Free math worksheets

- Think of \(2x\) as the sum of two equal angles. - Apply the sine addition formula before combining like terms.

1. Write \(\sin(2x)=\sin(x+x)\). 2. Apply \(\sin(A+B)=\sin(A)\cos(B)+\cos(A)\sin(B)\). 3. Then \(\sin(2x)=\sin(x)\cos(x)+\cos(x)\sin(x)=2\sin(x)\cos(x)\).

\(\sin(2x)=2\sin(x)\cos(x)\)

- Recall a double-angle identity for cosine. - The final angle should be twice the original angle.

1. Recognize the cosine double-angle identity \(\cos(2x)=\cos^2(x)-\sin^2(x)\). 2. Therefore, the expression is \(\cos(2x)\).

\(\cos(2x)\)

- Express \(60^\circ\) as twice a familiar angle. - Use the sine double-angle identity with exact special-angle values.

1. Write \(60^\circ=2\cdot30^\circ\). 2. Use \(\sin(2x)=2\sin(x)\cos(x)\). 3. Then \(\sin(60^\circ)=2\cdot\frac{1}{2}\cdot\frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2}\).

\(\frac{\sqrt{3}}{2}\)

- Choose the cosine double-angle form that uses only cosine. - Check the sign against the quadrant of \(120^\circ\).

1. Write \(120^\circ=2\cdot60^\circ\). 2. Use \(\cos(2x)=2\cos^2(x)-1\). 3. Then \(\cos(120^\circ)=2\left(\frac{1}{2}\right)^2-1=-\frac{1}{2}\).

\(-\frac{1}{2}\)

- Which double-angle identity uses one sine value and one cosine value? - Substitute the exact fractions before simplifying.

1. Use \(\sin(2x)=2\sin(x)\cos(x)\). 2. Substitute the given values: \(\sin(2x)=2\cdot\frac{3}{5}\cdot\frac{4}{5}=\frac{24}{25}\).

\(\frac{24}{25}\)

- Use the double-angle identity for cosine. - Recall the condition for an even function. - Treat \(\cos(x)\) as a variable restricted to \([-1,1]\). - Compare the vertex value with the values at both endpoints of that interval.

1. Use \(\cos(2x)=2\cos^2(x)-1\): \(f(x)=2\cos^2(x)+2\cos(x)-1\). Completing the square gives \(f(x)=2\left(\cos(x)+\frac{1}{2}\right)^2-\frac{3}{2}\). 2. Since cosine is even, \(f(-x)=\cos(-2x)+2\cos(-x)=f(x)\). Therefore, the graph has y-axis symmetry. 3. Let \(u=\cos(x)\), where \(-1\le u\le1\). Then \(f=2\left(u+\frac{1}{2}\right)^2-\frac{3}{2}\). The minimum occurs at \(u=-\frac{1}{2}\), giving \(-\frac{3}{2}\). For the maximum, compare the endpoints: at \(u=-1\), the value is \(-1\), and at \(u=1\), the value is \(3\). Thus the global maximum is \(3\).

a) \(f(x)=2\left(\cos(x)+\frac{1}{2}\right)^2-\frac{3}{2}\). b) Yes. Since \(f(-x)=f(x)\), the graph has y-axis symmetry. c) Global minimum: \(-\frac{3}{2}\). Global maximum: \(3\).

- Choose the cosine double-angle form that uses the value already given. - Squaring removes the sign issue from the quadrant. - Keep the calculation exact.

1. Use the cosine double-angle form \(\cos(2x)=1-2\sin^2(x)\), which requires no sign choice for \(\cos(x)\). 2. Substitute \(\sin(x)=\frac{5}{13}\): \(\cos(2x)=1-2\cdot\frac{25}{169}\). 3. Therefore, \(\cos(2x)=\frac{119}{169}\).

\(\frac{119}{169}\)

- Use a cosine double-angle form that depends only on cosine. - The negative sign on \(\cos(x)\) disappears when you square it. - Check that the final value lies in \([-1,1]\).

