√(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 ...
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)
3 answers
Can you please Answer c) please. That is my main problem
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.
1 answer