maths (trigonometry)
posted by marley .
please solve these if possible
Q1. If sinx +siny=3(cosycosx) then the value of sin3x/sin3y.
Q2. If sina ,cosa,and tan a are in g.p.then cos cubea+cos square a is equal to

Q1.
From the given relation, we can write, 3 cos x + sin x = 3 cos y  sin y
put r cos ƒ¿=3 and r sin ƒ¿ = 1 , then we get , r = ã 10, tanƒ¿= 1/3
so r cos ( x  ƒ¿) = r cos (x+ƒ¿)
x =  y or x = }(y+ƒ¿)
clearly x =  y satisfies the equation
therefore, x =  y
so 3x = 3y
sin 3x = sin (3y) =  sin 3y
sin 3x / sin 3y = 1 
ƒ¿= Greek letter Alpha

Q2)Because sina, cosa, tana  g.p.
cos^2(a)=sina*tana=sin^2(a)/cosa
cos^2(a)=(1cos^2(a))/cosa
cos^3(a)=1cos^2(a)