- 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 \(1-\sin^2(x)=\cos^2(x)\).

\(\cos^2(x)\)

- Recall the Pythagorean identity obtained by dividing \(\sin^2(x)+\cos^2(x)=1\) by \(\cos^2(x)\). - The requested form should contain cosine but no sine or tangent outside a square.

1. The Pythagorean identity related to tangent is \(1+\tan^2(x)=\frac{1}{\cos^2(x)}\).

\(\frac{1}{\cos^2(x)}\)

- What two numbers have the given square? - Which sign does sine have in Quadrant III?

1. From the squared value, \(\sin(\theta)=\pm\frac{3}{5}\). 2. Sine is negative in Quadrant III, so \(\sin(\theta)=-\frac{3}{5}\).

\(\sin(\theta)=-\frac{3}{5}\)

- Which familiar trigonometric function is paired with secant? - How do you recover a number from its reciprocal?

1. Secant is the reciprocal of cosine, so \(\cos(\theta)=\frac{1}{\sec(\theta)}\). 2. Thus, \(\cos(\theta)=-\frac{1}{2}\).

\(\cos(\theta)=-\frac{1}{2}\)

- 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 \(\tan(\alpha)=\frac{4}{3}\)

- Substitute into the Pythagorean identity before taking any square root. - The acute-angle condition determines which square root of \(\cos^2(\alpha)\) is valid. - After cosine is known, tangent can be written as a quotient of sine and cosine.

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

a) \(\left(\frac{4}{5}\right)^2+\cos^2(\alpha)=1\), so \(\cos^2(\alpha)=\frac{9}{25}\) b) \(\cos(\alpha)=\frac{3}{5}\) and \(\tan(\alpha)=\frac{4}{3}\)

- 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) \(2\)

- Keep the squared cosine value separate from the sign choice until you use the acute-angle condition. - For tangent, use the ratio of the two trigonometric values rather than introducing an angle measure.

1. Substitute into the Pythagorean identity: \(\frac{4}{9}+\cos^2(\beta)=1\), so \(\cos^2(\beta)=\frac{5}{9}\). 2. Because \(\beta\) is acute, \(\cos(\beta)=\frac{\sqrt{5}}{3}\). 3. Then \(\tan(\beta)=\frac{2/3}{\sqrt{5}/3}=\frac{2\sqrt{5}}{5}\).

a) \(\frac{4}{9}+\cos^2(\beta)=1\), so \(\cos(\beta)=\frac{\sqrt{5}}{3}\) b) \(\tan(\beta)=\frac{2\sqrt{5}}{5}\)

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

\(\cos(\alpha)\)

- Substitute the given sine value into the Pythagorean identity before choosing a sign. - Solving for \(\cos^2(\alpha)\) gives two possible square roots. - Use the sign of cosine in Quadrant II to select the correct one.

1. Substitute into \(\sin^2(\alpha)+\cos^2(\alpha)=1\): \(\frac{25}{64}+\cos^2(\alpha)=1\). 2. Therefore, \(\cos^2(\alpha)=\frac{39}{64}\), so \(\cos(\alpha)=\pm\frac{\sqrt{39}}{8}\). 3. An obtuse angle is in Quadrant II, where cosine is negative. Thus, \(\cos(\alpha)=-\frac{\sqrt{39}}{8}\).

\(\frac{25}{64}+\cos^2(\alpha)=1\), so \(\cos(\alpha)=-\frac{\sqrt{39}}{8}\)

- 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) In the right triangle, \(OH^2+HP^2=OP^2\). Since \(OH=\cos(\alpha)\), \(HP=\sin(\alpha)\), and the radius \(OP=1\), it follows that \(\sin^2(\alpha)+\cos^2(\alpha)=1\).

- Replace cotangent with its given definition. - How can you write \(1\) over the same denominator as the other term? - Look for an identity that simplifies the new numerator.

1. Write \(1+\cot^2(\theta)=1+\frac{\cos^2(\theta)}{\sin^2(\theta)}\). 2. Combine the terms: \(\frac{\sin^2(\theta)+\cos^2(\theta)}{\sin^2(\theta)}=\frac{1}{\sin^2(\theta)}=\csc^2(\theta)\).

a) \(\csc^2(\theta)\) b) \(1+\frac{\cos^2(\theta)}{\sin^2(\theta)}=\frac{1}{\sin^2(\theta)}=\csc^2(\theta)\), using \(\sin^2(\theta)+\cos^2(\theta)=1\).

- Decide the sign of the horizontal coordinate of \(P\). - A segment length is positive, but a coordinate can be negative. - What happens when a negative coordinate is squared?

1. The point \(P\) lies to the left of the origin, so its horizontal coordinate is negative. The circle has radius \(1\), giving \(\cos(\alpha)=-OH\). 2. The height of \(P\) is \(HP>0\), so \(\sin(\alpha)=HP\). 3. In the right triangle, \(OH^2+HP^2=OP^2=1\). Squaring removes the negative sign, so \(\cos^2(\alpha)+\sin^2(\alpha)=1\).

a) \(\cos(\alpha)=-OH\), because \(P\) is in Quadrant II. b) \(OH^2+HP^2=OP^2=1\), so the Pythagorean identity remains true even though cosine is negative.

- Think about what a reciprocal does to a ratio. - Which of the given trigonometric values is paired with each reciprocal function? - Check by multiplying each answer by its matching original value.

