Unit Circle Trigonometry Worksheets (Interesting Problems) - Go ... - Free Printable

Here are the step-by-step solutions for each problem on the worksheet. We will use the Unit Circle to find the exact values.

1. $\cos(\pi/2)$
* On the unit circle, the angle $\pi/2$ corresponds to the point $(0, 1)$.
* Cosine is the x-coordinate.
* The x-coordinate is $0$.
* Answer: 0

2. $\csc(\pi/4)$
* First, find $\sin(\pi/4)$. The angle $\pi/4$ is at point $(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$.
* Sine is the y-coordinate, so $\sin(\pi/4) = \frac{\sqrt{2}}{2}$.
* Cosecant ($\csc$) is the reciprocal of sine ($1/\sin$).
* $\csc(\pi/4) = \frac{1}{\frac{\sqrt{2}}{2}} = \frac{2}{\sqrt{2}}$.
* Rationalizing the denominator: $\frac{2\sqrt{2}}{2} = \sqrt{2}$.
* Answer: $\sqrt{2}$

3. $\tan(3\pi/4)$
* The angle $3\pi/4$ is in Quadrant II. The reference angle is $\pi/4$.
* In Quadrant II, tangent is negative.
* $\tan(\pi/4) = 1$, so $\tan(3\pi/4) = -1$.
* Alternatively, using coordinates $(-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$: $\tan = \frac{y}{x} = \frac{\frac{\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}} = -1$.
* Answer: -1

4. $\sin(-\pi/3)$
* Negative angles go clockwise. $-\pi/3$ is in Quadrant IV.
* The reference angle is $\pi/3$.
* In Quadrant IV, sine (y-coordinate) is negative.
* $\sin(\pi/3) = \frac{\sqrt{3}}{2}$.
* Therefore, $\sin(-\pi/3) = -\frac{\sqrt{3}}{2}$.
* Answer: $-\frac{\sqrt{3}}{2}$

5. $\cot(-3\pi/2)$
* Angle $-3\pi/2$ rotates clockwise three-quarters of the way around the circle. This lands on the positive y-axis, which is the same position as $\pi/2$.
* The coordinates are $(0, 1)$.
* Cotangent is $\frac{x}{y}$ (or $\frac{\cos}{\sin}$).
* $\cot(-3\pi/2) = \frac{0}{1} = 0$.
* Answer: 0

6. $\cos(5\pi/4)$
* The angle $5\pi/4$ is in Quadrant III (past $\pi$).
* The reference angle is $\pi/4$.
* In Quadrant III, cosine (x-coordinate) is negative.
* $\cos(\pi/4) = \frac{\sqrt{2}}{2}$.
* Therefore, $\cos(5\pi/4) = -\frac{\sqrt{2}}{2}$.
* Answer: $-\frac{\sqrt{2}}{2}$

7. $\sec(\pi/4)$
* First, find $\cos(\pi/4)$. From problem #2, we know the x-coordinate is $\frac{\sqrt{2}}{2}$.
* Secant ($\sec$) is the reciprocal of cosine ($1/\cos$).
* $\sec(\pi/4) = \frac{1}{\frac{\sqrt{2}}{2}} = \frac{2}{\sqrt{2}}$.
* Rationalizing: $\sqrt{2}$.
* Answer: $\sqrt{2}$

8. $\tan(-\pi/6)$
* Angle $-\pi/6$ is in Quadrant IV.
* The reference angle is $\pi/6$.
* In Quadrant IV, tangent is negative.
* $\tan(\pi/6) = \frac{\sin(\pi/6)}{\cos(\pi/6)} = \frac{1/2}{\sqrt{3}/2} = \frac{1}{\sqrt{3}}$.
* Rationalizing $\frac{1}{\sqrt{3}}$ gives $\frac{\sqrt{3}}{3}$.
* Since it's negative: $-\frac{\sqrt{3}}{3}$.
* Answer: $-\frac{\sqrt{3}}{3}$

9. $\tan(-11\pi/2)$
* Let's simplify the angle. One full rotation is $2\pi$ or $\frac{4\pi}{2}$.
* $-11\pi/2$ is equivalent to rotating clockwise.
* $-11\pi/2 + 6\pi$ (which is $\frac{12\pi}{2}$) $= \frac{\pi}{2}$.
* So, $-11\pi/2$ lands on the same spot as $\pi/2$ (the positive y-axis).
* At $\pi/2$, the coordinates are $(0, 1)$.
* Tangent is $\frac{y}{x} = \frac{1}{0}$. Division by zero is undefined.
* Answer: Undefined

10. $\csc(\pi/4)$
* This is the same as problem #2.
* $\sin(\pi/4) = \frac{\sqrt{2}}{2}$.
* Reciprocal is $\sqrt{2}$.
* Answer: $\sqrt{2}$

11. $\tan(6\pi)$
* $6\pi$ represents 3 full rotations around the circle ($2\pi \times 3$).
* You end up back at the starting point: $(1, 0)$.
* Tangent is $\frac{y}{x} = \frac{0}{1}$.
* Answer: 0

12. $\sin(-\pi)$
* Angle $-\pi$ rotates halfway around the circle clockwise.
* This lands on the negative x-axis at point $(-1, 0)$.
* Sine is the y-coordinate.
* The y-coordinate is $0$.
* Answer: 0

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Final Answer:
1. 0
2. $\sqrt{2}$
3. -1
4. $-\frac{\sqrt{3}}{2}$
5. 0
6. $-\frac{\sqrt{2}}{2}$
7. $\sqrt{2}$
8. $-\frac{\sqrt{3}}{3}$
9. Undefined
10. $\sqrt{2}$
11. 0
12. 0