Answer:
1. [tex]\cot B=\frac{5}{12}[/tex]
2. [tex]\sin2A=\frac{24}{25}[/tex]
3. [tex]\cos(\frac{5\pi}{6}+B )=-\frac{5}{26}(\sqrt{3}+1)[/tex]
4. [tex]\tan(A-\frac{\pi}{4} )=-\frac{1}{7}}[/tex]
Step-by-step explanation:
If [tex]\sin A=\frac{3}{5}[/tex] and [tex]\cos B=\frac{5}{13}[/tex], then we can use the Pythagorean identity to find [tex]\cos A[/tex] and [tex]\sin B[/tex].
[tex]\sin^2A+\cos^2A=1[/tex]
[tex](\frac{3}{5} )^2+\cos^2 A=1[/tex]
[tex]\frac{9}{25}+\cos ^2A=1[/tex]
[tex]\cos^2 A=1-\frac{9}{25}[/tex]
[tex]\cos^2 A=\frac{16}{25}[/tex]
[tex]\cos A=\pm \sqrt{\frac{16}{25}}[/tex]
Since A is in quadrant I,
[tex]\cos A=\sqrt{\frac{16}{25}}[/tex]
[tex]\cos A=\frac{4}{5}[/tex]
Also;
[tex]\sin^2B+\cos^2B=1[/tex]
[tex](\frac{5}{13} )^2+\sin^2 B=1[/tex]
[tex]\frac{25}{169}+\sin ^2B=1[/tex]
[tex]\sin^2 B=1-\frac{25}{169}[/tex]
[tex]\sin^2 B=\frac{144}{169}[/tex]
[tex]\sin B=\pm \sqrt{\frac{144}{169}}[/tex]
Since A is in quadrant I,
[tex]\sin B=\sqrt{\frac{144}{169}}[/tex]
[tex]\sin B=\frac{12}{13}[/tex]
This implies that;
[tex]\cot B=\frac{\cos B}{\sin B}[/tex]
[tex]\cot B=\frac{\frac{5}{13} }{\frac{12}{13} }[/tex]
[tex]\cot B=\frac{5}{12}[/tex]
[tex]\sin2A=2\sin A \cos A[/tex]
[tex]\sin2A=2\times \frac{3}{5} \times \frac{4}{5}[/tex]
[tex]\sin2A=\frac{24}{25}[/tex]
[tex]\cos(\frac{5\pi}{6}+B )=\cos(\frac{5\pi}{6})\cos(B )-\sin(\frac{5\pi}{6})\sin(B)[/tex]
This implies that;
[tex]\cos(\frac{5\pi}{6}+B )=\cos(\frac{5\pi}{6})\times \frac{5}{13}-\sin(\frac{5\pi}{6})\times \frac{5}{13}[/tex]
[tex]\cos(\frac{5\pi}{6}+B )=-\frac{\sqrt{3}}{2}\times \frac{5}{13}-\frac{1}{2})\times \frac{5}{13}[/tex]
[tex]\cos(\frac{5\pi}{6}+B )=-\frac{5}{26}(\sqrt{3}+1)[/tex]
[tex]\tan(A-\frac{\pi}{4} )=\frac{\tan A-\tan \frac{\pi}{4} }{1+\tan A \tan \frac{\pi}{4}}[/tex]
But; [tex]\tan A=\frac{\sin A}{\cos A}[/tex]
[tex]\tan A=\frac{\frac{3}{5} }{\frac{4}{5} }=\frac{3}{4}[/tex]
[tex]\tan(A-\frac{\pi}{4} )=\frac{\frac{3}{4}-\tan \frac{\pi}{4} }{1+\frac{3}{4} \tan \frac{\pi}{4}}[/tex]
Simplify;
[tex]\tan(A-\frac{\pi}{4} )=\frac{\frac{3}{4}-1 }{1+\frac{3}{4}}[/tex]
[tex]\tan(A-\frac{\pi}{4} )=-\frac{1}{7}}[/tex]
