Question

SinA-sin5A+sin9A-sin13A / cosA-Cos5A-Cos9A+cos13A = cot4A

Answer 1

1.
`\sin A-\sin 5A= 2\sin (A-5A)/(2) \cos (A+5A)/(2)=-2\sin2A\cos(3A)`;

2.
`\sin 9A-\sin 13A=2\sin (9A-13A)/(2) \cos (9A+13A)/(2)=-2\sin 2A\cos 11A`;

3.
`\sin A-\sin 5A+\sin 9A-\sin 13A=-2\sin 2A\cos 3A-2\sin 2A\cos 11A=`
`=-2\sin 2A(\cos 3A+\cos 11A)`.

4.
`\cos A-\cos 5A=-2\sin (A+5A)/(2) \sin (A-5A)/(2) =2\sin 3A\sin 2A`;

5.
`\cos 13A-\cos 9A=-2\sin (13A+9A)/(2) \sin (13A-9A)/(2) =-2\sin 11A\sin 2A`;

6.
`\cos A-\cos 5A-\cos 9A+\cos 13A=2\sin 3A\sin 2A-2\sin 11A\sin 2A=`
`=2\sin 2A(\sin 3A-\sin 11A)`.

7.
`(\sin A-\sin 5A+\sin 9A-\sin 13A)/(\cos A-\cos 5A-\cos 9A+\cos 13A) = (-2\sin 2A(\cos 3A+\cos 11A))/(2\sin 2A(\sin 3A-\sin 11A)) =`
`= -(\cos 3A+\cos 11A)/(\sin 3A-\sin 11A) =-(2\cos7A\cos4A)/(2\cos7A\sin(-4A))=-(\cos4A)/(-\sin4A)=\cot4A`.