How is $\sin 45^\circ=\frac{1}{\sqrt 2}$?

Question

I've been reading about the proof of $\sin 45^\circ=\dfrac{1}{\sqrt 2}$ in my book. They did it as following, let $\triangle ABC$ be an isosceles triangle as shown,

Since the triangle is isosceles with base and perpendicular equal the opposite angle $\angle C$ and $\angle B$ must be equal. Putting $\angle A=90^\circ$ and using $\angle A + \angle B + \angle C = 180^0$, we get $\angle A= \angle B= 45^\circ$. Now using Pythagoras theorem gives $\sin 45^\circ = \dfrac {1}{\sqrt 2}$. I am satisfied with this explanation. But on https://proofwiki.org they used a square to prove this formula as shown:

In that proof over proofwiki they say that the $\angle A$ is $45^\circ$ because the diagonal $AC$ is the bisector of $\angle A$. They do not use the property of isosceles triangle. My questions are,

• Why is the diagonal of a square bisector of the angle from which it originates? How can we prove this mathematically?
• Are there some other geometric proofs of $\sin 45^\circ=\dfrac{1}{\sqrt 2}$?

Answer 1

You can prove that the two isosceles triangles are congruent by SSS, and hence the angles adding up to 90 are equal.

Answer 2

The angle bisector is exactly the line of points equidistant from the two lines it bisects. The distance of $C$ to $AB$ is $BC$ and the distance of $C$ to $AD$ is $CD$. Since $ABCD$ is a square, $BC=CD$ and thus $C$ lies on the angle bisector. Now since the angle bisector also goes through $A$, it must be exactly the line $AC$.

Concerning other proofs, here's another, albeit similar, proof: It is clear from the same construction that $\sin{45^\circ} = \cos{45^\circ}$. Now since

$$\left(\sin\theta\right)^2+\left(\cos\theta\right)^2=1$$

we get:

$$2\left(\sin{45^\circ}\right)^2=1\Rightarrow\sin{45^{\circ}}=\frac{1}{\sqrt2}$$

observing that it must be positive.

Answer 3

Why is the diagonal of a square bisector of the angle from which it originates? How can we prove this mathematically?

Let $ABCD$ be a square. We know that two triangles $\triangle{ABD},\triangle{CBD}$ are congruent because $AB=CB,AD=CD,\angle{BAD}=\angle{BCD}$. Now we have $$\angle{ADB}=\angle{CDB}.$$

Answer 4

It's quite elementary, my dear Watson. $ABCD$ is a square. More generally, it's a rhombus, and the following applies to that also: $AB ≅ AD$, $BC ≅ DC$, $CA ≅ CA$, thus $△ABC ≅ △ADC$; and from this, it follows $∠BAC = ∠DAC$.

This proves that the $90^∘$ right angle $∠BAD$ is bisected, thereby showing that both $∠BAC$ and $∠DAC$ are both $45^∘$ angles.

You would do better to consider the fact that since $$\cos^2 x + \sin^2 x = 1, \quad \cos^2 x - \sin^2 x = \cos{2x},$$ then $$2\cos^2 x = 1 + \cos{2x},$$ from which it follows that $$\cos x = ±\sqrt{\frac{1 + \cos{2x}}{2}},$$ and likewise when $x$ is divided by $2$: $$\cos \frac{x}{2} = ±\sqrt{\frac{1 + \cos{2x}}{2}},$$ For angles $x$ between $0^∘$ and $90^∘$, inclusive, $\cos x ≥ 0$, so that we always have $$\cos \frac{x}{2} = +\sqrt{\frac{1 + \cos{2x}}{2}} = \frac{1}{2}\sqrt{2 + 2\cos{2x}}.$$

Therefore, we can iterate and interpolate, starting with $$\cos 0^∘ = 1, \quad \cos 90^∘ = 0.$$

In the first round, we get: $$\cos 45^∘ = \frac{1}{2}\sqrt{2 + 2·0} = \frac{1}{2}\sqrt{2}.$$

In the second round, making use of complements ($\cos{135^∘} = -\cos{45^∘}$), we get: $$\cos 22½^∘ = \frac{1}{2}\sqrt{2 + \sqrt{2}}, \quad \cos 67½^∘ = \frac{1}{2}\sqrt{2 - \sqrt{2}}.$$

In the third round, again making use of complements ($\cos 112½^∘ = -\cos 67½^∘$ and $\cos 157½^∘ = -\cos 22½^∘$), we get: $$ \cos 11¼^∘ = \frac{1}{2}\sqrt{2 + \sqrt{2 + \sqrt{2}}}, \quad \cos 33¾^∘ = \frac{1}{2}\sqrt{2 + \sqrt{2 - \sqrt{2}}},\\ \cos 56¼^∘ = \frac{1}{2}\sqrt{2 - \sqrt{2 - \sqrt{2}}}, \quad \cos 78¾^∘ = \frac{1}{2}\sqrt{2 - \sqrt{2 + \sqrt{2}}}. $$

In the fourth round, once more making use of complements ($\cos 101¼^∘ = -\cos 78¾^∘$, $\cos 123¾^∘ = -\cos 56¼^∘$, $\cos 146¼^∘ = -\cos 33¾^∘$ and $\cos 168¾^∘ = -\cos 11¼^∘$), we get: $$ \cos 5⅝^∘ = \frac{1}{2}\sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2}}}}, \quad \cos 16⅞^∘ = \frac{1}{2}\sqrt{2 + \sqrt{2 + \sqrt{2 - \sqrt{2}}}},\\ \cos 28⅛^∘ = \frac{1}{2}\sqrt{2 + \sqrt{2 - \sqrt{2 - \sqrt{2}}}}, \quad \cos 39⅜^∘ = \frac{1}{2}\sqrt{2 + \sqrt{2 - \sqrt{2 + \sqrt{2}}}},\\ \cos 50⅝^∘ = \frac{1}{2}\sqrt{2 - \sqrt{2 - \sqrt{2 + \sqrt{2}}}}, \quad \cos 61⅞^∘ = \frac{1}{2}\sqrt{2 - \sqrt{2 - \sqrt{2 - \sqrt{2}}}},\\ \cos 73⅛^∘ = \frac{1}{2}\sqrt{2 - \sqrt{2 + \sqrt{2 - \sqrt{2}}}}, \quad \cos 84⅜^∘ = \frac{1}{2}\sqrt{2 - \sqrt{2 + \sqrt{2 + \sqrt{2}}}}. $$

And so on...

It goes without saying that $0^∘$ and $90^∘$ fit within this scheme with: $$ 1 = \cos 0^∘ = \frac{1}{2}\sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + ⋯}}}}},\\ 0 = \cos 90^∘ = \frac{1}{2}\sqrt{2 - \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + ⋯}}}}}. $$