Use multiplication of power series to find the Taylor series for f(x)=sinxsin3x centered at c=π/6

Write the first six terms of the series with coefficients as reduced fraction multiples of √3 .
Please show all work thank you!

y = sin x ==> y(π/6) = 1/2.
y' = cos x ==> y'(π/6) = √3/2.
y'' = -sin x ==> y''(π/6) = -1/2.
y''' = -cos x ==> y'''(π/6) = -√3/2.
y'''' = sin x ==> y''''(π/6) = 1/2.
y''''' = cos x ==> y'''''(π/6) = √3/2.

So, sin x = 1/2 + (√3/2)(x - π/6) + (-1/2)(x - π/6)^2/2! + (-√3/2)(x - π/6)^3/3! + (1/2)(x - π/6)^4/4! + (√3/2)(x - π/6)^5/5! + ...

= 1/2 + (√3/2)(x - π/6) - (1/4)(x - π/6)^2 - (√3/12)(x - π/6)^3 + (1/48)(x - π/6)^4 + (√3/240)(x - π/6)^5 + ...
-----------------------
sin(3x) = sin(3(x - π/6) + π/2)
...........= cos(3(x - π/6)), via addition identity
...........= 1 - (3(x - π/6))^2/2! + (3(x - π/6))^4/4! - ...
...........= 1 - (9/2)(x - π/6)^2 + (27/8) (x - π/6)^4 - ...

So, sin x sin(3x)
= [1/2 + (√3/2)(x - π/6) - (1/4)(x - π/6)^2 - (√3/12)(x - π/6)^3 + (1/48)(x - π/6)^4 + (√3/240)(x - π/6)^5 + ...] * [1 - (9/2)(x - π/6)^2 + (27/8) (x - π/6)^4 - ...]

= 1/2 + (√3/2)(x - π/6) + (-9/4 - 1/4)(x - π/6)^2 + (-9√3/4 - √3/12) (x - π/6)^3
+ (27/16 + 9/8 + 1/48) (x - π/6)^4 + (27√3/16 + 9√3/24 + √3/240) (x - π/6)^5 + ...

= 1/2 + (√3/2)(x - π/6) - (5/2)(x - π/6)^2 - (7√3/3)(x - π/6)^3 + (17/6)(x - π/6)^4
+ (31√3/15) (x - π/6)^5 + ...

(Double checked with Wolfram Alpha.)

I hope this helps!