S. D. V. Stephens

notes / Integrated Mathematics / chapter 15: Trigonometry

15.3 Angle Addition Identities

The identity for the sine of a sum is read off a triangle; the identities for the cosine and tangent of a sum, and for differences, follow from it and from the basic identities.

Deriving the identities

Problem 15.3.1. Given the following triangle with , , and :

Right triangle ABC with the right angle at B and angle alpha at C; D lies on BC with AD = 1, E is the foot of the perpendicular from D to AC, and angle DAE is beta.

(a) Why does and ?

(b) Find in terms of and .

(c) Find in terms of and .

(d) Show that and .

(e) Show that .

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(a) Since and .

(b) In right triangle , and , so

(c) . Also and , so

(d) In right triangle (right angle at , hypotenuse , angle at ), and since sine is opposite over hypotenuse (in triangle ).

(e) From (d), , and , so . We have , so , and . Therefore

Problem 15.3.2. Use the newly found identity with previous ones to find . Test your identity with , .

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Just apply it as , and at the end note but . So we have Let , . Then , and .

Problem 15.3.3. Previously we found .

(a) Use with the identity for to express .

(b) Test your new identity with , .

(c) Find an identity for .

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(a)

(b) and

(c) We can undergo a similar process as in part (a):

Problem 15.3.4. Suppose that , , are defined. Find in terms of and .

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Since , we have (and ), so this is

* Requirements: ; ; , defined.

Using the identities

Problem 15.3.5. (a) Find . (b) Find .

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(a)

(b)

Problem 15.3.6. Find the value of .

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Let and , so we want Now , so

Trying : Oops, back at the same spot.

Instead, (since ), taken as , giving And So that is, The answer is : since is its range and has range , for all .

Problem 15.3.7. Compute if .

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First, Then

Exercises

Exercise 15.3.8. (a) (b) (c) (d)

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(a)

(b)

(c)

(d) .

Exercise 15.3.9. Show .

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and since we have

Exercise 15.3.10. Compute .

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, and with , Then, with and ,

Exercise 15.3.11. Expand with the angle addition identity.

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Since :

Exercise 15.3.12. Prove that .

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Since , :

Exercise 15.3.13. Find all acute angles such that .

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. With , , so .

This doesn’t seem like the right path. Find all such that : So between we have as solutions for .

Wait, acute means , so and are the only solutions.

Exercise 15.3.14. and are acute angles such that . If , compute the measure of .

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, so Then :

Exercise 15.3.15. Suppose that and . What is ?

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We want . Squaring both givens, Adding, and pairing , :

Exercise 15.3.16. (a) Let and . Why must and be acute?

(b) Write in terms of , , .

(c) Express in terms of , . Why is it important here that , are acute?

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(a) If , the acuteness of is trivial (and likewise for ).

(b) and , so we have but and , so

(c)