Special Triangles, Unit Circle in Trigonometry - Pre-Calculus

$\cos(2\pi-x)=\cos(x)$. Angle $2\pi-x$ is in quadrant IV, where cosine is positive.

$\sin(\pi+x)=-\sin(x)$. Adding $\pi$ reflects across origin, changing sine's sign.

$\frac{\sqrt{2}}{2}$. At $45°$, sine equals opposite/hypotenuse = $1/\sqrt{2} = \sqrt{2}/2$.

$\sqrt{3}$. At $60°$, tangent equals opposite/adjacent = $\sqrt{3}/1 = \sqrt{3}$.

$\tan(2\pi-x)=-\tan(x)$. Since $\sin$ is negative and $\cos$ is positive, $\tan$ is negative.

$1:1:\sqrt{2}$. In a 45-45-90 triangle, the hypotenuse equals leg times $\sqrt{2}$.

$1:\sqrt{3}:2$. In a 30-60-90 triangle, sides are in ratio short leg : long leg : hypotenuse.

$\frac{1}{2}$. At $30°$, sine equals opposite/hypotenuse = $1/2$ in the 30-60-90 triangle.

$\frac{\sqrt{3}}{2}$. At $30°$, cosine equals adjacent/hypotenuse = $\sqrt{3}/2$ in the 30-60-90 triangle.

$\frac{\sqrt{3}}{3}$. At $30°$, tangent equals opposite/adjacent = $1/\sqrt{3} = \sqrt{3}/3$.

$\frac{\sqrt{3}}{2}$. At $60°$, sine equals opposite/hypotenuse = $\sqrt{3}/2$ in the 30-60-90 triangle.

$\cos(\pi-x)=-\cos(x)$. Supplementary angles have opposite cosine values in quadrants I and II.

$\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)$. Unit circle point at $45°$ has coordinates $(\cos 45°, \sin 45°)$.

$\frac{1}{2}$. At $60°$, cosine equals adjacent/hypotenuse = $1/2$ in the 30-60-90 triangle.

$\cos(\pi+x)=-\cos(x)$. Adding $\pi$ reflects across origin, changing cosine's sign.

$\left(\frac{\sqrt{3}}{2},\frac{1}{2}\right)$. Unit circle point at $30°$ has coordinates $(\cos 30°, \sin 30°)$.

$\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)$. Unit circle point at $60°$ has coordinates $(\cos 60°, \sin 60°)$.

$\sin(\pi-x)=\sin(x)$. Supplementary angles have the same sine value in quadrants I and II.

$\tan(\pi+x)=\tan(x)$. Since both $\sin$ and $\cos$ change sign, $\tan$ stays the same.

$\tan(\pi-x)=-\tan(x)$. Since $\sin$ stays same and $\cos$ changes sign, $\tan$ changes sign.

$\sin(2\pi-x)=-\sin(x)$. Angle $2\pi-x$ is in quadrant IV, where sine is negative.

$1$. At $45°$, tangent equals opposite/adjacent = $1/1 = 1$.

$\cos(2\pi-x)=\cos(x)$. Clockwise rotation by $x$ from $2\pi$ keeps x-coordinate (cosine) same.

$\sin(\pi-x)=\sin(x)$. Reflection across y-axis keeps y-coordinate (sine) unchanged.

$P(x)=(\cos(x),\sin(x))$. Point at angle $x$ on unit circle has coordinates (cos, sin).

$\cos(\pi-x)=-\cos(x)$. Reflection across y-axis negates x-coordinate (cosine).

$\tan(\pi-x)=-\tan(x)$. Since $\tan = \sin/\cos$ and cosine changes sign, tangent negates.

$\sin(2\pi-x)=-\sin(x)$. Clockwise rotation by $x$ from $2\pi$ negates y-coordinate (sine).

$\tan(\pi+x)=\tan(x)$. Both sine and cosine negate, so their ratio remains unchanged.