Free math worksheets

- 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 \(\cos(2x)\).

\(\cos(2x)\)

- Look for a double-angle identity whose right side is a product of sine and cosine. - The rewritten expression should contain one trigonometric function with input \(2x\).

1. The sine double-angle identity is \(\sin(2x)=2\sin(x)\cos(x)\), so the expression is \(\sin(2x)\).

\(\sin(2x)\)

- Look for the half-angle identity that comes from the cosine double-angle formula. - The plus sign selects the squared cosine form rather than the squared sine form.

1. The half-angle identity \(\cos^2\left(\frac{x}{2}\right)=\frac{1+\cos(x)}{2}\) gives the required form.

\(\cos^2\left(\frac{x}{2}\right)\)

- Recall the double-angle formula written entirely in terms of tangent. - The denominator contains \(1\) and the square of \(\tan(x)\); determine the sign between them.

1. In the tangent double-angle identity, the denominator is \(1-\tan^2(x)\).

\(1-\tan^2(x)\)

- 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 \(\sin(2x)=\sin(x)\cos(x)+\cos(x)\sin(x)=2\sin(x)\cos(x)\).

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

- Choose the double-angle identity that uses the two trigonometric values already given. - Substitute both values before simplifying the product.

1. Use \(\sin(2x)=2\sin(x)\cos(x)\). 2. Substitute the given values: \(\sin(2x)=2\cdot\frac{5}{13}\cdot\frac{12}{13}=\frac{120}{169}\).

\(\frac{120}{169}\)

- Use the cosine double-angle form that does not require finding sine first. - Square the given cosine value before applying the coefficient and subtraction.

1. Use \(\cos(2x)=2\cos^2(x)-1\). 2. Substitute \(\cos(x)=\frac{7}{9}\): \(\cos(2x)=2\cdot\frac{49}{81}-1=\frac{98}{81}-\frac{81}{81}=\frac{17}{81}\).

\(\frac{17}{81}\)

- 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: \(\sin(2x)=2\cdot\frac{3}{5}\cdot\frac{4}{5}=\frac{24}{25}\).

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

- Choose the cosine double-angle form that uses the value already given. - You do not need to determine another trigonometric function first. - Keep the calculation exact.

1. Use the cosine double-angle form \(\cos(2x)=1-2\sin^2(x)\). 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. - Pay attention to what happens to the sign when the cosine value is squared. - 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, \(\cos(2x)=\frac{98}{625}-\frac{625}{625}=-\frac{527}{625}\).

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

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

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

- Use trigonometry in one isosceles triangular section to find the side length. - Apply the double-angle identity for sine in part b). - Determine what happens to the cosine argument as \(n\) grows.

1. The central angle subtended by one side is \(\frac{2\pi}{n}\). Bisecting the corresponding isosceles triangle gives side length \(s_n=2R\sin\left(\frac{\pi}{n}\right)\). Therefore, \(p_n=2nR\sin\left(\frac{\pi}{n}\right)\). 2. Replacing \(n\) by \(2n\), \(p_{2n}=4nR\sin\left(\frac{\pi}{2n}\right)\). Thus, \(v_n=\frac{2\sin(\pi/(2n))}{\sin(\pi/n)}\). 3. Using \(\sin(2x)=2\sin x\cos x\) with \(x=\frac{\pi}{2n}\), \(\sin\left(\frac{\pi}{n}\right)=2\sin\left(\frac{\pi}{2n}\right)\cos\left(\frac{\pi}{2n}\right)\). Therefore, \(v_n=\frac{1}{\cos\left(\frac{\pi}{2n}\right)}\). 4. Since \(\frac{\pi}{2n}\to0\), \(\lim_{n\to\infty}v_n=\frac{1}{\cos0}=1\). The relative increase \(\frac{p_{2n}-p_n}{p_n}=v_n-1\) approaches \(0\), so doubling the number of sides produces a vanishing relative improvement as the polygonal perimeter approaches the circle's circumference.

a) \(p_n=2nR\sin\left(\frac{\pi}{n}\right)\). b) \(v_n=\frac{1}{\cos\left(\frac{\pi}{2n}\right)}\). c) The limit is \(1\); the relative perimeter increase from doubling the number of sides approaches \(0\).

