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Verification of triangles side length by Pythagoras theorem -

$1^{2} + 1^{2} = \sqrt{2}^{2}$ $\Rightarrow$ $1 + 1 = 2$ $\Rightarrow$ $LHS \ = \ RHS$ $\Rightarrow$ $Hence \ Verified$

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In the above picture, there is a right-isosceles triangle, with base angles of 45 and vertex angle of 90 degrees.

$\sin = \frac{opp}{hyp}$

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Angle opposite to a 45 degree angle is always 1, in the above triangle, and the hypotenuse is $\sqrt{2}$

$\sin(45^{\circ}) = \frac{opp}{hyp} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$

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The last simplification was done to remove the radical from the denominator, it can easily be done by multiplying both numerator and denominator by $\sqrt{2}$ .

$\frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{1 \times \sqrt{2}}{\sqrt{2}^{2}} = \frac{\sqrt{2}}{2}$

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$\textbf{\Huge TRIGONOMETRY SWAG!!!!!}$

It's a handy value to remember for many math problems. The value can be found by drawing out a Right-isosceles triangle and labeling the sides. With the mnemonic of SOH-CAH-TOA, we find $\sin{(45^{\circ})}$ as $\frac{1}{\sqrt{2}}$

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Same can be done to find $\sin{(30^{\circ})}$ or $\sin{(60^{\circ})}$ by drawing out an equilateral triangle and halving it into two right-angled-triangles