. - Rewrite \(\cos^2(x)\) using the Pythagorean identity. - Compare the constant term and the coefficient of \(\sin^2(x)\).

1. Multiply by \(\cos^2(x)\): \(a+b\sin^2(x)=\cos^2(x)\). 2. Use \(\cos^2(x)=1-\sin^2(x)\). 3. Matching the constant and \(\sin^2(x)\) terms gives \(a=1\) and \(b=-1\).

\(a=1\) and \(b=-1\)