Show that for real x that {[cos x + 2 sin x + 1]/[cos x + sin x] } cannot have a value between 1 and 2.
Let y = [(cos x+2 sin x + 1)/(cos x + sin x) ]
y(cos x + sin x) = (cos x + 2 sin x + 1)
sin x(y-2) + cos x(y-1)=1 , I just feel that this isn't the way to do this..
Any other way to do this?
cosx+2sinx+1 = (cosx+sinx) + (sinx+1)
So, your quotient can be written
1+(1+sinx)/(cosx+sinx)
Now, since cosx < 1, cosx+sinx < 1+sinx
So, (1+sinx)/(cosx+sinx) > 1
and the rest follows.
Hope you accept the fact that the max of cosx + sinx is √2
and its minimum is -√2
These occur at x = 45° and x = 225° (π/4 and 5π/4)
see: http://www.wolframalpha.com/input/?i=y+%3D+sinx%2Bcosx
we can write it as
y = (cosx + sinx + sinx + 1)/(cosx + sinx)
= 1 + (sinx + 1)/(sinx + cosx)
the graph of y = (cos x+2 sin x + 1)/(cos x + sin x)
suggests that the statement is true
see: http://www.wolframalpha.com/input/?i=plot+y+%3D+(cos+x%2B2+sin+x+%2B+1)%2F(cos+x+%2B+sin+x)
we have critical values at x = -45°, 135°, 315° , causing asymptotes for those values. The denominator would be zero.
Look at the first graph
consider the intervals from
a) -45° to 135° , we know that cosx+sinx has a max of √2 at 45° and is positive for that interval
and y = 1 + (sinx + 1)/(sinx + cosx)
= 1 + (positive)
y>1
at 135°, the function is undefined, or y is infinitely large ----> y > 1
for x between 135 and 315, cosx + sinx has a min of -√2 and is negative throughout the interval.
let's take some samples for y = 1 + (sinx + 1)/(sinx + cosx)
x = 140 , y = -12.3 , which gives us y < 1
x = 180, y = -2 , which gives us y < 1
x = 200, y = .4886 , which gives us y < 1
x = 225, y = .79.. , y < 1
x = 310, y = -.89.. , y < 1
according to my reasoning, at x = 320, we should get y > 1
x = 320 , y = 1 + (sin320 + 1)/(cos320+sin320) = 3.898.. , true enough
Looks like mine was overkill
Many thanks to both of you!
Yes, there is another way to prove that the expression cannot have a value between 1 and 2 for real values of x.
Let's start by assuming that there exists a real number x such that the expression:
y = [(cos x + 2 sin x + 1)/(cos x + sin x)]
has a value between 1 and 2. This means that 1 < y < 2.
First, notice that the denominator in the expression (cos x + sin x) cannot be zero, since division by zero is undefined. Therefore, cos x + sin x ≠ 0.
Now, let's consider the numerator (cos x + 2 sin x + 1). We can rewrite it as:
(cos x + sin x) + sin x + 1
Next, let's divide both the numerator and the denominator of the expression y by (cos x + sin x). This gives us:
y = 1 + (sin x + 1)/(cos x + sin x)
Now, observe that the range of values sin x can take is [-1, 1], and the range of values cos x can take is [−1, 1]. This implies that sin x + 1 and cos x + sin x are both limited to the range [0, 2].
Considering the expression (sin x + 1)/(cos x + sin x), we can conclude that it is always less than or equal to 1 for any real value of x, since the numerator is no greater than the denominator.
Therefore, we can rewrite y as:
y = 1 + [(sin x + 1)/(cos x + sin x)]
Since (sin x + 1)/(cos x + sin x) is always less than or equal to 1, we have:
y = 1 + [(sin x + 1)/(cos x + sin x)] ≤ 1 + 1 = 2
This implies that y is always less than or equal to 2.
Now, since we assumed that 1 < y < 2, and we have just shown that y ≤ 2, it follows that our assumption was incorrect. Therefore, the expression [(cos x + 2 sin x + 1)/(cos x + sin x)] cannot have a value between 1 and 2 for any real value of x.