Simpler Derivation of $\sin \frac{\pi}{4} = \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$,

Question

As far I've studied the Basic Trigonometry in School, those are below - $$ \frac{1}{\sin \theta} = \csc \theta$$ $$\frac{1}{\cos \theta} = \sec \theta$$ $$\frac{1}{\tan \theta} = \cot \theta$$

And Angle Relations like -

$$\sin \theta = cos(90 - \theta)$$ $$\tan \theta = \cot (90 - \theta)$$ $$\sec \theta = \csc(90 - \theta)$$

And Vice-versa,

And few Trigonometry ratios,

like - $$\sin ^2 \theta + \cos ^2 \theta = 1$$ $$\sec ^2 \theta - \tan ^2 \theta = 1$$ $$\csc^2 \theta - \cot ^2 \theta = 1$$

Now, to prove - $\sin \frac{\pi}{4} = \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$

I've no clue what's going on, Why Right Angled Triangles have $\pi$ involved in them and What is the relation between a Right Angled Triangle and a Circle (constant ratio of $\frac{ circumference}{diameter}$).

As far I've understood the question, it says that For a right Angled Triangle, having a angle = $\frac{\pi}{4} = $0.78539 (approx.),

gets the ratio of Side Opposite to $\theta$ and Hypotenuse and the ratio of Side Adjacent to $\theta$ and Hypotenuse = $\frac{1}{\sqrt{2}}$

Also, If it is correct, then Can I calculate the Value of $\pi$ without Drawing Circles and measuring the Diameter? (mean fully theoretical way?)

I've found some similar links like this - real analysis - how do i prove that $\sin(\pi/4)=\cos(\pi/4)$? - Mathematics Stack Exchange

But the proof was too more advanced for me to Understand

Thanks in Advance!

Answer 1

The better way to show this is by definition of trigonometric circle.

Notably for an angle of 45 degrees $cos\theta$ is the side of a square with diagonal with length equal to 1 thus it’s equal to $\frac{\sqrt{2}}{2}$.

To better visualize take a look to the following figure:

Answer 2

You have a right triangle with an angle given to be $\pi /4$.

You know that the angle sum of every triangle is $\pi $

With a right angle and an angle of $\pi /4$ you have counted for $ 3\pi /4$ out of $\pi$. The remaining $\pi /4$ implies that your triangle is an isosceles triangle. Now you have symmetry and the rest is obvious.

Answer 3

When you are considering this consider a geometrical proof. Consider a $\triangle ABC$ in which $\angle B$ is $90^\circ$ and $AB = BC$.

Now suppose $BC= AB = a$ Then by Pythagoras theorem, $AC = AB^2 + BC^2$ = $a^2+ a^2 = 2a^2$ and thus $AC = a\sqrt2$

Now using the ratio's $sin\frac {\pi}{4}$ = $sin 45^\circ$ = $\frac{BC}{AC}$ = $\frac{a}{a\sqrt2}$ = $\frac{1}{\sqrt2}$ = $cos\frac{\pi}{4}$.

And you are done.