One x, two x, three x

$\sin x = -13\cos x \Rightarrow \frac{\displaystyle \sin x}{\displaystyle \cos x} = \tan x = -13 \\ \frac{\displaystyle \sin x + \sin 2x + \sin 3x}{\displaystyle \cos x + \cos 2x + \cos 3x} \\ = \frac{\displaystyle (\sin 3x + \sin x) + \sin 2x}{\displaystyle (\cos3x + \cos x) + \cos 2x} \\ = \frac{\displaystyle 2\sin 2x\cos x + \sin 2x}{\displaystyle 2\cos 2x\cos x + \cos 2x} \\ = \frac{\displaystyle \sin2x(2\cos x + 1)}{\displaystyle \cos2x(2\cos x + 1)} \\ = \tan 2x \\ = \frac{\displaystyle 2\tan x}{\displaystyle 1-\tan^2 x} = \frac{\displaystyle 2 \cdot (-13)}{\displaystyle 1 - (-13)^2} = \boxed{\frac{\displaystyle 13}{\displaystyle 84}}$

This problem can be solved by using the basic trigonometric formulae. $\frac{\sin x+\sin 2x +\sin 3x}{\cos x+\cos 2x +\cos 3x}$ $=\frac{(\sin x+\sin 3x) +\sin 2x}{(\cos x+\cos 3x) +\cos 2x}$ $=\frac{2\sin(\frac{x+3x}{2}) \cos(\frac{x-3x}{2})+\sin 2x}{2\cos(\frac{x+3x}{2}) \cos(\frac{x-3x}{2})+\cos2x}$ $=\frac{2\sin(\frac{4x}{2}) \cos(\frac{-2x}{2})+\sin 2x}{2\cos(\frac{4x}{2}) \cos(\frac{-2x}{2})+\cos2x}$ $=\frac{2(\sin 2x)(\cos (-x))+\sin 2x}{2(\cos 2x)(\cos (-x))+\cos 2x}$ $=\frac{\sin 2x(2\cos (-x)+1)}{\cos 2x(2\cos (-x)+1)}$ $=\frac{\sin 2x}{\cos 2x}$ $=\tan 2x$ $=\frac{2\tan x}{1-\tan^{2}x}$ $=\frac{2\frac{\sin x}{\cos x}}{1-\frac{\sin^{2}x}{\cos^{2}x}}$ Now substitutins $\sin x=-13\cos x$ ,we get $=\frac{2 \times -13}{1-(-13)^{2}}$ $=\frac{-26}{1-169}$ $=\frac{-26}{-168}$ $=\frac{13}{84}$ Now, $\frac{a}{b}=\frac{13}{84}$ $\Rightarrow a+b=13+84$ $\Rightarrow a+b=97$ Therefore, the answer to this question is $\boxed{97}$ .

$\cfrac{ \sin x + \sin 2x + \sin 3x }{\cos x + \cos 2x + \cos 3x} = \cfrac{ \sin x + 2 \sin x \cos x + \sin x (2\cos 2x +1) }{\cos x + 2\cos ^2 x -1 + \cos x (2 \cos 2x -1) } =$

$\cfrac{ \sin x (1 + 2\cos x + 2\cos 2x +1) }{\cos x + 2\cos ^2 x -1 + 2 \cos x \cos 2x -\cos x } = \cfrac{ 2\sin x (1 + \cos x + \cos 2x) }{2\cos ^2 x -1 + 2 \cos x \cos 2x } =$

$\cfrac{ 2\sin x (1 + \cos x + 2\cos ^2 x -1) }{\cos 2x + 2 \cos x \cos 2x } = \cfrac{ 2\sin x (\cos x + 2\cos ^2 x ) }{\cos 2x (1 + 2 \cos x) } =$

$\cfrac{ 2\sin x \cos x (1 + 2\cos x) }{\cos 2x (1 + 2 \cos x) } = \cfrac{ 2\sin x \cos x }{\cos 2x } = \cfrac{ \sin 2x }{\cos 2x } = \tan 2x$

