Exercise 2.6

]> Exercise 2.6

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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 = \frac{1}{\sin θ}

so that on taking complements we have as well,

\sec θ = \frac{1}{\cos θ}

We also get

\tan θ = \sin θ \sec θ = \frac{\sin θ}{\cos θ}

and the complementary result

\cot θ = \frac{\cos θ}{\sin θ}