Free math worksheets

- Look for the structure of a sine sum rather than expanding anything. - Match the order of the sine and cosine factors to the addition formula. - The final answer should contain one trigonometric function, not a sum.

1. Compare the expression with \(\sin(A+B)=\sin(A)\cos(B)+\cos(A)\sin(B)\). 2. Taking \(A=x\) and \(B=30^\circ\) gives \(\sin(x+30^\circ)\).

\(\sin(x+30^\circ)\)

- Recall the addition formula specifically for sine, not cosine. - Check that each term contains one sine factor and one cosine factor.

1. The sine angle-addition identity gives \(\sin(A+B)=\sin(A)\cos(B)+\cos(A)\sin(B)\).

\(\sin(A)\cos(B)+\cos(A)\sin(B)\)

- Recall the cosine formula for a difference of two angles. - Pay particular attention to the sign between the two product terms; cosine subtraction has a different sign pattern from the angle itself.

1. The cosine angle-subtraction identity gives \(\cos(A-B)=\cos(A)\cos(B)+\sin(A)\sin(B)\).

\(\cos(A)\cos(B)+\sin(A)\sin(B)\)

- Focus on the denominator sign in the tangent subtraction formula. - Compare the sign joining the angles with the sign between the product term and \(1\).

1. In the tangent subtraction identity, the denominator is \(1+\tan(A)\tan(B)\).

\(1+\tan(A)\tan(B)\)

- Compare the order of the two products with a sine angle-combination pattern. - Decide whether the minus sign between the products corresponds to a sum or a difference of angles. - After combining the angles, simplify the coefficient of \(t\).

1. Use \(\sin(A-B)=\sin(A)\cos(B)-\cos(A)\sin(B)\). 2. With \(A=5t\) and \(B=2t\), the expression is \(\sin(5t-2t)\). 3. Therefore, the expression simplifies to \(\sin(3t)\).

Identity: \(\sin(A-B)=\sin(A)\cos(B)-\cos(A)\sin(B)\); result: \(\sin(3t)\).

- Express \(75^\circ\) as a sum of two familiar special angles. - Choose the sine addition formula. - Keep the result exact rather than converting to a decimal.

1. Write \(75^\circ=45^\circ+30^\circ\). 2. Use \(\sin(A+B)=\sin(A)\cos(B)+\cos(A)\sin(B)\). 3. Then \(\sin(75^\circ)=\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2}\cdot\frac{1}{2}=\frac{\sqrt{6}+\sqrt{2}}{4}\).

\(\frac{\sqrt{6}+\sqrt{2}}{4}\)

- Split \(105^\circ\) into two special angles. - Pay attention to the subtraction sign in the cosine addition formula. - The sign of the final value should agree with the quadrant of \(105^\circ\).

1. Write \(105^\circ=60^\circ+45^\circ\). 2. Use \(\cos(A+B)=\cos(A)\cos(B)-\sin(A)\sin(B)\). 3. Then \(\cos(105^\circ)=\frac{1}{2}\cdot\frac{\sqrt{2}}{2}-\frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}{2}=\frac{\sqrt{2}-\sqrt{6}}{4}\).

\(\frac{\sqrt{2}-\sqrt{6}}{4}\)

- Use a decomposition of \(75^\circ\) involving \(45^\circ\) and \(30^\circ\). - The tangent addition formula has a denominator that can affect the sign. - Simplify the exact fraction without using a decimal approximation.

1. Write \(75^\circ=45^\circ+30^\circ\). 2. Use \(\tan(A+B)=\frac{\tan(A)+\tan(B)}{1-\tan(A)\tan(B)}\). 3. Substitute the special-angle values: \(\tan(75^\circ)=\frac{1+\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}}}\). 4. Simplifying gives \(\tan(75^\circ)=2+\sqrt{3}\).

\(2+\sqrt{3}\)

- Identify which addition formula directly uses all four given values. - Keep a common denominator when combining the two products. - Check that the result lies between \(-1\) and \(1\).

