Can you multiply the double angle forumla?

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The discussion clarifies that sin(4x) cannot be directly expressed as 4sin(x)cos(x) using the double angle formula. Instead, the correct approach involves recognizing sin(4x) as sin(2(2x)), allowing the application of the double angle formula, which states sin(2A) = 2sin(A)cos(A). By substituting A with 2x, the expression can be rewritten correctly. Careful adherence to the formula patterns is essential for accurate transformations. Understanding these nuances is crucial for solving trigonometric equations effectively.
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Homework Statement


I have sin4x, can I use the double angle forumla and make it 4sinxcosx?


Homework Equations


double angle forumla


The Attempt at a Solution


sin4x----> 4sinxcosx
 
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...try graphing both and see if they're the same.
 
musiclover55 said:

Homework Statement


I have sin4x, can I use the double angle forumla and make it 4sinxcosx?
In a word, no.
The double-angle formula says that sin(2A) = 2*sin(A)cos(A).

When you use formulas, you need to be very careful to follow the pattern being used.
If you write sin(4x) as sin(2(2x)), you can use the double-angle formula, replacing A in the above by 2x.
musiclover55 said:

Homework Equations


double angle forumla


The Attempt at a Solution


sin4x----> 4sinxcosx
 
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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