LESSON 9: Trigonometric Identities

In Section 10.3, we saw the utility of the Pythagorean Identities in Theorem 10.8 along with the Quotient and Reciprocal Identities in Theorem 10.6. Not only did these identities help us compute the values of the circular functions for angles, they were also useful in simplifying expressions involving the circular functions. In this section, we introduce several collections of identities which have uses in this course and beyond. Our first set of identities is the ‘Even / Odd’ identities.1 Theorem 10.12. Even / Odd Identities: For all applicable angles θ, ˆ cos(−θ) = cos(θ) ˆ sec(−θ) = sec(θ) ˆ sin(−θ) = − sin(θ) ˆ csc(−θ) = − csc(θ) ˆ tan(−θ) = − tan(θ) ˆ cot(−θ) = − cot(θ) In light of the Quotient and Reciprocal Identities, Theorem 10.6, it suffices to show cos(−θ) = cos(θ) and sin(−θ) = − sin(θ). The remaining four circular functions can be expressed in terms of cos(θ) and sin(θ) so the proofs of their Even / Odd Identities are left as exercises. Consider an angle θ plotted in standard position. Let θ0 be the angle coterminal with θ with 0 ≤ θ0 < 2π. (We can construct the angle θ0 by rotating counter-clockwise from the positive x-axis to the terminal side of θ as pictured below.) Since θ and θ0 are coterminal, cos(θ) = cos(θ0) and sin(θ) = sin(θ0). 

We now consider the angles −θ and −θ0. Since θ is coterminal with θ0, there is some integer k so that θ = θ0 + 2π · k. Therefore, −θ = −θ0 − 2π · k = −θ0 + 2π · (−k). Since k is an integer, so is (−k), which means −θ is coterminal with −θ0. Hence, cos(−θ) = cos(−θ0) and sin(−θ) = sin(−θ0). Let P and Q denote the points on the terminal sides of θ0 and −θ0, respectively, which lie on the Unit Circle. By definition, the coordinates of P are (cos(θ0),sin(θ0)) and the coordinates of Q are (cos(−θ0),sin(−θ0)). Since θ0 and −θ0 sweep out congruent central sectors of the Unit Circle, it 1As mentioned at the end of Section 10.2, properties of the circular functions when thought of as functions of angles in radian measure hold equally well if we view these functions as functions of real numbers. Not surprisingly, the Even / Odd properties of the circular functions are so named because they identify cosine and secant as even functions, while the remaining four circular functions are odd. 

follows that the points P and Q are symmetric about the x-axis. Thus, cos(−θ0) = cos(θ0) and sin(−θ0) = − sin(θ0). Since the cosines and sines of θ0 and −θ0 are the same as those for θ and −θ, respectively, we get cos(−θ) = cos(θ) and sin(−θ) = − sin(θ), as required. The Even / Odd Identities are readily demonstrated using any of the ‘common angles’ noted in Section 10.2. Their true utility, however, lies not in computation, but in simplifying expressions involving the circular functions. In fact, our next batch of identities makes heavy use of the Even / Odd Identities. 

Theorem 10.13. 

Sum and Difference Identities for Cosine: For all angles α and β, ˆ cos(α + β) = cos(α) cos(β) − sin(α) sin(β) ˆ cos(α − β) = cos(α) cos(β) + sin(α) sin(β) We first prove the result for differences. As in the proof of the Even / Odd Identities, we can reduce the proof for general angles α and β to angles α0 and β0, coterminal with α and β, respectively, each of which measure between 0 and 2π radians. Since α and α0 are coterminal, as are β and β0, it follows that α − β is coterminal with α0 − β0. Consider the case below where α0 ≥ β0.

Since the angles P OQ and AOB are congruent, the distance between P and Q is equal to the distance between A and B. 2 The distance formula, Equation 1.1, yields p (cos(α0) − cos(β0))2 + (sin(α0) − sin(β0))2 = p (cos(α0 − β0) − 1)2 + (sin(α0 − β0) − 0)2 Squaring both sides, we expand the left hand side of this equation as (cos(α0) − cos(β0))2 + (sin(α0) − sin(β0))2 = cos2 (α0) − 2 cos(α0) cos(β0) + cos2 (β0) + sin2 (α0) − 2 sin(α0) sin(β0) + sin2 (β0) = cos2 (α0) + sin2 (α0) + cos2 (β0) + sin2 (β0) −2 cos(α0) cos(β0) − 2 sin(α0) sin(β0) 2 In the picture we’ve drawn, the triangles P OQ and AOB are congruent, which is even better. However, α0 − β0 could be 0 or it could be π, neither of which makes a triangle. It could also be larger than π, which makes a triangle, just not the one we’ve drawn. You should think about those three cases.

From the Pythagorean Identities, cos2 (α0) + sin2 (α0) = 1 and cos2 (β0) + sin2 (β0) = 1, so (cos(α0) − cos(β0))2 + (sin(α0) − sin(β0))2 = 2 − 2 cos(α0) cos(β0) − 2 sin(α0) sin(β0) Turning our attention to the right hand side of our equation, we find (cos(α0 − β0) − 1)2 + (sin(α0 − β0) − 0)2 = cos2 (α0 − β0) − 2 cos(α0 − β0) + 1 + sin2 (α0 − β0) = 1 + cos2 (α0 − β0) + sin2 (α0 − β0) − 2 cos(α0 − β0) Once again, we simplify cos2 (α0 − β0) + sin2 (α0 − β0) = 1, so that (cos(α0 − β0) − 1)2 + (sin(α0 − β0) − 0)2 = 2 − 2 cos(α0 − β0) Putting it all together, we get 2 − 2 cos(α0) cos(β0) − 2 sin(α0) sin(β0) = 2 − 2 cos(α0 − β0), which simplifies to: cos(α0 − β0) = cos(α0) cos(β0) + sin(α0) sin(β0). Since α and α0, β and β0 and α − β and α0 − β0 are all coterminal pairs of angles, we have cos(α − β) = cos(α) cos(β) + sin(α) sin(β). For the case where α0 ≤ β0, we can apply the above argument to the angle β0 − α0 to obtain the identity cos(β0 − α0) = cos(β0) cos(α0) + sin(β0) sin(α0). Applying the Even Identity of cosine, we get cos(β0 − α0) = cos(−(α0 − β0)) = cos(α0 − β0), and we get the identity in this case, too. To get the sum identity for cosine, we use the difference formula along with the Even/Odd Identities cos(α + β) = cos(α − (−β)) = cos(α) cos(−β) + sin(α) sin(−β) = cos(α) cos(β) − sin(α) sin(β)

Theorem 10.19. 

Half Angle Formulas: For all applicable angles θ, ˆ cos θ 2 = ± r 1 + cos(θ) 2 ˆ sin θ 2 = ± r 1 − cos(θ) 2 ˆ tan θ 2 = ± s 1 − cos(θ) 1 + cos(θ) where the choice of ± depends on the quadrant in which the terminal side of θ 2 lies

Theorem 10.21. 

Sum to Product Formulas: For all angles α and β, ˆ cos(α) + cos(β) = 2 cos α + β 2 cos α − β 2 ˆ cos(α) − cos(β) = −2 sin α + β 2 sin α − β 2 ˆ sin(α) ± sin(β) = 2 sin α ± β 2 cos α ∓ β 2