Question

Find the zeros in one cycle of the given functions, and the value of \(y\) at the given value of \(x\).$$y=\sin \left(2 x+15^{\circ}\right) \quad x=15^{\circ}$$

Short answer

The zeros in one cycle are found when the argument of the sine function, \(2x + 15^\circ\), is an integer multiple of \(\pi\). The value of \(y\) at \(x=15^\circ\) is \(\frac{\sqrt{2}}{2}\).

Step-by-step solution

- Determine the zeros in one cycle

To find the zeros of the function within one cycle, set the function equal to zero and solve for x. For the sine function, this occurs when the angle in the sine function is an integer multiple of \(\pi\). The function \(y = \sin(2x + 15^\circ)\) has a period of \(\frac{\pi}{2}\) because the coefficient of \(x\) inside the sine function is 2. Solve for \(x\) when \(2x + 15^\circ = k\pi\), where \(k\) is an integer and \(x\) is restricted to \(0 \leq x < \frac{\pi}{2}\).

- Calculate the value of y at x = 15 degrees

To find the value of \(y\) when \(x = 15^\circ\), substitute \(15^\circ\) into the equation for \(x\) and solve for \(y\). The equation becomes \(y = \sin(2(15^\circ) + 15^\circ)\), which simplifies to \(y = \sin(45^\circ)\). The sine of \(45^\circ\) is \(\frac{\sqrt{2}}{2}\), so \(y = \frac{\sqrt{2}}{2}\).

Key concepts

Solving Trigonometric Equations

Trigonometric equations can appear daunting at first, but with a systematic approach, you can unlock their solutions with ease. When facing any trigonometric equation, such as finding the zeros of a sine function, your first step is to establish the basic form of the equation. In this case, the sine function is set to zero, which leads us to solve for the inner angle that makes the whole function amount to zero.

In practice, you identify the angles for which the sine value is zero. Remember that sine values are zero at integer multiples of \(\pi\). Hence, for an equation like \(y = \sin(2x + 15^\circ)\), you take the angle inside the sine function and set it to \(k\pi\), where \(k\) is an integer. When the equation includes a coefficient other than 1 for the variable (as in this case, a 2 for \(x\)), it indicates a 'stretched' or 'compressed' sine function which affects the periodicity of the zeros. The zeros occur more frequently if the sine function is 'compressed' and less frequently if it's 'stretched'.

By rearranging the equation to solve for \(x\), we obtain specific values or a range of values that satisfy the original equation. This process exemplifies how we approach solving trigonometric equations, allowing students to find specific values, like zeros, within one or multiple cycles of the trigonometric function.

Period of a Sine Function

The concept of the period of a sine function is fundamental to understanding the behavior of sinusoidal patterns. The period refers to the length of one complete cycle of the sine curve on a graph. It is the distance on the \(x\)-axis between two consecutive points where the function begins to repeat its pattern.

The standard sine function \(y = \sin(x)\) has a period of \(2\pi\) radians, which translates to 360 degrees. But what happens when the sine function is modified? In the given exercise, we have \(y = \sin(2x + 15^\circ)\). Here, the presence of the coefficient 2 in front of \(x\) indicates a change in period. To find the new period, we divide the standard period \(2\pi\) by the coefficient of \(x\), giving us \(\frac{\pi}{2}\).

Understanding periodicity is crucial when solving trigonometric equations, as it helps to predict where the function will be zero (its zeros) and the values it will take at specific angles. This comprehension also aids in sketching the function and predicting its behavior without necessarily computing every single value.

Sine Function Values

The values of the sine function are fundamental to trigonometry. Their importance is highlighted when finding specific points on a trigonometric graph or solving equations. At certain standard angles, such as \(0^\circ\), \(30^\circ\), \(45^\circ\), \(60^\circ\), and \(90^\circ\), the sine values are well-known and often memorized.

For example, \(\sin(45^\circ)\) is a commonly known value of \(\frac{\sqrt{2}}{2}\). This knowledge comes in handy, as demonstrated in the exercise where we substitute \(x = 15^\circ\) into the given function \(y = \sin(2x + 15^\circ)\), which results in \(y = \sin(45^\circ)\). Recognizing these standard sine values speeds up the process and eliminates the need for calculators every time. It is equally useful for solving more complex trigonometric equations, where these values can be landmarks guiding you to the solution.

Understanding and recalling the sine values at various standard angles is key not just for solving particular problems but for a deeper grasp of how trigonometric functions behave across different quadrants in the coordinate system. The sine function starts at zero, rises to its maximum at \(\frac{\pi}{2}\) or \(90^\circ\), descends back to zero at \(\pi\) or \(180^\circ\), and then goes negative before completing its cycle and starting anew.