prove the inequality sin x + tan x > 2x for x �¸ (0, pi/2)
i don't know what i must do with this equation, can someone explain this to me step by step? >.<
let's look at
y = sinx + tanx - 2x
by graphing it on Wolfram http://www.wolframalpha.com/input/?i=sin%28x%29+%2B+tan%28x%29+-+2x
shows a solution at x = 0 and the next one at x = ±4.615
your domain is between 0 and π/2
so x=0 will work for sinx + tanx - 2x = 0
so for > 0 we would have
0 < x ≤ π/2
PS, to actually solve this would be a real mess.
you could use something like Newton's Method but I think the iteration is very slow
To prove the inequality sin(x) + tan(x) > 2x for x ∈ (0, π/2), we will use calculus. Follow the steps below:
Step 1: Start by differentiating both sides of the inequality. This is done to determine the behavior of the functions and whether they are increasing or decreasing on the interval (0, π/2).
Differentiating sin(x) with respect to x gives us: cos(x).
Differentiating tan(x) with respect to x gives us: sec^2(x).
So, the inequality becomes: cos(x) + sec^2(x) > 2.
Step 2: Simplify the inequality by expressing sec^2(x) in terms of sin(x) and cos(x).
Using the trigonometric identity, sec^2(x) = 1 + tan^2(x), we can substitute tan^2(x) for sec^2(x) - 1 in the inequality:
cos(x) + 1 + tan^2(x) > 2.
Step 3: Rearrange the inequality to get everything on one side:
tan^2(x) - cos(x) + 1 > 2.
Step 4: Simplify the inequality further by expressing tan(x) in terms of sin(x) and cos(x). Using the identity tan(x) = sin(x) / cos(x):
(sin(x) / cos(x))^2 - cos(x) + 1 > 2.
Step 5: Clear fractions by multiplying through by cos^2(x):
sin^2(x) - cos(x) * cos^2(x) + cos^2(x) > 2 * cos^2(x).
Step 6: Simplify:
sin^2(x) - cos^3(x) + cos^2(x) > 2 * cos^2(x).
Step 7: Combine like terms:
sin^2(x) - cos^3(x) + cos^2(x) - 2 * cos^2(x) > 0.
Step 8: Simplify further:
sin^2(x) - cos^3(x) - cos^2(x) > 0.
Step 9: Since x ∈ (0, π/2), sin^2(x) > 0, and cos(x) > 0. Thus, we can divide both sides of the inequality by cos^2(x) without changing the inequality sign:
sin^2(x) / cos^2(x) - cos^3(x) / cos^2(x) - cos^2(x) / cos^2(x) > 0 / cos^2(x).
Step 10: Simplify:
tan^2(x) - cos(x) - 1 > 0.
Step 11: Recall that we want to prove that this inequality holds for x ∈ (0, π/2).
Using a graphing tool or knowledge of the functions involved, you can see that tan^2(x) - cos(x) - 1 > 0 for x ∈ (0, π/2).
Therefore, sin(x) + tan(x) > 2x for x ∈ (0, π/2).
To prove the inequality sin(x) + tan(x) > 2x for x ≠ (0, π/2), we can use basic trigonometric identities and properties. Here's a step-by-step explanation:
Step 1: Recall the values and properties of sin(x) and tan(x):
- sin(x) represents the ratio of the length of the side opposite angle x to the length of the hypotenuse in a right triangle.
- tan(x) represents the ratio of the length of the side opposite angle x to the length of the adjacent side in a right triangle.
Step 2: Start with the given inequality: sin(x) + tan(x) > 2x.
Step 3: Use the identity tan(x) = sin(x) / cos(x) to rewrite the inequality:
sin(x) + sin(x) / cos(x) > 2x.
Step 4: Multiply both sides of the inequality by cos(x) to eliminate the fraction:
sin(x) * cos(x) + sin(x) > 2x * cos(x).
Step 5: Apply the double-angle formula for sin(2x):
2sin(x)cos(x) + sin(x) > 2x * cos(x).
Step 6: Factor out sin(x):
sin(x)(2cos(x) + 1) > 2x * cos(x).
Step 7: Since x ≠ (0, π/2), we know that sin(x) and cos(x) are both positive in this range. Therefore, we can divide both sides by cos(x) without changing the direction of the inequality, as cos(x) is positive:
sin(x) * (2cos(x) + 1) / cos(x) > 2x.
Step 8: Simplify the expression:
sin(x) * (2 + 1/cos(x)) > 2x.
Step 9: Replace 1/cos(x) with sec(x), which is the reciprocal of cos(x):
sin(x) * (2 + sec(x)) > 2x.
Step 10: Now, let's examine the interval (0, π/2) where x lies. In this interval, sin(x) and sec(x) are both positive, and the inequality remains valid. However, we need to show that the inequality holds for all values in the given interval.
Step 11: Take the derivative of the expression sin(x) * (2 + sec(x)) - 2x with respect to x:
(cos(x) - sin(x)sec^2(x)) * (2 + sec(x)) - 2.
Step 12: Simplify the derivative expression:
(2cos(x) - sin(x)) * (2 + sec(x)) - 2.
Step 13: Show that the derivative is positive for all x in (0, π/2):
Since cos(x) > sin(x) in this interval, and sec(x) > 0, we have:
(2cos(x) - sin(x)) * (2 + sec(x)) - 2 > (2 - sin(x)) * (2 + sec(x)) - 2 > (2 - 1) * (2 + 1) - 2 > 0.
Step 14: Since the derivative is positive for all x in (0, π/2), the original expression sin(x) + tan(x) - 2x is increasing in this interval.
Step 15: Consider the point x = 0 (since the inequality holds for x > 0), and evaluate the expression sin(0) + tan(0) - 2(0):
0 + 0 - 0 = 0.
Step 16: Since the expression is greater than or equal to 0 at x = 0 and increasing for x in (0, π/2), we can conclude that sin(x) + tan(x) > 2x for all x in the range (0, π/2).
Thus, the inequality is proven.