MHB Use double or half angle formulas

  • Thread starter Thread starter Elissa89
  • Start date Start date
  • Tags Tags
    Angle Formulas
Elissa89
Messages
52
Reaction score
0
Ok, So with this problem it says to use double angle or half angle formula. I have the formulas in my notes just not sure how to apply them to the problem I feel like I should be using the double formula though. Here's the problem:

cos(2*theta)+sin^2=0
 
Mathematics news on Phys.org
Yes the double-angle formula for $\cos$ would be useful. There are a few different forms but consider that $\cos (2 \theta) = \cos^2 (\theta) - \sin^2 (\theta)$. Substituting in we get,

$\cos^2 (\theta) - \sin^2 (\theta) + \sin^2 (\theta) = \cos^2 (\theta) = 0$.

Do you have any ideas on how to solve for $\theta$?
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.
Back
Top