1. Use \(\sin(\alpha+\beta)=\sin(\alpha)\cos(\beta)+\cos(\alpha)\sin(\beta)\). 2. Substitute the given values: \(\frac{3}{5}\cdot\frac{12}{13}+\frac{4}{5}\cdot\frac{5}{13}\). 3. The result is \(\frac{36}{65}+\frac{20}{65}=\frac{56}{65}\).

\(\frac{56}{65}\)

- Start from the cosine formula for a sum of two angles. - Treat \(x\) and \(\frac{\pi}{3}\) as the two inputs to that formula. - Substitute the exact unit-circle values only after writing the full expansion.

1. Use \(\cos(A+B)=\cos(A)\cos(B)-\sin(A)\sin(B)\). 2. Substitute \(A=x\) and \(B=\frac{\pi}{3}\): \(\cos\left(x+\frac{\pi}{3}\right)=\cos(x)\cos\left(\frac{\pi}{3}\right)-\sin(x)\sin\left(\frac{\pi}{3}\right)\). 3. Since \(\cos\left(\frac{\pi}{3}\right)=\frac{1}{2}\) and \(\sin\left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{2}\), the result is \(\frac{1}{2}\cos(x)-\frac{\sqrt{3}}{2}\sin(x)\).

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

- Express \(15^\circ\) as a difference of special angles. - Use the sine subtraction formula rather than a decimal approximation. - The result should be positive and smaller than \(\frac{1}{2}\).

1. Write \(15^\circ=45^\circ-30^\circ\). 2. Use \(\sin(A-B)=\sin(A)\cos(B)-\cos(A)\sin(B)\). 3. Then \(\sin(15^\circ)=\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}\cdot\frac{1}{2}=\frac{\sqrt{6}-\sqrt{2}}{4}\).

\(\frac{\sqrt{6}-\sqrt{2}}{4}\)

- Use the tangent addition formula directly with the two given tangent values. - Simplify the numerator and denominator separately before dividing. - Check that the denominator of the addition formula is nonzero.

1. Use \(\tan(\alpha+\beta)=\frac{\tan(\alpha)+\tan(\beta)}{1-\tan(\alpha)\tan(\beta)}\). 2. The numerator is \(\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\). 3. The denominator is \(1-\frac{1}{6}=\frac{5}{6}\). 4. Therefore, \(\tan(\alpha+\beta)=1\).

\(1\)

- Expand the shifted sine using an angle-addition identity. - Identify which resulting term is even and which is odd. - Set the coefficient of the odd term equal to \(0\).

1. An even function must satisfy \(f(-x)=f(x)\) for every \(x\). 2. Use the angle-addition identity: \(f(x)=\sin(c)\cos\left(\frac{x}{2}\right)+\cos(c)\sin\left(\frac{x}{2}\right)\). 3. The cosine term is even and the sine term is odd. For the entire function to be even, the coefficient of the odd term must be \(0\): \(\cos(c)=0\). 4. On \([0, 2\pi]\), this gives \(c=\frac{\pi}{2}\) or \(c=\frac{3\pi}{2}\). 5. These choices produce \(f(x)=\cos\left(\frac{x}{2}\right)\) and \(f(x)=-\cos\left(\frac{x}{2}\right)\), respectively, and both are even.

\(c\in\left\{\frac{\pi}{2}, \frac{3\pi}{2}\right\}\)

- Expand \(\sin(x+\phi)\) with the angle-addition identity. - Compare the sine and cosine coefficients. - Use the signs of both coefficients to choose the correct quadrant. - Square the amplitude equation to solve for \(k\).

