Show that B = √5 and ѳ = tan ⁻¹ ½

Joel · #1 $Old$ November 5th, 2009, 02:30 AM

Hi All,
Been given this one, and honestly I am looking forward to seeing you guys tackle it.
Consider the equation, 2 cosx + sinx = 1
Let 2cosx + sinx = B(cos(x – ѳ), where B>0 and 0 < ѳ < 360.
Show that B = √5 and ѳ = tan ⁻¹ ½

Thanks Heaps
Joel

HallsofIvy · #2 $Old$ November 5th, 2009, 03:42 AM

Quote:

Originally Posted by Joel $View Post$

Hi All,
Been given this one, and honestly I am looking forward to seeing you guys tackle it.
Consider the equation, 2 cosx + sinx = 1
Let 2cosx + sinx = B(cos(x – ѳ), where B>0 and 0 < ѳ < 360.
Show that B = √5 and ѳ = tan ⁻¹ ½

Thanks Heaps
Joel

Use the "difference formula": cos(a-b)= cos(a)cos(b)+ sin(a)sin(b). $cos(x-\theta)= cos(\theta)cos(x)+ sin(\theta)sin(x)$ so we would like to have $cos(\theta)= 2$ , $sin(\theta)= 1$ .

But that's impossible because that would mean that $cos^2(\theta)+ sin^2(\theta)= 2^2+ 1= 5$ when we know it must be 1. So factor out a $\sqrt{5}$ :

$\sqrt{5}(\frac{2}{\sqrt{5}}cos(x)+ \frac{1}{\sqrt{5}}sin(x))= \sqrt{5}(cos(\theta)cos(x)+ sin(\theta)sin(x))$ if and only if $cos(\theta)= \frac{2}{\sqrt{5}}$ and $sin(\theta)= \frac{1}{\sqrt{5}}$ . Now we have $cos^2(\theta)+ sin^2(\theta)= \frac{4}{5}+ \frac{1}{5}= 1$ so that is possible.

And, of course, we must have $tan(\theta)= \frac{sin(\theta)}{cos(\theta)}$ $= \frac{ \frac{1}{\sqrt{5}} }{ \frac{2}{\sqrt{5}} }= \frac{1}{2}$ .

Debsta · #3 $Old$ November 5th, 2009, 03:48 AM

Recall that cos(x-A) = cosx cos A + sin x sinA
Then B cos(x-A) = B cosA cosx + B sinA sinx
This equals 2 cos x + sin x, so B cosA = 2 and B sin A =1 (equating the terms)
Then you have 2 simul equations to solve for B and A. (to give B= sqr 5 and A = tan^-1(1/2)