Inverse Expressions and Finding the exact solution

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March 18th 2009, 02:33 PM
Byud

Inverse Expressions and Finding the exact solution

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.
March 18th 2009, 03:04 PM
stapel

Quote:

Originally Posted by Byud

1. Find the Exact Value of the Expression
Cos(tan^-1(4/3))

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

2. Find all exact solutions (in simplest form) to the following equations.

a) 2cosxsinx +cosx = 0

How would you start the solution to 2XY + X = 0? Do the same factorization here.

Quote:

Originally Posted by Byud

b)sin2x = tanx

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

I tried on each, and failed miserably on each.

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.)

:D
March 18th 2009, 03:10 PM
skeeter

Quote:

Originally Posted by Byud

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.

$y = \tan^{-1}\left(\frac{4}{3}\right)$

$\tan{y} = \frac{4}{3} = \frac{opp}{adj}$ ... hypotenuse = 5

$\cos{y} = \frac{adj}{hyp} = \frac{3}{5}$

solutions $0 \leq x < 2\pi$

$2\cos{x}\sin{x} + \cos{x} = 0$

$\cos{x}(2\sin{x} + 1) = 0$

$\cos{x} = 0$ ... $x = \frac{\pi}{2}$ , $x = \frac{3\pi}{2}$

$\sin{x} = -\frac{1}{2}$ ... $x = \frac{7\pi}{6}$ , $x = \frac{11\pi}{6}$

$\sin(2x) = \tan{x}$

$2\sin{x}\cos{x} = \frac{\sin{x}}{\cos{x}}$

$2\sin{x}\cos{x} - \frac{\sin{x}}{\cos{x}} = 0$

$\frac{2\sin{x}\cos^2{x}}{\cos{x}} - \frac{\sin{x}}{\cos{x}} = 0$

$2\sin{x}\cos^2{x} - \sin{x} = 0$

$\sin{x}(2\cos^2{x} - 1) = 0$

$\sin{x} = 0$ ... $x = 0$ , $x = \pi$

$\cos{x} = \pm \frac{\sqrt{2}}{2}$ ... $x = \frac{\pi}{4}$ , $x = \frac{3\pi}{4}$ , $x = \frac{5\pi}{4}$ , $x = \frac{7\pi}{4}$
March 18th 2009, 03:44 PM
Byud

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

$\cos{u} = \frac{1}{\sqrt{1+\tan{u^2}}}$

$\cos{u} = \frac{1}{\sqrt{1+\frac{4}{3}^2}}$

$\cos{u} = \frac{1}{\sqrt{1+\frac{16}{9}}}$

$\cos{u} = \frac{1}{\sqrt{\frac{25}{9}}}$

$\cos{u} = \frac{1}{\frac{5}{3}}$

$\cos{u} = \frac{3}{5}$
CHEER!