Master Trigonometric Identities with Double Angle Techniques

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The discussion focuses on simplifying the expression √(1 + cos(6θ)) using double angle identities. The suggestion is to rewrite 6θ as 2(3θ) to apply the double angle formula effectively. Participants are encouraged to clarify their understanding of the double angle technique to ensure accurate application. The conversation emphasizes the importance of mastering trigonometric identities for solving complex problems. Proper utilization of these identities can enhance problem-solving skills in trigonometry.
Mrencko
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I am doig trigonometric identities and i got this one, (all will be in the picture the solution and my work) i used the double angle for this but i am afraid i didn't get the exact idea, just guessing, good guessing, so i want to know how is the proper way to reach the solution
 
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Here is a little bigger view for folks.

upload_2015-5-6_18-28-10.png


So, I take it that you want to simplify ##\ \sqrt{1+\cos(6\theta)\,}\ ## using the double angle identity for cosine.

Write 6θ as 2(3θ) .
 
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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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