Let's Practice What We Just Discovered! Given: $\sin (\theta) = \cos (90^\circ - \theta$\]Select All The True Equations:a. $\cos (15^\circ) = \sin (15^\circ$\]b. $\cos (75^\circ) = \sin (15^\circ$\]c. $\cos (75^\circ) = \cos

Introduction

In our previous discussion, we explored the relationship between sine and cosine functions, particularly the identity $\sin (\theta) = \cos (90^\circ - \theta)$. This identity is a fundamental concept in trigonometry, and it has numerous applications in various fields, including mathematics, physics, and engineering. In this article, we will practice what we just discovered by applying this identity to solve some problems.

The Identity: $\sin (\theta) = \cos (90^\circ - \theta)$

The identity $\sin (\theta) = \cos (90^\circ - \theta)$ states that the sine of an angle $\theta$ is equal to the cosine of the complementary angle $90^\circ - \theta$. This means that if we know the value of $\sin (\theta)$, we can find the value of $\cos (90^\circ - \theta)$, and vice versa.

Problem 1: Selecting True Equations

We are given the following equations:

a. $\cos (15^\circ) = \sin (15^\circ)$ b. $\cos (75^\circ) = \sin (15^\circ)$ c. $\cos (75^\circ) = \cos (15^\circ)$

Using the identity $\sin (\theta) = \cos (90^\circ - \theta)$, we can determine which of these equations are true.

Equation a: $\cos (15^\circ) = \sin (15^\circ)$

Let's analyze equation a. We know that $\sin (15^\circ) = \cos (90^\circ - 15^\circ) = \cos (75^\circ)$. Therefore, equation a is equivalent to $\cos (15^\circ) = \cos (75^\circ)$. Since the cosine function is periodic with a period of $360^\circ$, we can write $\cos (15^\circ) = \cos (75^\circ + 360^\circ k)$, where $k$ is an integer. However, this does not necessarily mean that $\cos (15^\circ) = \sin (15^\circ)$. In fact, $\cos (15^\circ) \neq \sin (15^\circ)$, since the sine and cosine functions have different values at the same angle.

Equation b: $\cos (75^\circ) = \sin (15^\circ)$

Now, let's analyze equation b. We know that $\sin (15^\circ) = \cos (90^\circ - 15^\circ) = \cos (75^\circ)$. Therefore, equation b is equivalent to $\cos (75^\circ) = \cos (75^\circ)$, which is true.

Equation c: $\cos (75^\circ) = \cos (15^\circ)$

Finally, let's analyze equation c. We know that $\cos (75^\circ) = \cos (15^\circ + 60^\circ)$. Using the angle addition formula for cosine, we can write $\cos (75^\circ) = \cos (15^\circ) \cos (60^\circ) - \sin (15^\circ) \sin (60^\circ)$. Since $\cos (60^\circ) = \frac{1}{2}$ and $\sin (60^\circ) = \frac{\sqrt{3}}{2}$, we can simplify this expression to $\cos (75^\circ) = \frac{1}{2} \cos (15^\circ) - \frac{\sqrt{3}}{2} \sin (15^\circ)$. Therefore, equation c is not necessarily true.

Conclusion

In conclusion, we have analyzed three equations using the identity $\sin (\theta) = \cos (90^\circ - \theta)$. We found that equation b is true, while equations a and c are not necessarily true. This exercise demonstrates the importance of understanding the relationships between trigonometric functions and how to apply them to solve problems.

Practice Problems

Here are some practice problems to help you reinforce your understanding of the identity $\sin (\theta) = \cos (90^\circ - \theta)$:

1. Prove that $\sin (30^\circ) = \cos (60^\circ)$ using the identity $\sin (\theta) = \cos (90^\circ - \theta)$.
2. Find the value of $\cos (45^\circ)$ using the identity $\sin (\theta) = \cos (90^\circ - \theta)$.
3. Prove that $\sin (60^\circ) = \cos (30^\circ)$ using the identity $\sin (\theta) = \cos (90^\circ - \theta)$.

