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anemone
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If $\dfrac{\cos \alpha}{\cos \beta}+\dfrac{\sin \alpha}{\sin \beta}=-1$, evaluate $\dfrac{\cos^3 \beta}{\cos \alpha}+\dfrac{\sin^3 \beta}{\sin \alpha}$.
anemone said:If $\dfrac{\cos \alpha}{\cos \beta}+\dfrac{\sin \alpha}{\sin \beta}=-1$, evaluate $\dfrac{\cos^3 \beta}{\cos \alpha}+\dfrac{\sin^3 \beta}{\sin \alpha}$.
$\therefore \dfrac{m}{n}=-1-p$ $\therefore-n=\dfrac{m}{1+p}$ $\therefore m=-n(1+p)$ | $\therefore p+\dfrac{m}{\sin \beta}=-1$ $\sin \beta=-\left( \dfrac{m}{1+p} \right)$ $\sin^2 \beta=\left( \dfrac{m}{1+p} \right)^2$ $\sin^2 \beta=n^2$ $1-\sin^2 \beta=1-n^2$ $\cos^2 \beta=1-n^2$ |
$\therefore \dfrac{\cos^2 \alpha}{\cos^2 \beta}=p^2$ $\dfrac{1-\sin^2 \alpha}{1-\sin^2 \beta}=p^2$ $\dfrac{1-m^2}{1-n^2}=p^2$ $1-m^2=p^2(1-n^2)$ $1-(n(1+p))^2=p^2(1-n^2)$ $p^2+n^2+2pn^2=1$ |
The purpose of this equation is to find the sum of two trigonometric expressions, involving cosine and sine functions, with variables alpha and beta.
To evaluate this sum, you can use the trigonometric identity cos^3x = (1/4)(3cosx + cos3x) and sin^3x = (1/4)(3sinx - sin3x). Then, substitute these identities into the equation and simplify using basic algebraic manipulations.
Yes, this equation can be solved for specific values of alpha and beta. However, it is important to note that the solution may depend on the values chosen for these variables.
No, this sum is not always defined. It is only defined when the denominators, cos alpha and sin alpha, are not equal to zero.
This sum is significant in trigonometry as it demonstrates the use of trigonometric identities to simplify expressions and solve equations involving cosine and sine functions. It also highlights the importance of understanding the domains of trigonometric functions in order to properly evaluate expressions.