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

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

• Consider a right-angled isosceles triangle. – Lord Shark the Unknown Jan 17 '18 at 15:08
• @LordSharktheUnknown yep, then what? I 've tried that way also, but that didn't work for me – Abhas Kumar Sinha Jan 17 '18 at 15:11
• You are not being consistent. You start by saying that you know that $\sin(90^\circ- x)=\cos x$, but then you link an answer that uses the very same formula saying that it is too advanced. – user228113 Jan 17 '18 at 15:12
• I didn't understand what it said, and it was graphical way to solve @G.Sassatelli – Abhas Kumar Sinha Jan 17 '18 at 15:13
• @G.Sassatelli $$\sin (90 - \theta) = \cos \theta$$ is not same as $$\sin(\frac{\pi}{2}) = \cos \theta$$ – Abhas Kumar Sinha Jan 17 '18 at 15:16

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:

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.

• How $\pi$ = 180? Both are different, one is rational and other is irrational – Abhas Kumar Sinha Jan 17 '18 at 15:24
• You are correct \pi is not 180, because one of them is rational and the other one is irrational.. What we say is $\pi$ radian is 180 degrees. – Mohammad Riazi-Kermani Jan 17 '18 at 16:39
• sine ratios are usually taken in degrees, if they won't be taken in degrees then that won't work. So, how one is taken in radians? – Abhas Kumar Sinha Jan 18 '18 at 14:16
• @AbhasKumarSinha Yes, in real world the degree measure is more common. In mathematics they go with Radian which makes some formulas much easier. For example a right angle is 90 degrees and at the same time it is $\pi/4$ radians. Different modes for the same angle. – Mohammad Riazi-Kermani Jan 18 '18 at 14:27

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.