Solutions

Here are the solutions to the practice problems:

1. We know that $\sin (30^\circ) = \cos (90^\circ - 30^\circ) = \cos (60^\circ)$. Therefore, $\sin (30^\circ) = \cos (60^\circ)$.
2. We know that $\sin (45^\circ) = \cos (90^\circ - 45^\circ) = \cos (45^\circ)$. Therefore, $\cos (45^\circ) = \frac{1}{\sqrt{2}}$.
3. We know that $\sin (60^\circ) = \cos (90^\circ - 60^\circ) = \cos (30^\circ)$. Therefore, $\sin (60^\circ) = \cos (30^\circ)$.

Final Thoughts

Q: What is the identity $\sin (\theta) = \cos (90^\circ - \theta)$?

A: The identity $\sin (\theta) = \cos (90^\circ - \theta)$ states that the sine of an angle $\theta$ is equal to the cosine of the complementary angle $90^\circ - \theta$. This means that if we know the value of $\sin (\theta)$, we can find the value of $\cos (90^\circ - \theta)$, and vice versa.

Q: How do I use the identity $\sin (\theta) = \cos (90^\circ - \theta)$ to solve problems?

A: To use the identity $\sin (\theta) = \cos (90^\circ - \theta)$ to solve problems, you need to:

1. Identify the angle $\theta$ and its complementary angle $90^\circ - \theta$.
2. Use the identity to rewrite the sine or cosine function in terms of the other function.
3. Simplify the expression using trigonometric identities and formulas.

Q: What are some common applications of the identity $\sin (\theta) = \cos (90^\circ - \theta)$?

A: The identity $\sin (\theta) = \cos (90^\circ - \theta)$ has numerous applications in various fields, including:

1. Trigonometry: The identity is used to solve trigonometric equations and identities.
2. Physics: The identity is used to describe the motion of objects in terms of sine and cosine functions.
3. Engineering: The identity is used to design and analyze systems that involve periodic motion, such as oscillators and filters.
4. Computer Science: The identity is used in algorithms and data structures that involve trigonometric functions.

Q: Can I use the identity $\sin (\theta) = \cos (90^\circ - \theta)$ to solve problems with angles greater than $90^\circ$?

A: Yes, you can use the identity $\sin (\theta) = \cos (90^\circ - \theta)$ to solve problems with angles greater than $90^\circ$. However, you need to be careful when dealing with angles in the second and third quadrants, where the sine and cosine functions have different signs.

Q: How do I prove the identity $\sin (\theta) = \cos (90^\circ - \theta)$?

A: To prove the identity $\sin (\theta) = \cos (90^\circ - \theta)$, you can use the following steps:

1. Draw a right triangle with an angle $\theta$ and a complementary angle $90^\circ - \theta$.
2. Use the definitions of sine and cosine to write expressions for the sine and cosine functions in terms of the triangle.
3. Simplify the expressions using trigonometric identities and formulas.
4. Show that the two expressions are equal.

Q: Can I use the identity $\sin (\theta) = \cos (90^\circ - \theta)$ to solve problems with complex numbers?

A: Yes, you can use the identity $\sin (\theta) = \cos (90^\circ - \theta)$ to solve problems with complex numbers. However, you need to be careful when dealing with complex numbers, as the sine and cosine functions have different properties in the complex plane.

Q: How do I apply the identity $\sin (\theta) = \cos (90^\circ - \theta)$ to solve problems with parametric equations?

A: To apply the identity $\sin (\theta) = \cos (90^\circ - \theta)$ to solve problems with parametric equations, you need to:

1. Identify the parametric equations and the angle $\theta$.
2. Use the identity to rewrite the sine or cosine function in terms of the other function.
3. Simplify the expression using trigonometric identities and formulas.
4. Solve the resulting equation.

Conclusion

In conclusion, the identity $\sin (\theta) = \cos (90^\circ - \theta)$ is a powerful tool for solving trigonometric problems. By understanding this identity and how to apply it, you can solve a wide range of problems in mathematics, physics, and engineering. Remember to practice regularly to reinforce your understanding of this important concept.