1. Expanding \(A\sin(x+\phi)\) gives \(A\cos(\phi)\sin(x)+A\sin(\phi)\cos(x)\). Therefore, \(A\cos(\phi)=k+2\) and \(A\sin(\phi)=k\). 2. Squaring and adding gives \(A(k)=\sqrt{(k+2)^2+k^2}=\sqrt{2k^2+4k+4}\). 3. For \(k=1\), \(A=\sqrt{10}\). Also, \(\cos(\phi)=\frac{3}{\sqrt{10}}\) and \(\sin(\phi)=\frac{1}{\sqrt{10}}\), so \(\phi\approx0.32\) radians. 4. Set \(\sqrt{2k^2+4k+4}=\sqrt{20}\). Squaring and simplifying gives \(k^2+2k-8=0\), so \(k=2\) or \(k=-4\).

a) \(A(k)=\sqrt{2k^2+4k+4}\) b) \(A=\sqrt{10}\), \(\phi\approx0.32\) radians c) \(k=2\) or \(k=-4\)

- Compare corresponding maximum points and zeros. - Recall how left and right shifts appear inside a function. - Determine the period of \(f(x)=\sin\left(\frac{\pi}{2}x\right)\). - For the general verification, use periodicity first and then expand the shifted sine with the sine addition identity.

1. Comparing corresponding maximum points and zeros shows that \(g\) is obtained by shifting \(f\) left \(1\) unit. Since the period is \(4\), this is equivalent to shifting right \(3\) units. 2. The least positive right shift is \(d=3\), so \(g(x)=\sin\left(\frac{\pi}{2}(x-3)\right)\). 3. At \(x=0\), \(g(0)=\sin\left(-\frac{3\pi}{2}\right)=1\) and \(k(0)=\cos(0)=1\). At \(x=1\), \(g(1)=\sin(-\pi)=0\) and \(k(1)=\cos\left(\frac{\pi}{2}\right)=0\). 4. Let \(u=\frac{\pi}{2}x\). By periodicity, \(\sin\left(u-\frac{3\pi}{2}\right)=\sin\left(u+\frac{\pi}{2}\right)\). 5. Apply the sine addition identity: \(\sin\left(u+\frac{\pi}{2}\right)=\sin(u)\cos\left(\frac{\pi}{2}\right)+\cos(u)\sin\left(\frac{\pi}{2}\right)=\cos(u)\). 6. Therefore, \(g(x)=\cos\left(\frac{\pi}{2}x\right)\) for every \(x\).

a) Shift left \(1\) unit or right \(3\) units. b) \(g(x)=\sin\left(\frac{\pi}{2}(x-3)\right)\), with \(d=3\) c) The values agree at \(x=0\) and \(x=1\), and the angle-addition identity shows \(g(x)=\cos\left(\frac{\pi}{2}x\right)\) for all \(x\).

- First recover the missing sine and cosine values using the quadrants. - The cosine subtraction formula contains a plus sign between its two products. - Track the negative sign from the Quadrant IV sine value.

1. Since \(\alpha\) is in Quadrant I, \(\sin(\alpha)=\frac{5}{13}\). 2. Since \(\beta\) is in Quadrant IV, \(\cos(\beta)=\frac{3}{5}\). 3. Use \(\cos(\alpha-\beta)=\cos(\alpha)\cos(\beta)+\sin(\alpha)\sin(\beta)\). 4. Thus, \(\cos(\alpha-\beta)=\frac{12}{13}\cdot\frac{3}{5}+\frac{5}{13}\cdot\left(-\frac{4}{5}\right)=\frac{16}{65}\).

\(\frac{16}{65}\)

- Expand the shifted sine using an addition identity. - Evaluate the special-angle coefficients exactly. - Compare the coefficient of \(\sin(x)\) on both sides.

1. Expand the left side: \(\sin\left(x+\frac{\pi}{6}\right)=\sin(x)\cos\left(\frac{\pi}{6}\right)+\cos(x)\sin\left(\frac{\pi}{6}\right)\). 2. Substitute the exact values to get \(\frac{\sqrt{3}}{2}\sin(x)+\frac{1}{2}\cos(x)\). 3. Therefore, \(k=\frac{\sqrt{3}}{2}\).

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

- Compare the cosine addition and cosine subtraction formulas. - The sign in the angle and the sign between the two products do not match in the same way as they do for sine. - After correcting the formula, substitute exact values at \(\frac{\pi}{3}\).

