Precalculus Flashcards: Special Triangles Unit Circle In Trigonometry

What this deck covers

This deck focuses on Special Triangles Unit Circle In Trigonometry, giving you a quick way to review the definitions, rules, and examples that matter most for Precalculus.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

Flashcard 1: What are the side ratios in a $\frac{\pi}{4}$-$\frac{\pi}{4}$-$\frac{\pi}{2}$ triangle?

Answer: $1:1:\sqrt{2}$. 45-45-90 triangle has equal legs and hypotenuse = leg × $\sqrt{2}$.

Flashcard 2: What identity expresses $\cos(2\pi-x)$ in terms of $\cos(x)$?

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

Flashcard 3: What is $\sin(2\pi-x)$ in terms of $\sin(x)$?

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

Flashcard 4: What identity expresses $\sin(\pi-x)$ in terms of $\sin(x)$?

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

Flashcard 5: What are $\sin(\pi\/6)$, $\cos(\pi\/6)$, and $\tan(\pi\/6)$?

Answer: $\sin(\pi/6)=\frac{1}{2}$, $\cos(\pi/6)=\frac{\sqrt{3}}{2}$, $\tan(\pi/6)=\frac{\sqrt{3}}{3}$. From 30-60-90 triangle: opposite/hyp = 1/2, adjacent/hyp = $\sqrt{3}$/2.

Flashcard 6: What is $\sin\left(\frac{\pi}{3}\right)$?

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

Flashcard 7: What identity expresses $\tan(\pi-x)$ in terms of $\tan(x)$?

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

Flashcard 8: What is $\sin(\pi-x)$ in terms of $\sin(x)$?

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

Flashcard 9: What is $\sin\left(\frac{\pi}{6}\right)$?

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

Flashcard 10: What is the unit-circle coordinate for angle $\frac{\pi}{4}$?

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

Flashcard 11: What are $\sin(\pi\/4)$, $\cos(\pi\/4)$, and $\tan(\pi\/4)$?

Answer: $\sin(\pi/4)=\frac{\sqrt{2}}{2}$, $\cos(\pi/4)=\frac{\sqrt{2}}{2}$, $\tan(\pi/4)=1$. From 45-45-90 triangle: both legs equal, so sine = cosine = $\frac{1}{\sqrt{2}}$.

Flashcard 12: What are $\sin(\pi\/3)$, $\cos(\pi\/3)$, and $\tan(\pi\/3)$?

Answer: $\sin(\pi/3)=\frac{\sqrt{3}}{2}$, $\cos(\pi/3)=\frac{1}{2}$, $\tan(\pi/3)=\sqrt{3}$. From 30-60-90 triangle: opposite/hyp = $\sqrt{3}$/2, adjacent/hyp = 1/2.

Flashcard 13: What identity expresses $\cos(\pi-x)$ in terms of $\cos(x)$?

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

Flashcard 14: What is $\tan(2\pi-\pi\/4)$?

Answer: $-1$. $2\pi - \pi/4 = 7\pi/4$; using $\tan(2\pi-x)=-\tan(x)$ gives $-\tan(\pi/4)=-1$.

Flashcard 15: What is $\cos\left(\frac{\pi}{4}\right)$?

Answer: $\frac{\sqrt{2}}{2}$. At $45°$, cosine equals adjacent/hypotenuse = $1/\sqrt{2} = \sqrt{2}/2$.

Flashcard 16: What is $\cos\left(\frac{\pi}{6}\right)$?

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

Flashcard 17: What identity expresses $\tan(\pi+x)$ in terms of $\tan(x)$?

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

Flashcard 18: What is $\cos(\pi+x)$ in terms of $\cos(x)$?

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

Flashcard 19: What is $\tan(\pi-x)$ in terms of $\tan(x)$?

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

Flashcard 20: What is $\tan(2\pi-x)$ in terms of $\tan(x)$?

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

Flashcard 21: What is $\cos(\pi-x)$ in terms of $\cos(x)$?

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

Flashcard 22: What is the unit-circle coordinate for angle $\frac{\pi}{3}$?

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

Flashcard 23: What is $\sin(2\pi-\pi\/3)$?

Answer: $-\frac{\sqrt{3}}{2}$. $2\pi - \pi/3 = 5\pi/3$; using $\sin(2\pi-x)=-\sin(x)$ gives $-\sin(\pi/3)$.

Flashcard 24: What is the unit-circle definition of $\sin(x)$ and $\cos(x)$ using the point $P(x)$?

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

Flashcard 25: What is $\tan\left(\frac{\pi}{3}\right)$?

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

Flashcard 26: What is $\tan\left(\frac{\pi}{4}\right)$?

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

Flashcard 27: What is $\cos(2\pi-x)$ in terms of $\cos(x)$?

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

Flashcard 28: What identity expresses $\tan(2\pi-x)$ in terms of $\tan(x)$?

Answer: $\tan(2\pi-x)=-\tan(x)$. Since sine negates and cosine stays same, tangent negates.

Flashcard 29: What is $\sin\left(\frac{\pi}{4}\right)$?

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

Flashcard 30: What identity expresses $\sin(\pi+x)$ in terms of $\sin(x)$?

Answer: $\sin(\pi+x)=-\sin(x)$. Rotation by $\pi$ negates both coordinates, so sine becomes negative.

Flashcard 31: What are the side ratios of a $30^ -60^ -90^ $ triangle with shortest leg $1$?

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

Flashcard 32: What identity expresses $\sin(2\pi-x)$ in terms of $\sin(x)$?

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

Flashcard 33: What is $\tan\left(\frac{\pi}{6}\right)$?

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

Flashcard 34: What is $\sin(\pi+x)$ in terms of $\sin(x)$?

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

Flashcard 35: What are the side ratios in a $\frac{\pi}{6}$-$\frac{\pi}{3}$-$\frac{\pi}{2}$ triangle?

Answer: $1:\sqrt{3}:2$. 30-60-90 triangle has sides opposite to 30°:60°:90° in ratio 1:$\sqrt{3}$:2.

Flashcard 36: What is $\tan(\pi+x)$ in terms of $\tan(x)$?

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

Flashcard 37: What is $\sin(\pi-\pi\/6)$?

Answer: $\frac{1}{2}$. $\pi - \pi/6 = 5\pi/6$; using $\sin(\pi-x)=\sin(x)$ gives $\sin(\pi/6)=1/2$.

Flashcard 38: What identity expresses $\cos(\pi+x)$ in terms of $\cos(x)$?

Answer: $\cos(\pi+x)=-\cos(x)$. Rotation by $\pi$ negates both coordinates, so cosine becomes negative.

Flashcard 39: What is $\cos\left(\frac{\pi}{3}\right)$?

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

Flashcard 40: What is the unit-circle coordinate for angle $\frac{\pi}{6}$?

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

Flashcard 41: What are the side ratios of a $45^ -45^ -90^ $ triangle with legs $1$ and $1$?

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