1. Use \(\cos(2x)=2\cos^2(x)-1\). 2. Substitute \(\cos(x)=-\frac{7}{25}\): \(\cos(2x)=2\cdot\frac{49}{625}-1\). 3. Therefore, \(\cos(2x)=\frac{98}{625}-\frac{625}{625}=-\frac{527}{625}\).

\(-\frac{527}{625}\)

- View \(22.5^\circ\) as half of a special angle. - Determine the sign from the quadrant of the half-angle. - Simplify the nested fraction before taking the square root.

1. Since \(22.5^\circ=\frac{45^\circ}{2}\), use \(\sin\left(\frac{\theta}{2}\right)=\pm\sqrt{\frac{1-\cos(\theta)}{2}}\). 2. The angle \(22.5^\circ\) is in Quadrant I, so use the positive sign. 3. Substitute \(\cos(45^\circ)=\frac{\sqrt{2}}{2}\): \(\sin(22.5^\circ)=\sqrt{\frac{1-\frac{\sqrt{2}}{2}}{2}}=\frac{\sqrt{2-\sqrt{2}}}{2}\).

\(\frac{\sqrt{2-\sqrt{2}}}{2}\)

- First determine the quadrant of \(\frac{\theta}{2}\). - The half-angle formula gives a magnitude; the quadrant determines the sign. - Substitute the given cosine value only after deciding the sign.

1. Since \(\pi<\theta<\frac{3\pi}{2}\), the half-angle satisfies \(\frac{\pi}{2}<\frac{\theta}{2}<\frac{3\pi}{4}\), so it lies in Quadrant II and its cosine is negative. 2. Use \(\cos\left(\frac{\theta}{2}\right)=\pm\sqrt{\frac{1+\cos(\theta)}{2}}\). 3. The magnitude is \(\sqrt{\frac{1-\frac{3}{5}}{2}}=\sqrt{\frac{1}{5}}=\frac{\sqrt{5}}{5}\). 4. Therefore, \(\cos\left(\frac{\theta}{2}\right)=-\frac{\sqrt{5}}{5}\).

\(-\frac{\sqrt{5}}{5}\)

- Rearrange the cosine double-angle identities to isolate \(\sin^2(x)\) and \(\cos^2(x)\). - The quadrant tells you which square-root signs to choose. - Check that the two squared values add to \(1\).

1. Use \(\cos^2(x)=\frac{1+\cos(2x)}{2}\): \(\cos^2(x)=\frac{1+\frac{5}{13}}{2}=\frac{9}{13}\). 2. Because \(x\) is in Quadrant I, \(\cos(x)=\frac{3}{\sqrt{13}}\). 3. Use \(\sin^2(x)=\frac{1-\cos(2x)}{2}\): \(\sin^2(x)=\frac{1-\frac{5}{13}}{2}=\frac{4}{13}\). 4. Because \(x\) is in Quadrant I, \(\sin(x)=\frac{2}{\sqrt{13}}\).

\(\sin(x)=\frac{2}{\sqrt{13}}\) and \(\cos(x)=\frac{3}{\sqrt{13}}\)

- Choose the cosine double-angle form that already contains \(\sin^2(x)\). - Treat the identity as an algebraic equation and isolate \(\sin^2(x)\).

1. Start with \(\cos(2x)=1-2\sin^2(x)\). 2. Rearrange to \(2\sin^2(x)=1-\cos(2x)\). 3. Divide by \(2\): \(\sin^2(x)=\frac{1-\cos(2x)}{2}\).

\(\sin^2(x)=\frac{1-\cos(2x)}{2}\)

- Compare zeros, maxima, minima, and period in the two curves. - Start the proof from the \(\cos^2(x)-\sin^2(x)\) form. - Use the Pythagorean identity to eliminate \(\sin^2(x)\).