1. The cosine subtraction identity is \(\cos(A-B)=\cos(A)\cos(B)+\sin(A)\sin(B)\), so the middle sign should be plus. 2. Therefore, \(\cos\left(x-\frac{\pi}{3}\right)=\frac{1}{2}\cos(x)+\frac{\sqrt{3}}{2}\sin(x)\).

The sign is incorrect. The correct expansion is \(\frac{1}{2}\cos(x)+\frac{\sqrt{3}}{2}\sin(x)\).

- Expand both compound-angle terms before combining them. - Compare the two middle terms after expansion. - Look for cancellation rather than evaluating any angles.

1. Expand both terms: \(\sin(x+y)=\sin(x)\cos(y)+\cos(x)\sin(y)\) and \(\sin(x-y)=\sin(x)\cos(y)-\cos(x)\sin(y)\). 2. Add the expansions. The \(\cos(x)\sin(y)\) terms cancel. 3. The result is \(2\sin(x)\cos(y)\).

\(2\sin(x)\cos(y)\)

- Compare the locations of peaks, zeros, and troughs of the two curves. - For the algebraic part, expand only the shifted sine. - Use exact values at \(\frac{\pi}{4}\).

1. The two graphs coincide, suggesting that \(f(x)=g(x)\) for every \(x\). 2. Expand \(f(x)\): \(\sin\left(x+\frac{\pi}{4}\right)=\sin(x)\cos\left(\frac{\pi}{4}\right)+\cos(x)\sin\left(\frac{\pi}{4}\right)\). 3. Since both special-angle values equal \(\frac{\sqrt{2}}{2}\), \(f(x)=\frac{\sqrt{2}}{2}\sin(x)+\frac{\sqrt{2}}{2}\cos(x)=g(x)\).

a) The graphs coincide, so they suggest \(f(x)=g(x)\) for every \(x\). b) \(\sin\left(x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\bigl(\sin(x)+\cos(x)\bigr)\).

- If two expressions are identical, their graphs must coincide everywhere. - In the sine subtraction formula, track which special-angle value multiplies \(\sin(x)\) and which multiplies \(\cos(x)\). - Compare the exact values of sine and cosine at \(\frac{\pi}{3}\).

1. The graphs do not coincide, so the claimed identity is false. 2. Use \(\sin(A-B)=\sin(A)\cos(B)-\cos(A)\sin(B)\). 3. Therefore, \(\sin\left(x-\frac{\pi}{3}\right)=\frac{1}{2}\sin(x)-\frac{\sqrt{3}}{2}\cos(x)\). 4. Amina interchanged the coefficients from \(\cos\left(\frac{\pi}{3}\right)\) and \(\sin\left(\frac{\pi}{3}\right)\).

a) The graph shows that the two expressions are not identical. b) \(\sin\left(x-\frac{\pi}{3}\right)=\frac{1}{2}\sin(x)-\frac{\sqrt{3}}{2}\cos(x)\).

- Add the two marked central angles to determine \(\angle AOP\). - Keep the sine-addition and cosine-addition formulas separate. - On the unit circle, the horizontal coordinate is cosine and the vertical coordinate is sine.

1. The diagram shows adjacent central angles of \(45^\circ\) and \(30^\circ\), so \(\angle AOP=75^\circ\). 2. Use the sine addition identity: \(\sin(75^\circ)=\sin(45^\circ)\cos(30^\circ)+\cos(45^\circ)\sin(30^\circ)=\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2}\cdot\frac{1}{2}=\frac{\sqrt{6}+\sqrt{2}}{4}\). 3. Use the cosine addition identity: \(\cos(75^\circ)=\cos(45^\circ)\cos(30^\circ)-\sin(45^\circ)\sin(30^\circ)=\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}\cdot\frac{1}{2}=\frac{\sqrt{6}-\sqrt{2}}{4}\). 4. On the unit circle, \(P=(\cos(75^\circ),\sin(75^\circ))\). 5. Therefore, \(P=\left(\frac{\sqrt{6}-\sqrt{2}}{4},\frac{\sqrt{6}+\sqrt{2}}{4}\right)\).

a) \(75^\circ\) b) \(\sin(75^\circ)=\frac{\sqrt{6}+\sqrt{2}}{4}\) and \(\cos(75^\circ)=\frac{\sqrt{6}-\sqrt{2}}{4}\) c) \(P=\left(\frac{\sqrt{6}-\sqrt{2}}{4},\frac{\sqrt{6}+\sqrt{2}}{4}\right)\)

- Begin by writing tangent as a quotient of sine and cosine at the difference angle. - Expand both subtraction expressions before trying to rewrite them in terms of tangent. - For the exact-value part, express \(15^\circ\) as a difference of two familiar special angles.

