Question

find sin 2a and cot 2a:

Answer 1

Answer:

$sin(2\alpha )=2(\frac{5}{12})(\frac{12}{13})=\frac{10}{13}\\cot(2\alpha ) = \frac{16511}{18720}$

Step-by-step explanation:

$\text{if } cos(\alpha)=\frac{12}{13}\\\text{That must mean we have a triangle with base 12, and hypotenuse 13.}\\\text{Using Pythagoras, we can determine the base of the triangle must be 5.}\\a^2+b^2=c^2 \text{, where c is the hypotenuse and a, b are the two other sides.}\\c^2-b^2=a^2\\\sqrt{c^2-b^2}=a\\\sqrt{13^2-12^2}=\sqrt{169-144}=\sqrt{25}=5\\\text{Therefore, }sin(\alpha) = \frac{5}{12}\\sin(2\alpha)=2sin(\alpha )cos(\alpha)\\\text{(From double angle formulae identities)}$

$sin(2\alpha )=2(\frac{5}{12})(\frac{12}{13})=\frac{10}{13}\\cos(2\alpha )=cos^2(\alpha)-sin^2(\alpha)\\cos(2\alpha )=(\frac{12}{13})^2-(\frac{5}{12})^2=\frac{16511}{24336}\\cot(2\alpha)=\frac{cos(2\alpha)}{sin(2\alpha)}=\frac{\frac{16511}{24336}}{\frac{10}{13}}=\frac{16511}{18720}$