1. The two graphs coincide, suggesting that \(f(x)=g(x)\) for every real \(x\). 2. From \(\cos(2x)=\cos^2(x)-\sin^2(x)\), replace \(\sin^2(x)\) by \(1-\cos^2(x)\). 3. Then \(\cos(2x)=\cos^2(x)-\left(1-\cos^2(x)\right)=2\cos^2(x)-1\).

a) The graphs coincide. b) \(\cos(2x)=2\cos^2(x)-1\).

- A square determines a magnitude but not automatically a sign. - Ask what information decides whether the half-angle sine is positive or negative. - The relevant quadrant is the quadrant of \(\frac{\theta}{2}\), not necessarily the quadrant of \(\theta\).

1. Squaring the half-angle relation determines only the magnitude of \(\sin\left(\frac{\theta}{2}\right)\). 2. The sign depends on the quadrant containing \(\frac{\theta}{2}\). 3. The correct identity is \(\sin\left(\frac{\theta}{2}\right)=\pm\sqrt{\frac{1-\cos(\theta)}{2}}\), with the sign chosen from the half-angle's quadrant.

The student omitted the sign choice. The correct identity is \(\sin\left(\frac{\theta}{2}\right)=\pm\sqrt{\frac{1-\cos(\theta)}{2}}\), with the sign determined by the quadrant of \(\frac{\theta}{2}\).

- Read the opposite, adjacent, and hypotenuse lengths directly from the triangle. - Use the same two trigonometric ratios in the sine and cosine double-angle identities. - Determine the quadrant of \(2\alpha\) from the signs of its sine and cosine rather than estimating the angle.

1. From the triangle, \(\sin(\alpha)=\frac{3}{5}\) and \(\cos(\alpha)=\frac{4}{5}\). 2. \(\sin(2\alpha)=2\sin(\alpha)\cos(\alpha)=2\cdot\frac{3}{5}\cdot\frac{4}{5}=\frac{24}{25}\). 3. \(\cos(2\alpha)=\cos^2(\alpha)-\sin^2(\alpha)=\frac{16}{25}-\frac{9}{25}=\frac{7}{25}\). 4. Both sine and cosine of \(2\alpha\) are positive, so \(2\alpha\) lies in Quadrant I.

a) \(\sin(\alpha)=\frac{3}{5}\), \(\cos(\alpha)=\frac{4}{5}\) b) \(\sin(2\alpha)=\frac{24}{25}\), \(\cos(2\alpha)=\frac{7}{25}\) c) Quadrant I

- The square root in a half-angle identity does not determine the sign by itself. - Use the diagram to identify the quadrant containing the half-angle radius. - Choose the signs of sine and cosine only after locating that quadrant.

1. Since \(\theta=240^\circ\), the half-angle is \(120^\circ\), and the diagram places \(Q\) in Quadrant II. 2. Use \(\sin\left(\frac{\theta}{2}\right)=\pm\sqrt{\frac{1-\cos(\theta)}{2}}\). In Quadrant II sine is positive, so \(\sin\left(\frac{\theta}{2}\right)=\sqrt{\frac{1+\frac{1}{2}}{2}}=\frac{\sqrt{3}}{2}\). 3. Use \(\cos\left(\frac{\theta}{2}\right)=\pm\sqrt{\frac{1+\cos(\theta)}{2}}\). In Quadrant II cosine is negative, so \(\cos\left(\frac{\theta}{2}\right)=-\sqrt{\frac{1-\frac{1}{2}}{2}}=-\frac{1}{2}\).

\(\sin\left(\frac{\theta}{2}\right)=\frac{\sqrt{3}}{2}\) \(\cos\left(\frac{\theta}{2}\right)=-\frac{1}{2}\)

- Rewrite both the numerator and denominator using double-angle identities. - Look for a common factor after substitution. - Keep the original domain restriction in mind when canceling.

1. Use \(1-\cos(2x)=2\sin^2(x)\) and \(\sin(2x)=2\sin(x)\cos(x)\). 2. Then \(\frac{1-\cos(2x)}{\sin(2x)}=\frac{2\sin^2(x)}{2\sin(x)\cos(x)}\). 3. Because the original denominator is nonzero, the cancellation is valid on the stated domain. 4. The result is \(\frac{\sin(x)}{\cos(x)}=\tan(x)\).

