📐 Level 1 - Topic 8: The Pythagorean Theorem and Trigonometry 📐🔗

1) The Pythagorean Theorem - A Cornerstone of Geometry 📐

The Pythagorean Theorem is one of the most famous and fundamental theorems in geometry. It describes a crucial relationship between the sides of a right triangle. You've likely encountered it before, but let's review it and see its powerful connection to trigonometry.

Pythagorean Theorem: In a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides (the legs).

Formula of the Pythagorean Theorem:

If \( a \) and \( b \) are the lengths of the legs of a right triangle, and \( c \) is the length of the hypotenuse, then:

\( a^2 + b^2 = c^2 \)

Equivalently, \( c = \sqrt{a^2 + b^2} \) (to find the hypotenuse if you know the legs)

Applications of the Pythagorean Theorem:

Finding Unknown Side Lengths: If you know the lengths of two sides in a right triangle, you can always find the length of the third side using the Pythagorean Theorem.
Checking if a Triangle is a Right Triangle: If you know the side lengths of a triangle, you can check if it's a right triangle by seeing if the Pythagorean Theorem holds true.
Distance Calculations: The theorem is used in distance formulas in coordinate geometry and in various applications involving distances in 2D and 3D space.
Basis for Trigonometry: As we'll see, it's intimately linked to fundamental trigonometric identities.

2) Pythagorean Theorem and Trigonometric Ratios - The Link 🔗

Now, let's explore how the Pythagorean Theorem connects to the trigonometric ratios we've been learning about (sine, cosine, tangent, etc.). Consider a right triangle with an acute angle \( \theta \). Let's label the sides as:

Opposite side = \( O \)
Adjacent side = \( A \)
Hypotenuse = \( H \)

According to the Pythagorean Theorem: \( O^2 + A^2 = H^2 \).

We also know the definitions of sine and cosine:

\( \sin(\theta) = \frac{O}{H} \)
\( \cos(\theta) = \frac{A}{H} \)

Let's see if we can combine these relationships. We want to somehow relate \( \sin(\theta) \) and \( \cos(\theta) \) using the Pythagorean Theorem.

3) The Pythagorean Identity - A Fundamental Trigonometric Identity 🌟

Let's start with the Pythagorean Theorem: \( O^2 + A^2 = H^2 \). We want to involve \( \sin(\theta) = \frac{O}{H} \) and \( \cos(\theta) = \frac{A}{H} \). A clever trick is to divide the entire Pythagorean equation by \( H^2 \) (assuming \( H \neq 0 \), which is always true for a triangle):

Divide both sides of \( O^2 + A^2 = H^2 \) by \( H^2 \):

\( \frac{O^2}{H^2} + \frac{A^2}{H^2} = \frac{H^2}{H^2} \)

\( \left(\frac{O}{H}\right)^2 + \left(\frac{A}{H}\right)^2 = 1 \)

Now, recognize that \( \frac{O}{H} = \sin(\theta) \) and \( \frac{A}{H} = \cos(\theta) \). Substituting these in, we get:

\( (\sin(\theta))^2 + (\cos(\theta))^2 = 1 \)

Pythagorean Identity: For any angle \( \theta \), the following identity holds true:

\( \sin^2(\theta) + \cos^2(\theta) = 1 \)

*Note: \( \sin^2(\theta) \) means \( (\sin(\theta))^2 \) and \( \cos^2(\theta) \) means \( (\cos(\theta))^2 \).*

This is called the Pythagorean Identity, and it is one of the most fundamental and important identities in all of trigonometry! It's true for *any* angle \( \theta \), not just acute angles in a right triangle (though we derived it from a right triangle context initially).

Key Takeaway: The Pythagorean Identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \) is a direct consequence of the Pythagorean Theorem and the definitions of sine and cosine in a right triangle. It links sine and cosine together in a fundamental way.

4) Using the Pythagorean Identity - Examples and Applications 🎯

Example 1: Finding \( \sin(\theta) \) when \( \cos(\theta) \) is known

If \( \cos(\theta) = \frac{3}{5} \) and \( \theta \) is an acute angle, find \( \sin(\theta) \).

Solution:

Use the Pythagorean Identity: \( \sin^2(\theta) + \cos^2(\theta) = 1 \).
Substitute \( \cos(\theta) = \frac{3}{5} \): \( \sin^2(\theta) + \left(\frac{3}{5}\right)^2 = 1 \)
\( \sin^2(\theta) + \frac{9}{25} = 1 \)
\( \sin^2(\theta) = 1 - \frac{9}{25} = \frac{25}{25} - \frac{9}{25} = \frac{16}{25} \)
\( \sin(\theta) = \pm \sqrt{\frac{16}{25}} = \pm \frac{4}{5} \).
Since \( \theta \) is acute, \( \sin(\theta) \) must be positive. So, \( \sin(\theta) = \frac{4}{5} \).

