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
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). $\displaystyle cos(x-\theta)= cos(\theta)cos(x)+ sin(\theta)sin(x)$ so we would like to have $\displaystyle cos(\theta)= 2$, $\displaystyle sin(\theta)= 1$.
But that's impossible because that would mean that $\displaystyle cos^2(\theta)+ sin^2(\theta)= 2^2+ 1= 5$ when we know it must be 1. So factor out a $\displaystyle \sqrt{5}$:
$\displaystyle \sqrt{5}(\frac{2}{\sqrt{5}}cos(x)+ \frac{1}{\sqrt{5}}sin(x))= \sqrt{5}(cos(\theta)cos(x)+ sin(\theta)sin(x))$ if and only if $\displaystyle cos(\theta)= \frac{2}{\sqrt{5}}$ and $\displaystyle sin(\theta)= \frac{1}{\sqrt{5}}$. Now we have $\displaystyle cos^2(\theta)+ sin^2(\theta)= \frac{4}{5}+ \frac{1}{5}= 1$ so that is possible.
And, of course, we must have $\displaystyle tan(\theta)= \frac{sin(\theta)}{cos(\theta)}$$\displaystyle = \frac{ \frac{1}{\sqrt{5}} }{ \frac{2}{\sqrt{5}} }= \frac{1}{2}$.
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)