1. Write \(\tan(A-B)=\frac{\sin(A-B)}{\cos(A-B)}\). 2. Expand the numerator and denominator: \(\tan(A-B)=\frac{\sin(A)\cos(B)-\cos(A)\sin(B)}{\cos(A)\cos(B)+\sin(A)\sin(B)}\). 3. Divide the numerator and denominator by \(\cos(A)\cos(B)\). This gives \(\tan(A-B)=\frac{\tan(A)-\tan(B)}{1+\tan(A)\tan(B)}\), wherever the expressions are defined. 4. Since \(15^\circ=45^\circ-30^\circ\), substitute \(\tan(45^\circ)=1\) and \(\tan(30^\circ)=\frac{1}{\sqrt{3}}\). 5. Simplifying gives \(\tan(15^\circ)=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}=2-\sqrt{3}\).

a) \(\tan(A-B)=\frac{\tan(A)-\tan(B)}{1+\tan(A)\tan(B)}\) b) \(\tan(15^\circ)=2-\sqrt{3}\)

- Expand \(\cos(x-b)\) and \(\cos(-x-b)\). - For an identity to hold for every \(x\), the coefficients of the variable trigonometric terms must satisfy the required conditions. - Test odd symmetry by adding \(f(-x)\) and \(f(x)\).

1. Even symmetry requires \(f(-x)=f(x)\). Using angle identities, \(f(-x)-f(x)=-2\sin(b)\sin(x)\). 2. This expression is \(0\) for every \(x\) exactly when \(\sin(b)=0\), so \(b=k\pi\) for some integer \(k\). The vertical shift \(a\) may be any nonzero real number. 3. Odd symmetry would require \(f(-x)+f(x)=0\) for every \(x\). But \(f(-x)+f(x)=2a+2\cos(b)\cos(x)\). 4. For this expression to vanish for every \(x\), both \(a=0\) and \(\cos(b)=0\) would be necessary. Since \(a\ne0\), \(f\) cannot be odd.

Even: \((a, k\pi)\), where \(a\in\mathbb{R}\setminus\{0\}\) and \(k\in\mathbb{Z}\) Odd: impossible when \(a\ne0\)

- Use the amplitude formula for a sine-cosine combination. - Match coefficients after expanding \(A\sin(x+\alpha)\). - The sine function reaches a maximum when its argument is \(\frac{\pi}{2}\). - Relate the phase angle to the coefficient ratio with tangent.

1. The amplitude of \(p\sin(x)+q\cos(x)\) is \(\sqrt{p^2+q^2}\). Thus the maximum is \(\sqrt{t^2+11}\), and the minimum is \(-\sqrt{t^2+11}\). 2. For \(t=\sqrt{5}\), \(A=\sqrt{5+11}=4\). Matching coefficients gives \(\cos(\alpha)=\frac{\sqrt{5}}{4}\) and \(\sin(\alpha)=\frac{\sqrt{11}}{4}\). Both are positive, so \(\alpha\approx0.98\) radians. 3. A maximum of \(A\sin(x+\alpha)\) occurs when \(x+\alpha=\frac{\pi}{2}\). Requiring the first maximum at \(x=\frac{\pi}{3}\) gives \(\alpha=\frac{\pi}{6}\). 4. Since \(\tan(\alpha)=\frac{\sqrt{11}}{t}\), \(\frac{1}{\sqrt{3}}=\frac{\sqrt{11}}{t}\). Therefore, \(t=\sqrt{33}\).

a) Maximum: \(\sqrt{t^2+11}\); minimum: \(-\sqrt{t^2+11}\) b) \(A=4\), \(\alpha\approx0.98\) radians c) \(t=\sqrt{33}\)

- An even graph has a maximum or minimum on the y-axis. - Identify whether the graph matches a cosine function or its reflection. - Read the value of the function at \(x=0\). - For part b, use an angle-subtraction formula and separate the even and odd components.

