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)
√(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 ...
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.
a) To express the equation in the form r cos(theta + alpha), we need to find the values of r and alpha.
Let's start by rewriting f(theta) = 4cos theta - 6sin theta:
f(theta) = r cos(theta + alpha)
We can rewrite r cos(theta + alpha) as r(cos theta cos alpha - sin theta sin alpha).
Comparing the terms, we have:
4cos theta = r cos alpha
-6sin theta = -r sin alpha
Dividing the second equation by -6, we get:
sin theta = (r/6) sin alpha
Dividing the first equation by 4, we get:
cos theta = (r/4) cos alpha
We can now square both equations and add them to eliminate alpha:
(sin theta)^2 + (cos theta)^2 = [(r/6) sin alpha]^2 + [(r/4) cos alpha]^2
Using the trigonometric identity sin^2 theta + cos^2 theta = 1, we have:
1 = (r^2/36)sin^2 alpha + (r^2/16)cos^2 alpha
Now, we can use the identity sin^2 alpha + cos^2 alpha = 1 to simplify the equation:
1 = (r^2/36)(1-cos^2 alpha) + (r^2/16)cos^2 alpha
Simplifying further, we get:
1 = (r^2/36) - (r^2/36)cos^2 alpha + (r^2/16)cos^2 alpha
Combining like terms, we have:
1 = (r^2/36) + [(r^2/16)-(r^2/36)]cos^2 alpha
To eliminate the cos^2 alpha term, we need the coefficients of r^2/16 and r^2/36 to be equal:
(r^2/16) = (r^2/36)
36r^2 = 16r^2
r^2 = 0
Since r is the magnitude of the coefficient of the trigonometric terms, it cannot be zero. Therefore, there is no solution in the form r cos(theta + alpha) for f(theta) = 4cos theta - 6sin theta.
b) Since there is no solution in the required form, we need to find the general solution of the equation directly.
The equation is 4cos theta - 6sin theta = 5.
To find the general solution, we need to rewrite it in terms of sin and cos:
4cos theta - 6sin theta = 5
Dividing the equation by sqrt(4^2 + (-6)^2) = sqrt(52) = 2sqrt(13), we have:
(4/2sqrt(13))cos theta - (6/2sqrt(13))sin theta = 5/(2sqrt(13))
Simplifying further, we get:
(2/2sqrt(13))cos theta - (3/2sqrt(13))sin theta = 5/(2sqrt(13))
Now, we can rewrite cos theta and sin theta as sin(theta + alpha) and cos(theta + alpha):
sin(theta + alpha) = (2/2sqrt(13))
cos(theta + alpha) = -(3/2sqrt(13))
Dividing the equations, we get:
tan(alpha) = (3/2)/(2/2) = 3/2
Using the inverse tangent function, we can find alpha:
alpha = arctan(3/2)
Therefore, the general solution of the equation 4cos theta - 6sin theta = 5 is:
theta = -alpha + 2kπ, where k is an integer.
c) To find the minimum value of the function 1/4 + f(theta), we need to substitute the general solution of the equation into the function and minimize it.
Substituting the general solution theta = -alpha + 2kπ into the function, we have:
1/4 + f(theta) = 1/4 + (4cos theta - 6sin theta)
Substituting -alpha + 2kπ for theta, we get:
1/4 + (4cos (-alpha + 2kπ) - 6sin (-alpha + 2kπ))
Simplifying further, we have:
1/4 + 4cos (-alpha + 2kπ) + 6sin (-alpha + 2kπ)
Using the multiple-angle identity for cosine and sine, we have:
1/4 + 4(cos(-alpha)cos(2kπ) + sin(-alpha)sin(2kπ)) + 6(sin(-alpha)cos(2kπ) - cos(-alpha)sin(2kπ))
Since cos(2kπ) = 1 and sin(2kπ) = 0 for all integers k, the equation becomes:
1/4 + 4(cos(-alpha) + 0) + 6(0 - sin(-alpha))
Simplifying further, we have:
1/4 + 4cos(-alpha) - 6sin(-alpha)
Using the trigonometric identity cos(-alpha) = cos alpha and sin(-alpha) = -sin alpha, we get:
1/4 + 4cos alpha + 6sin alpha
Since we want to find the minimum value of the function, we can take the derivative with respect to alpha and set it to zero:
d/dalpha (1/4 + 4cos alpha + 6sin alpha) = 0
Simplifying the derivative, we have:
-4sin alpha + 6cos alpha = 0
Dividing by 2, we get:
-2sin alpha + 3cos alpha = 0
Dividing by cos alpha, we have:
(-2sin alpha + 3cos alpha)/cos alpha = 0
Simplifying further, we get:
-2tan alpha + 3 = 0
Rearranging the equation, we have:
tan alpha = 3/2
Using the inverse tangent function, we can find alpha:
alpha = arctan(3/2)
Now, we can substitute alpha into the function to find the minimum value:
Minimum value = 1/4 + 4cos alpha + 6sin alpha
Substituting the value of alpha, we have:
Minimum value = 1/4 + 4cos arctan(3/2) + 6sin arctan(3/2)
Since cos arctan(3/2) = 2/√13 and sin arctan(3/2) = 3/√13, the minimum value becomes:
Minimum value = 1/4 + 4(2/√13) + 6(3/√13)
Simplifying further, we have:
Minimum value = 1/4 + 8/√13 + 18/√13
To rationalize the denominators, we multiply both numerator and denominator by √13:
Minimum value = (1/4)(√13/√13) + (8/√13)(√13/√13) + (18/√13)(√13/√13)
Simplifying, we get:
Minimum value = (√13 + 8√13 + 18√13)/13
Combining like terms, we have:
Minimum value = 27√13/13