Answer to Question #71326 in Trigonometry for Antonei Cepeda

Question

Find the exact values of the trigonometric functions of a/2 and 2a by using half-angle and double angle formulas.

1) sin(a) = 4/5, and a in quadrant 1

Answer 1

1) Use half-angle identity for sine:

$$sin \frac{a}{2}=±\sqrt \frac{1-cos a}{2}$$

Find $cos a:$

$$cos a=±\sqrt{1-sin^2 a}$$ $$cos a=±\sqrt{1-(\frac{4}{5})^2}=±\sqrt{1-\frac{16}{25}}=±\sqrt{\frac{9}{25}}=±\frac{3}{5}$$

Since $a$ is in the quadrant I, $cos a>0, cos a=\frac{3}{5}$

Hence,

$$sin \frac{a}{2}=±\sqrt{\frac{1-\frac{3}{5}}{2}}=±\sqrt{\frac{\frac{2}{5}}{2}}=±\sqrt{\frac{1}{5}}$$

Since $a$ is in the quadrant I, $sin \frac{a}{2}>0, sin \frac{a}{2}=\sqrt{\frac{1}{5}}$

2) Use half-angle identity for cosine:

$$cos \frac{a}{2}=±\sqrt \frac{1+cos a}{2}$$ $$cos \frac{a}{2}=±\sqrt{\frac{1+\frac{3}{5}}{2}}=±\sqrt{\frac{\frac{8}{5}}{2}}=±\sqrt{\frac{4}{5}}$$

Since $a$ is in the quadrant I, $cos \frac{a}{2}>0, cos \frac{a}{2}=\sqrt{\frac{4}{5}}$

3) Half-angle for tangent and cotangent:

$$tan \frac{a}{2}=sin \frac{a}{2}:cos \frac{a}{2}=\sqrt{\frac{1}{5}}:\sqrt{\frac{4}{5}}=\sqrt{\frac{1}{4}}=\frac{1}{2}$$ $$cot \frac{a}{2}=cos \frac{a}{2}:sin \frac{a}{2}=\sqrt{\frac{4}{5}}:\sqrt{\frac{1}{5}}=\sqrt{\frac{4}{1}}=2$$

4) Use double-angle identity for sine:

$$sin 2a=2sinacos a$$ $$sin 2a=2\cdot \frac{4}{5} \cdot \frac{3}{5}=\frac{24}{25}$$

5) Use double-angle identity for cosine:

$$cos 2a=1-2sin^2a=1-2\cdot({\frac{4}{5}})^2=1-\frac{32}{25}=-\frac{7}{25}$$

6) Double-angle for tangent and cotangent:

$$tan2a=sin2a:cos 2a=\frac{24}{25}:(-\frac{7}{25})=-\frac{24}{7}$$

$$cot2a=cos2a:sin 2a=-\frac{7}{25}:\frac{24}{25}=-\frac{7}{24}$$

Answer: $sin \frac{a}{2}=\sqrt{\frac{1}{5}}, cos \frac{a}{2}=\sqrt{\frac{4}{5}}, tan \frac{a}{2}=\frac{1}{2}, cot\frac{a}{2}=2,$

$sin 2a=\frac{24}{25}, cos 2a=-\frac{7}{25}, tan 2a=-\frac{24}{7}, cot 2a=-\frac{7}{24}$