Example 2: Finding \( \cos(\theta) \) when \( \sin(\theta) \) is known

If \( \sin(\theta) = -\frac{1}{2} \) and \( \theta \) is in the third quadrant, find \( \cos(\theta) \).

Solution:

Use the Pythagorean Identity: \( \sin^2(\theta) + \cos^2(\theta) = 1 \).
Substitute \( \sin(\theta) = -\frac{1}{2} \): \( \left(-\frac{1}{2}\right)^2 + \cos^2(\theta) = 1 \)
\( \frac{1}{4} + \cos^2(\theta) = 1 \)
\( \cos^2(\theta) = 1 - \frac{1}{4} = \frac{4}{4} - \frac{1}{4} = \frac{3}{4} \)
\( \cos(\theta) = \pm \sqrt{\frac{3}{4}} = \pm \frac{\sqrt{3}}{2} \).
Since \( \theta \) is in the third quadrant, cosine is negative. So, \( \cos(\theta) = -\frac{\sqrt{3}}{2} \).

Example 3: Simplifying Trigonometric Expressions

Simplify the expression: \( (1 - \cos^2(\theta)) \).

Solution:

Start with the Pythagorean Identity: \( \sin^2(\theta) + \cos^2(\theta) = 1 \).
Rearrange the identity to isolate \( \sin^2(\theta) \) on one side: Subtract \( \cos^2(\theta) \) from both sides.
\( \sin^2(\theta) = 1 - \cos^2(\theta) \)
Therefore, \( (1 - \cos^2(\theta)) \) is simply equal to \( \sin^2(\theta) \).

5) Practice Questions 🎯

5.1 Fundamental – Applying Pythagorean Identity

1. State the Pythagorean Theorem in words and give its formula.

2. State the Pythagorean Identity. Explain how it is derived from the Pythagorean Theorem.

3. If \( \sin(\theta) = \frac{5}{13} \) and \( \theta \) is acute, find \( \cos(\theta) \) using the Pythagorean Identity.

4. If \( \cos(\theta) = -\frac{12}{13} \) and \( \theta \) is in the second quadrant, find \( \sin(\theta) \) using the Pythagorean Identity.

5. Simplify the expression: \( \sin^2(\theta) + \cos^2(\theta) + 5 \).

6. Simplify the expression: \( 1 - \sin^2(\theta) \).

7. Simplify the expression: \( 7\cos^2(\theta) + 7\sin^2(\theta) \).

8. If \( \sin(\theta) = 0.6 \) and \( \cos(\theta) = 0.8 \), verify if the Pythagorean Identity holds true for this case.

9. If \( \sin(\theta) = x \), express \( \cos^2(\theta) \) in terms of \( x \).

10. True or False: For any angle \( \theta \), \( \sin^2(\theta) + \cos^2(\theta) \) is always equal to 1. Explain.

5.2 Challenging – Problem Solving & Extensions 💪🚀

1. Given that \( \tan(\theta) = \frac{3}{4} \) and \( \theta \) is in the first quadrant, find \( \sin(\theta) \) and \( \cos(\theta) \) using the Pythagorean Identity and the fact that \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \). (Hint: Divide Pythagorean Identity by \( \cos^2(\theta) \)).

2. Simplify the expression: \( \frac{\sin^2(\theta) + \cos^2(\theta)}{\cos^2(\theta)} \).

3. Simplify the expression: \( \sin^4(\theta) + \sin^2(\theta)\cos^2(\theta) \).

4. Is it possible for both \( \sin(\theta) \) and \( \cos(\theta) \) to be simultaneously greater than 1? Explain using the Pythagorean Identity.

5. (Conceptual) Consider a right triangle inscribed in a unit circle, with one vertex at the origin and another on the unit circle at angle \( \theta \). Explain how the Pythagorean Theorem in this triangle directly leads to the Pythagorean Identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \). (Relate the side lengths to \( \sin(\theta) \) and \( \cos(\theta) \)).

6) Summary 🎉

Pythagorean Theorem: In right triangles, \( a^2 + b^2 = c^2 \) where \( a, b \) are legs, \( c \) is hypotenuse.
Link to Trigonometry: Pythagorean Theorem relates to trigonometric ratios sine and cosine.
Pythagorean Identity: \( \sin^2(\theta) + \cos^2(\theta) = 1 \). Fundamental identity in trigonometry, derived from Pythagorean Theorem. True for all angles \( \theta \).
Using the Identity: Can find \( \sin(\theta) \) if \( \cos(\theta) \) is known (and quadrant of \( \theta \)), or vice-versa. Used to simplify trigonometric expressions.
Applications: Foundation for many trigonometric manipulations and problem-solving.

Congratulations! You've now discovered the powerful connection between the Pythagorean Theorem and trigonometry, and learned about the fundamental Pythagorean Identity! This identity will be incredibly useful as you progress further in trigonometry. Practice using it to solve problems and simplify expressions. You're building a strong trigonometric foundation! 📐🔗🌟