Asked by Becky
a) Express f(theta) = 4cos theta - 6sin theta in the form r cos(theta + alpha)
b) Hence find the general solution of the equation 4cos theta - 6sin theta =5
c) Hence, find the minimum value of the function 1/4+f( theta)
b) Hence find the general solution of the equation 4cos theta - 6sin theta =5
c) Hence, find the minimum value of the function 1/4+f( theta)
Answers
Answered by
oobleck
√(4^2+6^2) = √52
so you need
sin alpha = 4/√52 = 2/√13
cos alpha = 6/√52 = 3/√13
see what you can do with that ...
so you need
sin alpha = 4/√52 = 2/√13
cos alpha = 6/√52 = 3/√13
see what you can do with that ...
Answered by
Becky
Can you please Answer c) please. That is my main problem
Answered by
oobleck
c'mon, the whole point of part a was to prepare you to write
f(θ) = √52 (4/√52 cosθ - 6/√52 sinθ)
= √52 (sinα cosθ - cosα sinθ)
= √52 sin(α-θ)
You now know what the max/min values of f(θ) are, so you can easily answer part c.
f(θ) = √52 (4/√52 cosθ - 6/√52 sinθ)
= √52 (sinα cosθ - cosα sinθ)
= √52 sin(α-θ)
You now know what the max/min values of f(θ) are, so you can easily answer part c.