\(\frac{1-\cos(2x)}{\sin(2x)}=\tan(x)\)

- Recover the missing sine value from the quadrant and the Pythagorean identity. - Choose a half-angle form for tangent that uses \(\sin(\theta)\) and \(\cos(\theta)\). - Check the sign using the quadrant of \(\frac{\theta}{2}\).

1. Since \(\theta\) is in Quadrant II, \(\sin(\theta)=\frac{12}{13}\). 2. Use the half-angle form \(\tan\left(\frac{\theta}{2}\right)=\frac{\sin(\theta)}{1+\cos(\theta)}\). 3. Substitute the values: \(\tan\left(\frac{\theta}{2}\right)=\frac{12/13}{1-5/13}=\frac{12/13}{8/13}=\frac{3}{2}\). 4. This is positive, consistent with \(\frac{\theta}{2}\) lying in Quadrant I.

\(\frac{3}{2}\)

- An identity would require the two graphs to coincide at every input. - Rewrite \(2x\) as \(x+x\). - Apply the sine addition identity before combining terms.

1. The graphs do not coincide, so the two expressions are not equal for every \(x\). 2. Write \(\sin(2x)=\sin(x+x)\). 3. The sine addition identity gives \(\sin(2x)=\sin(x)\cos(x)+\cos(x)\sin(x)=2\sin(x)\cos(x)\). 4. The student's expression is missing the factor \(\cos(x)\).

a) The graph shows different values for the two functions at many inputs, so the claim is false. b) \(\sin(2x)=2\sin(x)\cos(x)\).

- Use a cosine double-angle form written in terms of tangent. - Solve first for \(\tan^2(x)\). - The acute-angle condition determines the sign of \(\tan(x)\).

1. Use \(\cos(2x)=\frac{1-\tan^2(x)}{1+\tan^2(x)}\). 2. Let \(t=\tan^2(x)\). Then \(-\frac{7}{25}=\frac{1-t}{1+t}\). 3. Cross-multiplying gives \(-7-7t=25-25t\), so \(18t=32\) and \(t=\frac{16}{9}\). 4. Since \(x\) is acute, \(\tan(x)>0\). Therefore, \(\tan(x)=\frac{4}{3}\).

\(\frac{4}{3}\)

- Recover \(\cos(\theta)\) and determine the quadrant of the half-angle before taking square roots. - Use both half-angle formulas rather than finding \(\theta\) itself. - For the verification, substitute the two half-angle values into the sine double-angle identity.

1. Because \(\theta\) is in Quadrant II, \(\cos(\theta)=-\frac{5}{13}\), and \(\frac{\theta}{2}\) lies in Quadrant I. 2. \(\sin\left(\frac{\theta}{2}\right)=\sqrt{\frac{1-\cos(\theta)}{2}}=\sqrt{\frac{1+5/13}{2}}=\frac{3}{\sqrt{13}}\). 3. \(\cos\left(\frac{\theta}{2}\right)=\sqrt{\frac{1+\cos(\theta)}{2}}=\sqrt{\frac{1-5/13}{2}}=\frac{2}{\sqrt{13}}\). 4. Therefore, \(\tan\left(\frac{\theta}{2}\right)=\frac{3}{2}\). 5. Finally, \(2\cdot\frac{3}{\sqrt{13}}\cdot\frac{2}{\sqrt{13}}=\frac{12}{13}=\sin(\theta)\), so the double-angle identity is verified.

a) \(\sin\left(\frac{\theta}{2}\right)=\frac{3}{\sqrt{13}}\), \(\cos\left(\frac{\theta}{2}\right)=\frac{2}{\sqrt{13}}\), and \(\tan\left(\frac{\theta}{2}\right)=\frac{3}{2}\) b) \(2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)=\frac{12}{13}=\sin(\theta)\)