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). so we would like to have , .
But that's impossible because that would mean that when we know it must be 1. So factor out a :
if and only if and . Now we have so that is possible.
And, of course, we must have .
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)