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Exercise 2.6

Derive the relations between these functions by using similar triangles.

Solution:

By similar triangles, the ratio of BD to BO, sin ⁡ θ to 1 is the same as the ratio of CB to OC, tan ⁡ θ to sec ⁡ θ , and of OB to AO (1 to csc ⁡ θ ), and of CO to AC ( s e c θ to ( tan ⁡ θ + cot ⁡ θ )), and of CD to BC (( sec ⁡ θ − cos ⁡ θ ) to tan ⁡ θ ), and of OE to BO ( sin ⁡ θ to 1)

This tells us

csc ⁡ θ = 1 sin ⁡ θ

so that on taking complements we have as well,

sec θ = 1 cos ⁡ θ

We also get

tan ⁡ θ = sin ⁡ θ sec ⁡ θ = sin ⁡ θ cos ⁡ θ

and the complementary result

cot ⁡ θ = cos ⁡ θ sin ⁡ θ

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