Question

a) Copy each equation. Fill in the missing values to make the equation true. i) \(4 \sin \left(x-30^{\circ}\right)=4 \cos (x-\square)\) ii) \(2 \sin \left(x-\frac{\pi}{4}\right)=2 \cos (x-\square)\) iii) \(-3 \cos \left(x-\frac{\pi}{2}\right)=3 \sin (x+\square)\) iv) \(\cos (-2 x+6 \pi)=\sin 2(x+\square)\) b) Choose one of the equations in part a and explain how you got your answer.

Short answer

i) 60^\text{degree} or \frac\text{\text)\} ii) 60^\text{degree} iii) iii) iii) iii) iii) iii) -60^\texttheta}\big. iii) iii) iii) - \text)}.

Step-by-step solution

- Use Trigonometric Identities

Recall that \(\textbf{sin}(\theta) = \textbf{cos}\big(\frac{\textbf{\text_pi}}{2} - \theta\big)\). This identity will help to solve parts i) and ii).

- Solve part i)

Given: \(4 \textbf{sin}\big(x-30^{\textdegree}\big) = 4 \textbf{cos}\big(x-\text{\textbox}\big)\) To match the sine and cosine, use \(\frac{\text{\text_pi}}{2}\): \(\textbf{sin}\big(x-30^{\textdegree}\big) = \textbf{cos}\big(\frac{\text{\text_pi}}{2}-(x-30^{\textdegree})\big)\) Simplify: \ \textbf{cos}\big(\frac{\text{\text_pi}}{2}-x + 30^{\textdegree}\big)=\textbf{cos}\big(x-60^{\textdegree}\big).

- Solve part ii)

Given: \(2 \textbf{sin}\big(x-\frac{\text{\text_pi}}{4}\big) = 2 \textbf{cos}(x-\text{\textbox})\) Using the identity: \(\textbf{sin}\big(x-\frac{\text{\text_pi}}{4}\big) = \textbf{cos}\big(\frac{\text{\text_pi}}{2}-(x-\frac{\text{\text_pi}}{4})\big)\) Solve: \( \textbf{cos}\big(\frac{\text{\text_pi}}{2}-x + \frac{\text{\text_pi}}{4}\big) = \textbf{cos}\big(x -\frac{\text\text{\textbox}}{2}\big).\)

- Solve part iii)

Given: \(-3 \textbf{cos}\big(x-\frac{\text{\text_pi}}{2}\big) = 3 \textbf{sin} ((x+\textbox)\big)\) Here: \(\textbf{cos}\big(x-\frac{\text{\text_pi}}{2}\big) = - \textbf{sin}\big(x-\frac{\text{\text_pi}}{2}\big)\) which means: \(\textbf{cos}\big(x-\frac{\text{\text_pi}}{2}\big) = \textbf{sin}\big(x-\frac{\text{\text_pi}}{2}+\text{\textbox}\big)\)

- Solve part iv)

Given: \(\textbf{cos}(-2x+6\text{\textpi}) = \textbf{sin}\big(2(x+\textbox)\big)\). Use: \(\textbf{cos}(-x) = \textbf{cos}\big(2(x+\textword\big)\).

Key concepts

Sin-Cos Relationship

In trigonometry, the relationship between sine and cosine is fundamental. Sine and cosine functions are cofunctions, meaning they are related by a phase shift. Specifically, \( \text{sin}(\theta) = \text{cos}\big(\frac{\text{\textpi}}{2} - \theta\big) \). This identity is useful when trying to express a sine function as a cosine function, or vice versa.
For example, in the problem: i) \(4 \text{sin}\big(x-30^{\textdegree}\big)=4 \text{cos}(x-\text{\textbox})\). We can use this identity to translate the sine term into a cosine term. This results in: \( \text{sin}\big(x-30^{\textdegree}\big)=\text{cos}\big(\frac{\text{\textpi}}{2}-(x-30^{\textdegree})\big)\). By simplifying further, we get \( \text{cos}\big(\frac{\text{\textpi}}{2}-x + 30^{\textdegree}\big)=\text{cos}\big(x-60^{\textdegree}\big).\). Thus, the missing value is 60 degrees.

Trigonometric Equations

Trigonometric equations involve functions of angles and are key in many areas of mathematics and science. Solving these equations often requires using identities such as the sine-cosine relationship. In another example from the problem: ii) \(2 \text{sin}\big(x-\frac{\text{\textpi}}{4}\big) = 2 \text{cos}(x-\text{\textbox}) \). To solve this, we utilize the key identity again: \( \text{sin}\big(x-\frac{\text{\textpi}}{4}\big) = \text{cos}\big(\frac{\text{\textpi}}{2}-(x-\frac{\text{\textpi}}{4})\big) \). Simplifying it to: \( \text{cos}\big(\frac{\text{\textpi}}{2}-x + \frac{\text{\textpi}}{4}\big) = \text{cos}(x - \frac{\text{\textpi}}{4}) \). Therefore, the missing value is \( \frac{\text{\textpi}}{4} \).

Angle Transformation

Angle transformations are crucial when solving trigonometric equations. This concept often requires changing the angle in a trigonometric function to make it easier to solve. For example, part iii) involves \( -3 \text{cos}\big(x-\frac{\text{\textpi}}{2}\big) = 3 \text{sin} (x+\text{\textbox}\big) \). Here, we can use the identity that relates cosine and sine with a phase shift: \( \text{cos}\big(x-\frac{\text{\textpi}}{2}\big) = -\text{sin}(x) \. By rewriting it: \( \text{cos}\big(x-\frac{\text{\textpi}}{2}\big) = \text{sin}\big(x - \frac{\text{\textbox}}{2} \big) \). Thus, the missing value in this case is \ \frac{\text{\textbox}}{2} \).

Precalculus

Precalculus lays the foundation for calculus, focusing on functions and their properties. Understanding trigonometric identities and equations is a vital part of precalculus. It's crucial to grasp these before moving into higher-level math. Part iv) of the exercise shows a more complex trigonometric equivalency: \( \text{cos}(-2x+6\text{\textpi}) = \text{sin}\big(2(x+\text{\textbox})\big) \). Using the identity for cosine of a negative angle \( \text{cos}(-x) = \text{cos}(x) \), and simplifying, we find \( \text{cos}(-2x+6\text{\textpi}) = \text{cos}(2x - 6\text{\textpi}) \). Simplify further to get: \( \text{sin}(2(x+3\text{\textpi})) \. Thus, the missing value is \ 3\text{\textpi} \).