Category Archives: Problem Solving and Proofs

The Proof of the Tangent Half-Angle Formula

In this post, we prove the following trigonometric identity:

\displaystyle \tan \frac{\theta}{2} = \frac{\sin\theta}{1 + \cos \theta} = \frac{1 - \cos \theta}{\sin \theta}.

Proof

Consider a semi-circle with “center” O and diameter AB and radius equal to 1 unit as shown below.  If we let \angle BOC =\theta, then by the Inscribed Angle Theorem, \angle CAB = \frac{\theta}{2}.

Draw CD perpendicular to OB as shown in the second figure. We can compute for the sine and cosine of \theta which equal to the lengths of CD and OD, respectively. In effect, BD = 1 - \cos \theta and AD = 1 + \cos \theta. Continue reading

Limit by epsilon-delta proof: Example 2

This is the overdelayed continuation of the discussion on the \epsilon-\delta definition of limits. In this post, we discuss another example.

Prove that the \lim_{x \to 2} x^2 = 4.

Recall that the definition states that the limit of f(x) = L as x approaches a if for all \epsilon > 0, however small, there exists a \delta > 0 such that if 0 < |x - a| < \delta, then |f(x) - L| < \epsilon.

From the example 1, we have learned that we should manipulate |f(x)-L=|x^2 - 4|, to make one of the expressions look like |x-a|=|x-2|. Solving,  we have

|f(x) - L| = |x^2 - 4| = |(x+2)(x-2)| = |x+2||x-2|.

Note that we have accomplished our goal, going back to the definition, this means that if 0 < x - 2 < \delta, then |x+2||x-2| < \epsilon.

Now, it is not possible to divide both sides by x + 2 (making it |x-2| < \frac{\epsilon}{|x+2|}) because x varies. This means that we have to find a constant k such that |x + 2| < k. Continue reading

More examples of proof by contradiction

We have had good discussions on mathematical proofs, so I am planning to create a mathematical proof series that will discuss the basics such as direct proof, indirect proof, and proof by mathematical induction.  But before I do that, let me continue with more examples of proof by contradiction.

Proof by contradiction, as we have discussed, is a proof strategy where you assume the opposite of a statement, and then find a contradiction somewhere in your proof. Finding a contradiction means that your assumption is false and therefore the statement is true. Below are several more examples of this proof strategy.

Example 1:  \sqrt{2} irrational.

Example 2: \sqrt{6} is irrational. The proof of this is basically the same as example 1, so it is left as an exercise.

Example 3: Proof that there are infinitely many primes.

Example 4: Knights and Liars

Example 5: \sqrt{2} + \sqrt{3} is irrational. Continue reading

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