Problem 47 - Contemporary Precalculus

Question

Use graphs to determine whether the equation could possibly be an identity or definitely is not an identity. $$\cos \left(\frac{\pi}{2}+t\right)=-\sin t$$

Step-by-Step Solution

Answer

Answer: Yes, the equation is an identity.

Step 1: Left Side Graph creation

Recall the formula for the cosine function of a sum of two angles: $$\cos (a + b) = \cos a \cos b - \sin a \sin b$$ In this case, \(a = \frac{\pi}{2}\) and \(b = t\). Using this formula, we can rewrite the left side as: $$\cos \left(\frac{\pi}{2}+t\right) = \cos \left(\frac{\pi}{2}\right) \cos t - \sin \left(\frac{\pi}{2}\right) \sin t$$ Since \(\cos \left(\frac{\pi}{2}\right) = 0\) and \(\sin \left(\frac{\pi}{2}\right) = 1\), the left side simplifies to: $$0 \cos t - 1 \sin t= -\sin t$$ So we need to graph the function \(y = -\sin t\) for the left side of the equation.

Step 2: Right Side Graph creation

Now, let's analyze the right side of the equation. The given equation is: $$-\sin t$$ This is simply the negation of the basic sine function \(y = \sin t\), so we need to graph the function \(y = -\sin t\) for the right side of the equation.

Step 3: Comparing Graphs

We need to compare the two graphs that we have created and check if they are identical. The left side gave us the graph of \(y = -\sin t\), and the right side gave us the same graph, \(y = -\sin t\). Since the two graphs are identical, this means the original equation $$\cos \left(\frac{\pi}{2}+t\right)=-\sin t$$ is an identity.

Key Concepts

Graphical Analysis

Graphical analysis is a powerful method to visually understand whether two mathematical expressions are equivalent. To compare graphs, consider these points:

• Graph both sides of the equation separately.
• Observe the shape, position, and periodicity of each graph.
• If the graphs overlap completely and are identical across all values, the equation is an identity.

In our original exercise, we used graphical analysis to check the equation \(\cos \left(\frac{\pi}{2}+t\right)=-\sin t\).
We graphed the left and right sides separately, leading to the understanding that both graphs result in the same function \(y = -\sin t\). This overlap confirms the equation is an identity. Graphical analysis gives a clear visual confirmation beyond algebraic solutions.

Cosine Function

The cosine function is a crucial part of trigonometry and has specific properties that make it unique. Here are the vital aspects:

• It's an even function, meaning \(\cos(-t) = \cos(t)\).
• The cosine function has a period of \(2\pi\), meaning it repeats its values every \(2\pi\) radians.
• Key values: \(\cos(0) = 1\), \(\cos(\frac{\pi}{2}) = 0\), \(\cos(\pi) = -1\).

In the exercise, we used the identity for cosine of a sum, \(\cos(a + b) = \cos a \cos b - \sin a \sin b\), with \(a = \frac{\pi}{2}\) and \(b = t\).
The calculation simplifies to \(\cos(t) = 0\cos(t) - 1\sin(t) = -\sin(t)\). Thus, realizing the cosine of a shift by \(\frac{\pi}{2}\) matches the negative sine function is a magnificent identity of Euler's work.

Sine Function

The sine function, like cosine, is foundational in trigonometry. Some important characteristics include:

• It's an odd function, satisfying \(\sin(-t) = -\sin(t)\).
• The sine function is periodic with a period of \(2\pi\) and has values ranging between \(-1\) and \(1\).
• Key values: \(\sin(0) = 0\), \(\sin(\frac{\pi}{2}) = 1\), \(\sin(\pi) = 0\).

In the given problem, the right side analysis involved seeing \(y = -\sin t\), which is just the reflection of \(y = \sin t\) across the x-axis.
This illustrates the effect of negative multiplication which inverts wave peaks and valleys. Understanding sine's behavior gives intuition into wave patterns and its use in oscillations.