1. Take the reciprocal of cosine: \(\sec(\theta)=\frac{1}{4/5}=\frac{5}{4}\). 2. Take the reciprocal of sine: \(\csc(\theta)=\frac{1}{3/5}=\frac{5}{3}\).

\(\sec(\theta)=\frac{5}{4}\), the reciprocal of \(\cos(\theta)=\frac{4}{5}\). \(\csc(\theta)=\frac{5}{3}\), the reciprocal of \(\sin(\theta)=\frac{3}{5}\).

- What does the identity tell you about tangent squared when secant is known? - After finding a squared value, what choices remain for tangent itself? - Which quadrant sign is consistent with the original angle?

1. Rearrange the identity: \(\tan^2(\theta)=\sec^2(\theta)-1\). 2. Substitute: \(\tan^2(\theta)=\frac{169}{25}-1=\frac{144}{25}\). 3. Tangent is negative in Quadrant II, so \(\tan(\theta)=-\frac{12}{5}\).

\(\tan(\theta)=-\frac{12}{5}\)

- Use the exact sine and cosine of \(30^\circ\) to find cotangent and cosecant. - Compare the two sides of the stated equation numerically. - Can you write a corrected equation that works for all angles in its domain?

1. At \(30^\circ\), \(\cot(30^\circ)=\sqrt{3}\) and \(\csc(30^\circ)=2\). 2. The left side is \(1+(\sqrt{3})^2=4\), but the right side of Sam's claim is \(2\). Thus, the claim is false. 3. The missing square is essential: the correct identity is \(1+\cot^2(\theta)=\csc^2(\theta)\).

Sam is incorrect: \(4\ne2\) at \(30^\circ\). The correct identity is \(1+\cot^2(\theta)=\csc^2(\theta)\).

- Start with the identity named in part a and substitute the given tangent value before solving for cosine. - Solving for \(\cos^2(\beta)\) does not yet determine the sign of \(\cos(\beta)\). - After cosine is known, use the quotient definition of tangent to recover sine.

1. Substitute \(\tan(\beta)=\frac{2}{3}\) into \(1+\tan^2(\beta)=\frac{1}{\cos^2(\beta)}\): \(1+\frac{4}{9}=\frac{1}{\cos^2(\beta)}\). 2. Thus, \(\frac{13}{9}=\frac{1}{\cos^2(\beta)}\), so \(\cos^2(\beta)=\frac{9}{13}\). 3. Since \(\beta\) is acute, \(\cos(\beta)=\frac{3\sqrt{13}}{13}\). 4. From \(\tan(\beta)=\frac{\sin(\beta)}{\cos(\beta)}=\frac{2}{3}\), \(\sin(\beta)=\frac{2}{3}\cos(\beta)=\frac{2\sqrt{13}}{13}\).

a) \(1+\frac{4}{9}=\frac{1}{\cos^2(\beta)}\), so \(\cos^2(\beta)=\frac{9}{13}\) b) \(\cos(\beta)=\frac{3\sqrt{13}}{13}\) and \(\sin(\beta)=\frac{2\sqrt{13}}{13}\)

- 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, Priya's 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 \(1\).

- 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 \(1\).

\(1\)

- 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 \(b=-1\)

- Begin by undoing the reciprocal in the given cosecant value. - Use the quadrant to choose the correct sign when recovering cosine. - Once sine and cosine are known, use quotients and reciprocals for the remaining functions.

1. Since \(\csc(\theta)=-\frac{5}{3}\), \(\sin(\theta)=-\frac{3}{5}\). 2. In Quadrant IV, cosine is positive. Using \(\sin^2(\theta)+\cos^2(\theta)=1\) gives \(\cos(\theta)=\frac{4}{5}\). 3. Then \(\tan(\theta)=\frac{-3/5}{4/5}=-\frac{3}{4}\). 4. Taking reciprocals gives \(\sec(\theta)=\frac{5}{4}\), \(\csc(\theta)=-\frac{5}{3}\), and \(\cot(\theta)=-\frac{4}{3}\).

\(\sin(\theta)=-\frac{3}{5}\), \(\cos(\theta)=\frac{4}{5}\), \(\tan(\theta)=-\frac{3}{4}\), \(\csc(\theta)=-\frac{5}{3}\), \(\sec(\theta)=\frac{5}{4}\), \(\cot(\theta)=-\frac{4}{3}\)

- Which trigonometric ratio relates the horizontal component to the total force? - How can the Pythagorean identity give sine from a known cosine value? - Use the direction of the force to choose the sine sign. - How do you turn the sine ratio into a force in newtons?

1. The westward component is negative, so \(\cos(\theta)=\frac{-25}{65}=-\frac{5}{13}\). 2. Use the Pythagorean identity: \(\sin^2(\theta)=1-\cos^2(\theta)=1-\frac{25}{169}=\frac{144}{169}\). 3. The force points north, so sine is positive: \(\sin(\theta)=\frac{12}{13}\). 4. The northward component is \(65\cdot\frac{12}{13}=60\,\text{N}\).

a) \(\cos(\theta)=-\frac{5}{13}\) and \(\sin(\theta)=\frac{12}{13}\), using \(\sin^2(\theta)=1-\cos^2(\theta)\) b) \(60\,\text{N}\) north