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Ok, I have a few problems I just cant seem to understand.
1. Find the Exact Value of the Expression
Cos(tan^-1(4/3))
I know your supposed to write tangent in terms of cosine but im completely lost.
2. Find all exact solutions (in simplest form) to the following equations.
a) 2cosxsinx +cosx = 0
b)sin2x = tanx
I tried on each, and failed miserably on each. Any help is appericiated.
By definition, tan^(-1)(4/3) = (theta) means that tan(theta) = 4/3. So draw a right triangle, label the one angle as "theta", and fill in the "opposite" and "adjacent" values. Use the Pythagorean Theorem to find the length of the remaining side. Then take the cosine.Quote:
Originally Posted by Byud
How would you start the solution to 2XY + X = 0? Do the same factorization here.Quote:
Originally Posted by Byud
Convert the left-hand side to sines and cosines, using the double-angle identity they gave you. Convert the right-hand side to sines and cosines using the definition of tangent. See where that leads.... (Wink)Quote:
Originally Posted by Byud
How so? (You might have been "nearly there", but until we can see your work, we just can't know where you're needing the help.)Quote:
Originally Posted by Byud
:D
$\displaystyle y = \tan^{-1}\left(\frac{4}{3}\right)$Quote:
Originally Posted by Byud
$\displaystyle \tan{y} = \frac{4}{3} = \frac{opp}{adj}$ ... hypotenuse = 5
$\displaystyle \cos{y} = \frac{adj}{hyp} = \frac{3}{5}$
solutions $\displaystyle 0 \leq x < 2\pi$
$\displaystyle 2\cos{x}\sin{x} + \cos{x} = 0$
$\displaystyle \cos{x}(2\sin{x} + 1) = 0$
$\displaystyle \cos{x} = 0$ ... $\displaystyle x = \frac{\pi}{2}$ , $\displaystyle x = \frac{3\pi}{2}$
$\displaystyle \sin{x} = -\frac{1}{2}$ ... $\displaystyle x = \frac{7\pi}{6}$ , $\displaystyle x = \frac{11\pi}{6}$
$\displaystyle \sin(2x) = \tan{x}$
$\displaystyle 2\sin{x}\cos{x} = \frac{\sin{x}}{\cos{x}}$
$\displaystyle 2\sin{x}\cos{x} - \frac{\sin{x}}{\cos{x}} = 0$
$\displaystyle \frac{2\sin{x}\cos^2{x}}{\cos{x}} - \frac{\sin{x}}{\cos{x}} = 0$
$\displaystyle 2\sin{x}\cos^2{x} - \sin{x} = 0$
$\displaystyle \sin{x}(2\cos^2{x} - 1) = 0$
$\displaystyle \sin{x} = 0$ ... $\displaystyle x = 0$ , $\displaystyle x = \pi$
$\displaystyle \cos{x} = \pm \frac{\sqrt{2}}{2}$ ... $\displaystyle x = \frac{\pi}{4}$ , $\displaystyle x = \frac{3\pi}{4}$ , $\displaystyle x = \frac{5\pi}{4}$ , $\displaystyle x = \frac{7\pi}{4}$
Damn, that seems much easier now, I didnt even thinking about factoring out the trig functions, I was just trying to use those damned double angle formulas.
I did the first one the harder way, but I guess I did get it right!
my solution was...
Cos(tan^-1(4/3))
tan^-1(4/3) = u
$\displaystyle \cos{u} = \frac{1}{\sqrt{1+\tan{u^2}}}$
$\displaystyle \cos{u} = \frac{1}{\sqrt{1+\frac{4}{3}^2}}$
$\displaystyle \cos{u} = \frac{1}{\sqrt{1+\frac{16}{9}}}$
$\displaystyle \cos{u} = \frac{1}{\sqrt{\frac{25}{9}}}$
$\displaystyle \cos{u} = \frac{1}{\frac{5}{3}}$
$\displaystyle \cos{u} = \frac{3}{5}$
CHEER!