- 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: \(3\).

- 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 \(\cos(x)=\frac{3}{\sqrt{13}}\)

- 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.

Diego 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 \(\frac{\theta}{2}\).

- 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. Nia'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) \(\sin(2x)=2\sin(x)\cos(x)\).

- 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. The diagram places \(Q\), which represents \(\frac{\theta}{2}\), 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 \(\sin\left(\frac{\theta}{2}\right)=\sqrt{\frac{1+\frac{1}{3}}{2}}=\sqrt{\frac{2}{3}}=\frac{\sqrt{6}}{3}\). 3. Use \(\cos\left(\frac{\theta}{2}\right)=\pm\sqrt{\frac{1+\cos(\theta)}{2}}\). In Quadrant II cosine is negative, so \(\cos\left(\frac{\theta}{2}\right)=-\sqrt{\frac{1-\frac{1}{3}}{2}}=-\sqrt{\frac{1}{3}}=-\frac{\sqrt{3}}{3}\).

\(\sin\left(\frac{\theta}{2}\right)=\frac{\sqrt{6}}{3}\) \(\cos\left(\frac{\theta}{2}\right)=-\frac{\sqrt{3}}{3}\)

- Treat \(\tan(x)\) as one unknown quantity inside the tangent double-angle identity. - Clearing the denominator turns the identity equation into a quadratic in that unknown. - Check that any candidate does not make the original double-angle denominator zero.

1. Let \(t=\tan(x)\). The tangent double-angle identity gives \(\frac{2t}{1-t^2}=\frac{3}{4}\). 2. Cross-multiply: \(8t=3-3t^2\), so \(3t^2+8t-3=0\). 3. Factor: \((3t-1)(t+3)=0\). 4. Thus, \(t=\frac{1}{3}\) or \(t=-3\). Neither value makes \(1-t^2\) zero, so both are valid.

\(\tan(x)=\frac{1}{3}\) or \(\tan(x)=-3\)

- For the range, look for the nonzero x-intercept of the trajectory. - Use the symmetry of a parabola to locate the vertex from its two x-intercepts. - Rewrite the angle-dependent part of the range using the supplied double-angle identity. - For the speed comparison, focus on how \(v_0\) appears in the maximum-height expression.

1. Set \(f(x)=0\) and factor out \(x\). Besides \(x=0\), the nonzero root is \(R=\frac{2v_0^2}{g}\sin\alpha\cos\alpha\). 2. The roots are \(0\) and \(R\), so the vertex occurs at \(x=R/2=\frac{v_0^2\sin\alpha\cos\alpha}{g}\). Substituting this x-value into \(f\) gives \(h_{\max}=\frac{v_0^2\sin^2\alpha}{2g}\). 3. Using the double-angle identity, \(R=\frac{v_0^2}{g}\sin(2\alpha)\). 4. Since \(0^\circ<2\alpha<180^\circ\), \(\sin(2\alpha)\) is largest when \(2\alpha=90^\circ\). Therefore, the horizontal range is greatest at \(\alpha=45^\circ\). 5. For fixed \(\alpha\), \(h_{\max}\) is proportional to \(v_0^2\). Replacing \(v_0\) by \(2v_0\) multiplies the maximum height by \(4\).

a) \(R=\frac{2v_0^2}{g}\sin\alpha\cos\alpha\), and \(h_{\max}=\frac{v_0^2\sin^2\alpha}{2g}\). b) \(R=\frac{v_0^2}{g}\sin(2\alpha)\), which is greatest at \(\alpha=45^\circ\). c) For fixed \(\alpha\), the maximum height is proportional to \(v_0^2\). Doubling \(v_0\) quadruples the maximum height.

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

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

- 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) \(2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)=\frac{12}{13}=\sin(\theta)\)