1. The graph has a minimum of \(-1\) at \(x=0\), so it is \(-\cos(2x)\). 2. Since \(\sin\left(2x-\frac{\pi}{2}\right)=-\cos(2x)\), set \(2(x-c)=2x-\frac{\pi}{2}\). This gives \(c=\frac{\pi}{4}\). 3. In general, use the subtraction formula: \(\sin(b(x-c))=\sin(bx)\cos(bc)-\cos(bx)\sin(bc)\). 4. The function is even exactly when the coefficient of the odd term \(\sin(bx)\) is zero. Thus, \(\cos(bc)=0\), so \(bc=\frac{\pi}{2}+k\pi\), where \(k\) is an integer. 5. Therefore, \(c=\frac{\pi/2+k\pi}{b}\).

a) \(c=\frac{\pi}{4}\) b) \(bc=\frac{\pi}{2}+k\pi\), so \(c=\frac{\pi/2+k\pi}{b}\), where \(k \in \mathbb{Z}\)

- Expand both compound-angle factors first. - After expansion, look for a difference-of-squares structure. - Use the Pythagorean identity only after the product has been simplified.

1. Expand both sine factors: \(\sin(\alpha+\beta)=\sin(\alpha)\cos(\beta)+\cos(\alpha)\sin(\beta)\) and \(\sin(\alpha-\beta)=\sin(\alpha)\cos(\beta)-\cos(\alpha)\sin(\beta)\). 2. Multiply as a difference of squares: \(\sin^2(\alpha)\cos^2(\beta)-\cos^2(\alpha)\sin^2(\beta)\). 3. Replace \(\cos^2(\beta)\) with \(1-\sin^2(\beta)\) and \(\cos^2(\alpha)\) with \(1-\sin^2(\alpha)\). 4. The mixed terms cancel, leaving \(\sin^2(\alpha)-\sin^2(\beta)\).

\(\sin(\alpha+\beta)\sin(\alpha-\beta)=\sin^2(\alpha)-\sin^2(\beta)\)

- Expand both shifted sine expressions before comparing sides. - Group the \(\sin(x)\) terms and the \(\cos(x)\) terms separately. - For an identity, the corresponding coefficients must agree for every \(x\).

1. Expand both shifted sines. 2. The left side becomes \(\frac{\sqrt{2}}{2}(A+B)\sin(x)+\frac{\sqrt{2}}{2}(A-B)\cos(x)\). 3. Matching the sine coefficient with \(0\) gives \(A+B=0\). 4. Matching the cosine coefficient with \(2\) gives \(\frac{\sqrt{2}}{2}(A-B)=2\). 5. Solving the two equations gives \(A=\sqrt{2}\) and \(B=-\sqrt{2}\).

\(A=\sqrt{2}\) and \(B=-\sqrt{2}\)

- First simplify each bracket using angle addition and subtraction formulas. - Do not expand the outer squares until each bracket has a simpler form. - After factoring, look for a Pythagorean identity involving \(y\).

1. Expanding the sine terms gives \(\sin(x+y)+\sin(x-y)=2\sin(x)\cos(y)\). 2. Expanding the cosine terms gives \(\cos(x+y)-\cos(x-y)=-2\sin(x)\sin(y)\). 3. Squaring and adding gives \(4\sin^2(x)\cos^2(y)+4\sin^2(x)\sin^2(y)\). 4. Factor to get \(4\sin^2(x)\left(\cos^2(y)+\sin^2(y)\right)\). 5. By the Pythagorean identity, the result is \(4\sin^2(x)\).

\(4\sin^2(x)\)