Can I Add Sine and Cosine Functions with a Non-Factorable Scalar?

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To add sine and cosine functions with non-factorable scalars, expand the cosine term using the angle addition formula. For example, in the expression 5*cos(wt) + 6*cos(wt + π/4), first expand cos(wt + π/4) into its components. After expansion, group like terms to simplify the expression. The result can be expressed in the form R*cos(wt ± A) or R*sin(wt ± A). This method allows for the combination of the functions despite the presence of scalars.
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Hi guys,

I don't know if this should go here because it is an excerpt from a higher level problem. The part where I get stuck is when I try to add the cosine functions.

Is there any way to add sine and cosine functions that have a scalar in front that cannot be factored out? For example:

5*cos(wt) + 6*cos(wt + pi/4)

If there weren't any numbers in front of the functions then I could use the trig identity. What can I do with the numbers there? Thanks
 
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5cos(wt) + 6*cos(wt + π/4)


expand out cos(wt+π/4) then group the like terms. Then you can either put in the form Rcos(wt±A) or Rsin(wt±A)
 
cos(A+ B)= cops(A)cos(B)- sin(A)sin(B).
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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