Laws Of Sines, Cosines And Tangents; Identity Practice

Laws of Sines, Cosines and Tangents

• Law of Sines: $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$
• Law of Cosines: $a^2=b^2+c^2-2bc\cos A$, or $\cos A=\frac{b^2+c^2-a^2}{2bc}$
• Law of Tangents: $\frac{a-b}{a+b}=\frac{\tan \frac{1}{2}(A-B)}{\tan \frac{1}{2}(A+B)}$

Trigonometry Practice

1) Prove identities

a) $\sin \theta (\tan \theta +\cot \theta )=\sec \theta$

$\tan \theta =\frac{\sin \theta }{\cos \theta },\cot \theta =\frac{\cos \theta }{\sin \theta }$

$\sin \theta \left(\frac{\sin \theta }{\cos \theta }+\frac{\cos \theta }{\sin \theta }\right)=\sin \theta \left(\frac{{\sin }^2\theta +{\cos }^2\theta }{\sin \theta \cos \theta }\right)$

$=\sin \theta \left(\frac{1}{\sin \theta \cos \theta }\right)=\frac{1}{\cos \theta }=\sec \theta ✓$

(numerator: ${\sin }^2\theta +{\cos }^2\theta =1$)

b) $\frac{1}{\csc \theta +\cot \theta }=\csc \theta -\cot \theta$

$\frac{1}{\csc \theta +\cot \theta }=\frac{1}{\frac{1}{\sin \theta }+\frac{\cos \theta }{\sin \theta }}=\frac{1}{\frac{1+\cos \theta }{\sin \theta }}=\frac{\sin \theta }{1+\cos \theta }$

$=\frac{\sin \theta }{1+\cos \theta }\cdot \frac{1-\cos \theta }{1-\cos \theta }=\frac{(1-\cos \theta )\sin \theta }{1-{\cos }^2\theta }=\frac{(1-\cos \theta )\sin \theta }{{\sin }^2\theta }$

$=\frac{1-\cos \theta }{\sin \theta }=\frac{1}{\sin \theta }-\frac{\cos \theta }{\sin \theta }=\csc \theta -\cot \theta ✓$

c) $\frac{1}{1+\cos \theta }={\csc }^2\theta -\csc \theta \cot \theta$

$\frac{1}{1+\cos \theta }=\frac{1}{1+\cos \theta }\cdot \frac{1-\cos \theta }{1-\cos \theta }=\frac{1-\cos \theta }{1-{\cos }^2\theta }=\frac{1-\cos \theta }{{\sin }^2\theta }$

$=\frac{1}{{\sin }^2\theta }-\frac{\cos \theta }{{\sin }^2\theta }$

$\frac{1}{{\sin }^2\theta }={\csc }^2\theta$

$\frac{\cos \theta }{{\sin }^2\theta }=\frac{\cos \theta }{1-{\cos }^2\theta }$, or $\frac{\cos \theta }{\sin \theta }\cdot \frac{1}{\sin \theta }=\cot \theta \cdot \csc \theta$

$\therefore \frac{1}{1+\cos \theta }={\csc }^2\theta -\cot \theta \csc \theta ✓$

d) $\frac{1}{1+\sin \theta }={\sec }^2\theta -\sec \theta \tan \theta$

$\frac{1}{1+\sin \theta }\cdot \frac{1-\sin \theta }{1-\sin \theta }=\frac{1-\sin \theta }{1-{\sin }^2\theta }=\frac{1-\sin \theta }{{\cos }^2\theta }$

$=\frac{1}{{\cos }^2\theta }-\frac{\sin \theta }{{\cos }^2\theta }={\sec }^2\theta -\frac{1}{\cos \theta }\cdot \frac{\sin \theta }{\cos \theta }$

$={\sec }^2\theta -\sec \theta \tan \theta ✓$

e) $\frac{{\sin }^2(-\theta )-{\cos }^2(-\theta )}{\sin (-\theta )-\cos (-\theta )}=\cos \theta -\sin \theta$

Numerator $\Rightarrow {\sin }^2(-\theta )-{\cos }^2(-\theta )$

$\sin (-\theta )=-\sin \theta \Rightarrow {\sin }^2(-\theta )=(-\sin \theta {)}^2={\sin }^2\theta$

Likewise ${\cos }^2(-\theta )={\cos }^2\theta \Rightarrow \therefore {\sin }^2\theta -{\cos }^2\theta$

$=(\sin \theta -\cos \theta )(\sin \theta +\cos \theta )$

Denominator $\Rightarrow \sin (-\theta )-\cos (-\theta )=-\sin \theta -\cos \theta \Rightarrow -(\sin \theta +\cos \theta )$

Combine terms:

$\frac{(\sin \theta -\cos \theta )(\sin \theta +\cos \theta )}{-(\sin \theta +\cos \theta )}=-(\sin \theta -\cos \theta )=\cos \theta -\sin \theta ✓$