LESSON 8: The Pythagorean Identities

The Pythagorean Identities: 

1. cos2 (θ) + sin2 (θ) = 1. Common Alternate Forms: ˆ 1 − sin2 (θ) = cos2 (θ) ˆ 1 − cos2 (θ) = sin2 (θ) 

2. 1 + tan2 (θ) = sec2 (θ), provided cos(θ) 6= 0. Common Alternate Forms: ˆ sec2 (θ) − tan2 (θ) = 1 ˆ sec2 (θ) − 1 = tan2 (θ) 

3. 1 + cot2 (θ) = csc2 (θ), provided sin(θ) 6= 0. Common Alternate Forms: ˆ csc2 (θ) − cot2 (θ) = 1 ˆ csc2 (θ) − 1 = cot2 (θ)

Trigonometric identities play an important role in not just Trigonometry, but in Calculus as well. We’ll use them in this book to find the values of the circular functions of an angle and solve equations and inequalities. In Calculus, they are needed to simplify otherwise complicated expressions. In the next example, we make good use of the Theorems 10.6 and 10.8. Example 10.3.3. Verify the following identities. Assume that all quantities are defined. 1. 1 csc(θ) = sin(θ) 2. tan(θ) = sin(θ) sec(θ) 3. (sec(θ) − tan(θ))(sec(θ) + tan(θ)) = 1 4. sec(θ) 1 − tan(θ) = 1 cos(θ) − sin(θ) 5. 6 sec(θ) tan(θ) = 3 1 − sin(θ) − 3 1 + sin(θ) 6. sin(θ) 1 − cos(θ) = 1 + cos(θ) sin(θ) Solution. In verifying identities, we typically start with the more complicated side of the equation and use known identities to transform it into the other side of the equation. 

1. To verify 1 csc(θ) = sin(θ), we start with the left side. Using csc(θ) = 1 sin(θ) , we get: 1 csc(θ) = 1 1 sin(θ) = sin(θ), which is what we were trying to prove.

2. Starting with the right hand side of tan(θ) = sin(θ) sec(θ), we use sec(θ) = 1 cos(θ) and find: sin(θ) sec(θ) = sin(θ) 1 cos(θ) = sin(θ) cos(θ) = tan(θ), where the last equality is courtesy of Theorem 10.6. 

3. Expanding the left hand side of the equation gives: (sec(θ) − tan(θ))(sec(θ) + tan(θ)) = sec2 (θ) − tan2 (θ). According to Theorem 10.8, sec2 (θ) − tan2 (θ) = 1. Putting it all together, (sec(θ) − tan(θ))(sec(θ) + tan(θ)) = sec2 (θ) − tan2 (θ) = 1. 

4. While both sides of our last identity contain fractions, the left side affords us more opportunities to use our identities.5 Substituting sec(θ) = 1 cos(θ) and tan(θ) = sin(θ) cos(θ) , we get: sec(θ) 1 − tan(θ) = 1 cos(θ) 1 − sin(θ) cos(θ) = 1 cos(θ) 1 − sin(θ) cos(θ) · cos(θ) cos(θ) = 1 cos(θ) (cos(θ)) 1 − sin(θ) cos(θ) (cos(θ)) = 1 (1)(cos(θ)) − sin(θ) cos(θ) (cos(θ)) = 1 cos(θ) − sin(θ) , which is exactly what we had set out to show. 

5. The right hand side of the equation seems to hold more promise. We get common denominators and add: 3 1 − sin(θ) − 3 1 + sin(θ) = 3(1 + sin(θ)) (1 − sin(θ))(1 + sin(θ)) − 3(1 − sin(θ)) (1 + sin(θ))(1 − sin(θ)) = 3 + 3 sin(θ) 1 − sin2 (θ) − 3 − 3 sin(θ) 1 − sin2 (θ) = (3 + 3 sin(θ)) − (3 − 3 sin(θ)) 1 − sin2 (θ) = 6 sin(θ) 1 − sin2 (θ)

At this point, it is worth pausing to remind ourselves of our goal. We wish to transform this expression into 6 sec(θ) tan(θ). Using a reciprocal and quotient identity, we find 6 sec(θ) tan(θ) = 6 1 cos(θ) sin(θ) cos(θ) . In other words, we need to get cosines in our denominator. Theorem 10.8 tells us 1 − sin2 (θ) = cos2 (θ) so we get: 3 1 − sin(θ) − 3 1 + sin(θ) = 6 sin(θ) 1 − sin2 (θ) = 6 sin(θ) cos2(θ) = 6 1 cos(θ) sin(θ) cos(θ) = 6 sec(θ) tan(θ) 

6. It is debatable which side of the identity is more complicated. One thing which stands out is that the denominator on the left hand side is 1 − cos(θ), while the numerator of the right hand side is 1 + cos(θ). This suggests the strategy of starting with the left hand side and multiplying the numerator and denominator by the quantity 1 + cos(θ): sin(θ) 1 − cos(θ) = sin(θ) (1 − cos(θ)) · (1 + cos(θ)) (1 + cos(θ)) = sin(θ)(1 + cos(θ)) (1 − cos(θ))(1 + cos(θ)) = sin(θ)(1 + cos(θ)) 1 − cos2(θ) = sin(θ)(1 + cos(θ)) sin2 (θ) = ✘sin(✘θ✘)(1 + cos(θ)) ✘sin(✘θ✘) sin(θ) = 1 + cos(θ) sin(θ) 

Verifying trigonometric identities requires a healthy mix of tenacity and inspiration. You will need to spend many hours struggling with them just to become proficient in the basics. Like many things in life, there is no short-cut here – there is no complete algorithm for verifying identities. Nevertheless, a summary of some strategies which may be helpful (depending on the situation) is provided below and ample practice is provided for you in the Exercises. Strategies for Verifying Identities 

ˆ Try working on the more complicated side of the identity. 

ˆ Use the Reciprocal and Quotient Identities in Theorem 10.6 to write functions on one side of the identity in terms of the functions on the other side of the identity. Simplify the resulting complex fractions. 

ˆ Add rational expressions with unlike denominators by obtaining common denominators. 

ˆ Use the Pythagorean Identities in Theorem 10.8 to ‘exchange’ sines and cosines, secants and tangents, cosecants and cotangents, and simplify sums or differences of squares to one term. 

ˆ Multiply numerator and denominator by Pythagorean Conjugates in order to take advantage of the Pythagorean Identities in Theorem 10.8. 

ˆ If you find yourself stuck working with one side of the identity, try starting with the other side of the identity and see if you can find a way to bridge the two parts of your work.