Problem 133 - Algebra and Trigonometry Real Mathematics, Real People

Question

Find the exact value of each function for the given angle for \(f(\theta)=\sin \theta\) and \(g(\theta)=\cos \theta .\) Do not use a calculator. (a) \((f+g)(\theta)\) (b) \((g-f)(\theta)\) (c) \([g(\theta)]^{2}\) (d) \((f g)(\theta)\) (e) \(f(2 \theta)\) (f) \(g(-\boldsymbol{\theta})\) $$\theta=7 \pi / 2$$

Step-by-Step Solution

Answer

(a) 1, (b) -1, (c) 0, (d) 0, (e) 0, (f) 0

Step 1: Angles in the Unit Circle

Since sine and cosine have a period of \(2\pi\), \(7\pi / 2\) can be simplified to \(\pi / 2\), because \(7\pi / 2 = 2\pi*3 + \pi / 2 = \pi / 2\). Therefore, we can evaluate all functions at \(\pi / 2\). We know that \(\sin(\pi / 2) = 1\) and \(\cos(\pi / 2) = 0\).

Step 2: Solving (a): (f+g)(θ)

(f+g)(θ) = f(θ) + g(θ) = \(\sin(\pi / 2) + \cos(\pi / 2) = 1 + 0 = 1\).

Step 3: Solving (b): (g-f)(θ)

(g-f)(θ) = g(θ) - f(θ) = \(\cos(\pi / 2) - \sin(\pi / 2) = 0 - 1 = -1\).

Step 4: Solving (c): [g(θ)]²

[g(θ)]² = g(θ)*g(θ) = \(\cos^2(\pi / 2)\) = \(0^2 = 0\).

Step 5: Solving (d): (fg)(θ)

(fg)(θ) = f(θ)*g(θ) = \(\sin(\pi / 2) * \cos(\pi / 2) = 1 * 0 = 0\).

Step 6: Solving (e): f(2θ)

f(2θ) = \(\sin(2 * \pi / 2) = \sin(\pi) = 0\). Note that sin function becomes zero at π.

Step 7: Solving (f): g(-θ)

g(-θ) = \(\cos(-\pi / 2)\). Since cos(x) is an even function, \(\cos(-\pi / 2) = \cos(\pi / 2) = 0\).

Key Concepts

Sine and Cosine

Sine and cosine are fundamental trigonometric functions that are quite helpful in various mathematical and practical applications.They relate the angles and sides of right triangles, and also help in understanding the properties of waves, circles, and oscillations. Specifically, they measure the vertical and horizontal components related to the unit circle.

• Sine function (sin): It gives the length of the side opposite the angle, over the hypotenuse, in a right-angled triangle. On the unit circle, it represents the vertical coordinate of a point.
• Cosine function (cos): It describes the length of the adjacent side over the hypotenuse. On the unit circle, it corresponds to the horizontal coordinate.

Both of these functions are periodic with a period of \(2\pi\). This means they repeat their values every \(2\pi\), or every full circle.These periodic properties allow us to simplify complex angles easily and find sine and cosine values for angles like \(\frac{7\pi}{2}\) by understanding their behavior over the unit circle.

Unit Circle

The unit circle is a fantastic tool to understand trigonometric functions.It's a circle with radius 1, centered at the origin of the coordinate plane.The circle helps in visualizing how the sine and cosine values relate to angles.

• The x-coordinate on the unit circle gives the cosine value for a particular angle.
• The y-coordinate gives the sine value.
• Common angles, like \(\frac{\pi}{2}\), \(\pi\), \(3\pi/2\), and \(2\pi\), correspond to points on the circle that are easy to remember.

For our exercise, knowing the coordinates of \(\frac{\pi}{2}\) is crucial: the point is (0, 1) on the unit circle.Here, cosine is 0 and sine is 1, which aligns perfectly with how the unit circle works.

Angle Simplification

When dealing with angles, especially in problems like this one, angle simplification becomes vital.Since sine and cosine functions have a period of \(2\pi\), we use the fact that after each full cycle, the functions start repeating themselves.For any given angle \(\theta\), adding or subtracting multiples of \(2\pi\) doesn't change its sine or cosine.Thus, with \(\theta = \frac{7\pi}{2}\), we simplify by recognizing that it is \(\frac{3 \times 2\pi}{2} + \frac{\pi}{2} = \pi/2\).This simplification process not only makes calculations easier but also leverages the periodic nature of trigonometric functions, ensuring we can solve the problem efficiently without confusion.Practice will help students become faster and more confident in simplifying and using these angles.