Trig+Identities+Extended+9-28

There is no one right way to do a problem relating to trig problems. The key for each problem is to start with one side (usually the harder, "uglier" side), and then keep changing it until it equals the other side. A tip would be to find the similarities and differences between the two sides.

A example of a problem we discussed in class was;: math $ \frac{\cos (-x)}{\sin x \cot x}=-1 $ math The reciprocal and the quotient identities can both be used to solved this equation. math $ \frac{\cos (-x)}{\sin (x)\cot (x)}= \frac{\cos (-x)}{\sin(x)*\frac{\cos (-x)}{\sin (-x)}}= \frac{\cos (-x)\sin (-x)}{\sin (x)\cos (-x)}= \frac{\sin (-x)}{\sin (x)}= -\frac{\sin (x)}{\sin (x)}=-1 $ math or math $ \frac{\cos (-x)}{\sin (x)}*\frac{1}{\cot (x)}=\frac{\cos (-x)}{\sin (x)}*\frac{1}{\cot (x)}=-1 $ math Need another example? math $ \frac{2\sin ^{2}+3\cos (x)-3}{\sin ^{2}(x)}=\frac{2\cos (x)-1}{1+\cos (x)} $ math The first step was to make both sides cosine only. math $ \frac{2\sin ^{2}(x)+3\cos (x)-3}{\sin ^{2}(x)}=\frac{2(1-\cos ^{2}+3\cos (x))}{1-\cos ^{2}(x)}=\frac{2\\cos^{2}+3\cos (x)-3}{(1+\cos (x))(1-\cos x))}=\frac{-2\\cos (x) ^{2}+3\cos (x)-1}{(1+\cos (x))(1-\cos (x))}=\frac{(1-\cos (x)*(2\cos (x)-1))}{1-\cos (x)*1+\cos (x)}=\frac{2\cos (x)-1}{1+\cos (x)} $ math Need another example? math $ \tan\alpha +\cot\beta=\frac{\sin\alpha}{\cos\alpha}+\frac{\cos\beta}{\sin\beta}=\frac{\sin\alpha\sin\beta+\cos\alpha\cos\beta}{\cos\alpha\sin\beta}*\frac{\frac{1}{\cos\beta}*\frac{1}{\sin\alpha}}{\frac{1}{\cos\beta}*\frac{1}{\sin\alpha}}=\frac{\tan \beta +\cot\alpha }{\tan \beta \cot\alpha }=\frac{\tan \beta }{\tan \beta \cot \alpha } +\frac{\cot \alpha }{\tan \beta \cot \alpha }= \frac{1}{\cot \alpha}+\frac{1}{\tan\beta}=\tan\alpha +\cot\beta $ math

We also learned a new trig identity known as the angle sum identities. They are the followingg: math $ cis(\alpha)cis (\beta)=(\cos (\alpha)\cos (\beta)-\sin(\alpha) \sin (\beta))+i(\cos (\beta) \sin(\alpha)+ \cos(\alpha)\sin\beta)) $ math math $ cis(\alpha)cis(\beta)=cis(\alpha+\beta)=\cos(\alpha+\beta)+i\sin(\alpha+\beta) $ math math $ \cos(\alpha+\beta)=\cos(\alpha) \cos(\beta)-sin(\alpha)\sin(\beta) $ math math $ \sin(\alpha+\beta)=\cos(\beta)\sin(\alpha)+cos(\alpha) sin(\beta) $ math