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Well... this is not as easy as it should, since Sage doesn't (yet) do partial substitutions (e. replace a two-factors product in a multiple-factors expression).

But it can be done, if one can follow a bit of intuition :

sage: f=sqrt(3)/3*cos(x)+1/3*sin(x)
sage: f*3/2
1/2*sqrt(3)*cos(x) + 1/2*sin(x)

Aha ! ISTR that $\cos{\pi \over 3}={1 \over 2}$ and that $\sin{\pi \over 3}={\sqrt{3} \over 2}$. Our f*3/2* would then be :

sage: bool((f*3/2)==sin(pi/3)*cos(x)+cos(pi/3)*sin(x))
True

Aha again. Now, this looks like $\sin({\pi \over 3} +x)$. Let's check :

sage: bool(f*3/2==sin(pi/3+x))
True

Again a distant memory stirs in my obsolescent neurons :

sage: bool(sin(x)==cos(pi/2-x))
True

Therefore, since ${\pi \over 2}-({\pi \over 3} + x) = {\pi \over 6}-x$, we have :

sage: bool(f*3/2==cos(pi/6-x))
True
sage: bool(f==cos(pi/6-x)*2/3)
True

Q. E. bloody D...

Here, apart from the already mentioned problem of Sage having difficulties with partial substituition, Sage doesn't have methods for going from/to sums to products of trig/hyperbolic expressions. It can be done with subs using a relevant dictionary (using SR wildcards) such as one of those :

sage: S2PW
[cos($1) + cos($0) == 2*cos(1/2*$1 + 1/2*$0)*cos(-1/2*$1 + 1/2*$0),
 -cos($1) + cos($0) == -2*sin(1/2*$1 + 1/2*$0)*sin(-1/2*$1 + 1/2*$0),
 sin($1) + sin($0) == 2*cos(-1/2*$1 + 1/2*$0)*sin(1/2*$1 + 1/2*$0),
 -sin($1) + sin($0) == 2*cos(1/2*$1 + 1/2*$0)*sin(-1/2*$1 + 1/2*$0)]
sage: P2SW
[cos($1)*sin($0) == 1/2*sin($1 + $0) + 1/2*sin(-$1 + $0),
 cos($0)*sin($1) == 1/2*sin($1 + $0) - 1/2*sin(-$1 + $0),
 cos($1)*cos($0) == 1/2*cos($1 + $0) + 1/2*cos(-$1 + $0),
 sin($1)*sin($0) == -1/2*cos($1 + $0) + 1/2*cos(-$1 + $0)]
sage: SQSW
[cos($0)^2 == 1/2*cos(2*$0) + 1/2,
 sin($0)^2 == -1/2*cos(2*$0) + 1/2,
 tan($0)^2 == -(cos(2*$0) - 1)/(cos(2*$0) + 1),
 cos($0) + 1 == 2*cos(1/2*$0)^2,
 -cos(2*$0) + 1 == 2*sin($0)^2]

Those are not very difficult to derive in Sage, but adding them to Sage (possibly as an optional package) would be useful...

Well... this is not as easy as it should, since Sage doesn't (yet) do partial substitutions (e. replace a two-factors product in a multiple-factors expression).

But it can be done, if one can follow a bit of intuition :

sage: f=sqrt(3)/3*cos(x)+1/3*sin(x)
sage: f*3/2
1/2*sqrt(3)*cos(x) + 1/2*sin(x)

Aha ! ISTR that $\cos{\pi \over 3}={1 \over 2}$ and that $\sin{\pi \over 3}={\sqrt{3} \over 2}$. Our f*3/2* would then be :

sage: bool((f*3/2)==sin(pi/3)*cos(x)+cos(pi/3)*sin(x))
True

Aha again. Now, this looks like $\sin({\pi \over 3} +x)$. Let's check :

sage: bool(f*3/2==sin(pi/3+x))
True

Again a distant memory stirs in my obsolescent neurons :

sage: bool(sin(x)==cos(pi/2-x))
True

Therefore, since ${\pi \over 2}-({\pi \over 3} + x) = {\pi \over 6}-x$, we have :

sage: bool(f*3/2==cos(pi/6-x))
True
sage: bool(f==cos(pi/6-x)*2/3)
True

Q. E. bloody D...

Here, apart from the already mentioned problem of Sage having difficulties with partial substituition, Sage doesn't have methods for going from/to sums to products of trig/hyperbolic expressions. It can be done with subs using a relevant dictionary (using SR wildcards) such as one of those :

sage: S2PW
[cos($1) + cos($0) == 2*cos(1/2*$1 + 1/2*$0)*cos(-1/2*$1 + 1/2*$0),
 -cos($1) + cos($0) == -2*sin(1/2*$1 + 1/2*$0)*sin(-1/2*$1 + 1/2*$0),
 sin($1) + sin($0) == 2*cos(-1/2*$1 + 1/2*$0)*sin(1/2*$1 + 1/2*$0),
 -sin($1) + sin($0) == 2*cos(1/2*$1 + 1/2*$0)*sin(-1/2*$1 + 1/2*$0)]
sage: P2SW
[cos($1)*sin($0) == 1/2*sin($1 + $0) + 1/2*sin(-$1 + $0),
 cos($0)*sin($1) == 1/2*sin($1 + $0) - 1/2*sin(-$1 + $0),
 cos($1)*cos($0) == 1/2*cos($1 + $0) + 1/2*cos(-$1 + $0),
 sin($1)*sin($0) == -1/2*cos($1 + $0) + 1/2*cos(-$1 + $0)]
sage: SQSW
[cos($0)^2 == 1/2*cos(2*$0) + 1/2,
 sin($0)^2 == -1/2*cos(2*$0) + 1/2,
 tan($0)^2 == -(cos(2*$0) - 1)/(cos(2*$0) + 1),
 cos($0) + 1 == 2*cos(1/2*$0)^2,
 -cos(2*$0) + 1 == 2*sin($0)^2]

Those are not very difficult to derive in Sage, but adding them to Sage (possibly as an optional package) would be useful...