$\tan 2x = \cfrac {2\tan x}{1- \tan^2 x}$

We know that $\sin x = -13\cos x \Rightarrow \tan x = -13$

$\cfrac {2\tan x}{1- \tan^2 x} = \cfrac {-26}{1- 13^2} = \cfrac {26}{168} = \cfrac {13}{84}$

Solution is $13+84 = \boxed{97}$ .

numerator : sinx+sin2x+sin3x

by the formula - sinx+siny= sin(x+y)/2 cos(x-y)/2 { cos(-x)=cosx } for sinx+sin3x we get, 2sin2xcosx

so sin2x + 2sin2xcosx = sin2x(1+cosx)

Denominator : cosx+cos2x+cos3x

by the formula , cosx + cosy = 2cos(x+y)/2cos(x-y)/2 { cos(-x)=cosx } cosx+cos3x= 2cos2xcosx

cos2x+ 2cos2xcosx = cos2x(1+cosx)

now NR/ DR = sin2x(1+cosx)/cos2x(1+cosx)

we get , tan2x

WKT - tan2x= tanx/1-tan^2x

tanx = sinx/cosx

sinx=-13cosx (GVN)

there fore

tanx = -13 and Tan^2x = 169

substituting in tan2x we get 13/94 = a/b

there fore a+b = 97 (ANS)

We know that: sin(a+b)=sin(a) cos(b) +sin(b) cos(a) and cos(a+b)= cos(a) cos(b)-sin(a) sin(b). Since sin(-a) =-sin(a) and cos(-a)=cos(a) we also get: sin(a-b)=sin(a) cos(b) -sin(b) cos(a) and cos(a-b)= cos(a) cos(b)+sin(a) sin(b). Combing these inequalities we get that: sin(A) + sin(B) = 2* sin((A+B)/2)* cos((A-B)/2) and cos(A) + cos(B) = 2* cos((A+B)/2)* cos((A-B)/2), where A=a+b and B=a-b. So, sin(x) + sin(3x)= 2 sin(2x) cos(x) and sin(x)+sin(2x) +sin(3x)=(2 cos(x) +1) sin(2x). Also cos(x)+cos(3x)=2 cos(2x) cos(x) and cos(x) +cos(2x)+cos(3x)=(2 cos(x) +1) cos(2x). Finaly we get: sin(2x)/cos(2x) = (2 sin(x) cos(x)) / (cos(x)^2 -sin(x)^2) = -26*cos(x)^2/-168cos(x)^2 =26/168 = 13/84.

sin 2x (1+cos 2x)/cos 2x (1+cos 2x) = tan 2x = 2(-13)/(1-(-13)^2) = 13/84 = 97

Combining the first and last terms of numerator and denominator, we get sinx + sin3x + sin2x / cosx + cos3x + cos2x using the formula for sinc + sind = 2sin((C+D)/2)cos((C-D)/2) and cosc + cosd = 2cos((C+D)/2)cos((C-D)/2) we get 2sin2xcosx + sin2x / 2cos2xcosx + cos2x = sin2x(2cosx +1) / cos2x(2cosx+1), which on cancellation gives sin2x/cos2x which is equal to tan2x. Now using the formula again for tan2x, we have tan2x= 2tanx/1-tan^2 x. From the question we have sinx=-13cosx which gives tanx = -13. Plugging the values for tanx in the obtained expression we have expr = 26/168 = 13/84. Hence the answer = 13 + 84 = $\boxed{97}$

nice problem

[sin(x)+sin(2x)+sin(3x)]/[cos(x)+cos(2x)+cos(3x)] = tan(2x). Notice that sin(x)=-13cos(x). tan(2x)=[2sin(x)cos(x)]/[cos^2(x)-sin^2(x)] tan(2x)=-26cos^2(x)/168cos^2(x)= 13/84 a=13 and b=84, a+b=97