Free Math Worksheets

- 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.

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

- Compare corresponding maximum points and zeros. - Recall how left and right shifts appear inside a function. - Determine the period of \(f(x)=\sin\left(\frac{\pi}{2}x\right)\). - Matching two values is only a check; use \(\sin\left(u+\frac{\pi}{2}\right)=\cos(u)\) for a general argument.

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. For every \(x\), \(g(x)=\sin\left(\frac{\pi}{2}x-\frac{3\pi}{2}\right)=\sin\left(\frac{\pi}{2}x+\frac{\pi}{2}\right)=\cos\left(\frac{\pi}{2}x\right)\), using periodicity and \(\sin\left(u+\frac{\pi}{2}\right)=\cos(u)\).

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 identity shows \(g(x)=\cos\left(\frac{\pi}{2}x\right)\) for all \(x\).

- 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 \(a\ne0\)

- 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 \(k \in \mathbb{Z}\)