Exercises: Sine, Cosine and Tangent on the unit circle

1
Consider on the unit circle: For which angles between $0^{\circ}$ and $360^{\circ}$ you have $\sin (α) = 0.5$ ?
$α = 30^{\circ}$
$α = 45^{\circ}$
$α = 60^{\circ}$
$α = 150^{\circ}$
2
Determine all solutions of the following equations in the range $γ \in [- 180^{\circ}; 720^{\circ}]$ ( part (a) ) or $x \in [- 2 π; 6 π]$ (part (b) - (c) )
$\cos (γ) = \frac{1}{2} \sqrt{2}$

For this task you need the following basic knowledge: Trigonometry on the unit circle
The cosine is positive in the first quadrant and in the fourth quadrant.
1st angle
$\cos (γ) = \frac{1}{2} \sqrt{2}$
$γ_{1} = 45^{\circ}$
2nd angle
$\cos (360^{\circ} - γ_{1}) = \frac{1}{2} \sqrt{2}$
$\cos (360^{\circ} - 45^{\circ}) = \frac{1}{2} \sqrt{2}$
$γ_{2} = 315^{\circ}$
3rd angle
$\cos (360^{\circ} + γ_{1}) = \frac{1}{2} \sqrt{2}$
$\cos (360^{\circ} + 45^{\circ}) = \frac{1}{2} \sqrt{2}$
$γ_{3} = 405^{\circ}$
4th angle
$\cos (720^{\circ} - γ_{1}) = \frac{1}{2} \sqrt{2}$
$\cos (720^{\circ} - 45^{\circ}) = \frac{1}{2} \sqrt{2}$
$γ_{3} = 675^{\circ}$
5th angle
$\cos (0^{\circ} - γ_{1}) = \frac{1}{2} \sqrt{2}$
$\cos (0^{\circ} - 45^{\circ}) = \frac{1}{2} \sqrt{2}$
$γ_{4} = - 45^{\circ}$

$\sin (\frac{x}{2}) = 1$

For this task you need the following basic knowledge: Trigonometry on the unit circle
$\sin (\frac{x}{2}) = 1$
The sine is positive in the first and second quadrant.
$y \in [- π; 3 π]$
1st angle in radiant
$\frac{x}{2} = y$
$\sin (y) = 1$
$y_{1} = \frac{1}{2} π$
$x_{1} = π$
2nd angle in radiant
$\sin (2 π + y_{1}) = 1$
$\sin (2 π + \frac{π}{2}) = 1$
$γ_{2} = 2.5 π$
$x = 5 π$
3rd angle in radiant
$\sin (0 - y_{1}) = 1$
$\sin (- \frac{π}{2}) = 1$
$y_{3} = - \frac{1}{2} π$
$x = - π$

$\sin (x) = - 2$

For this task you need the following basic knowledge: Trigonometry on the unit circle
$\sin (x) = - 2$
Can never hold, since $- 1 \leq \sin (x) \leq 1$
3
For which angles $γ$ with $γ \in [0^{\circ}; 360^{\circ}]$ do we have $\cos (γ) = - \sin (γ)$

For this task you need the following basic knowledge: Trigonometry on the unit circle
$\cos (γ) = - \sin (γ)$
The angle(s) must be in the second and fourth quadrants.
$\sin (γ) = \cos (γ)$ holds if $γ = 45^{\circ}$ . However, the presign of sine and cosine are still the same.
To this angle $γ = 45^{\circ}$ , the corresponding angles in the other quadrants with opposite presigns are are
1st angle
$γ = 135^{\circ}$
$- (\frac{\sqrt{2}}{2}) = - \frac{\sqrt{2}}{2}$
2nd angle
$γ = 315^{\circ}$
$- (- \frac{\sqrt{2}}{2}) = \frac{\sqrt{2}}{2}$
$\frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2}$
4
This exercise is about calculating $\sin (60 °)$ , without knowing it before.
Draw a large coordinate system. (1 $unit of length \hat{=} 8 squares$ ). Construct the unit circle with the compass and apply a $60 °$ angle to the $x$ -axis with the set square. Construct the length $\sin (60 °)$ and measure it with the ruler.

For this task you need the following basic knowledge: Trigonometry on the unit circle
Draw the coordinate system.
Construct the unit circle with your compass.
Use a triangle to draw a $60 °$ angle with respect to the $x$ -axis. Mark the intersection of the resulting straight line with the unit circle.
Draw the perpendicular to the $x$ -axis that passes through the marked point.
Now you can measure the length of the sine with your ruler.
$\sin (60 °) \approx 0.87$

Calculate $\sin (60 °)$ exactly. First find the value for $\cos (60 °)$ . Construct an equilateral triangle for it.

In a right triangle with a $60 °$ angle, the third interior angle has $30 °$ . In the unit circle the hypotenuse has length $1$ .
Extend the triangle to an equilateral triangle by mirroring the given triangle at the edge with $\sin (60 °)$ . The triangle must be equilateral because it has three $60 °$ angles.
In an equilateral triangle all sides have the same length, i.e. length $1$ .
$\cos (60 °)$ is exactly half the side length of the triangle, i.e. $\frac{1}{2}$ .
Now use the Pythagorean theorem in the right triangle to calculate $\sin (60 °)$ .
Plug in the value for the cosine. ${(\cos (60 °))}^{2} + {(\sin (60 °))}^{2} = 1^{2}$
${(\frac{1}{2})}^{2} + {(\sin (60 °))}^{2} = 1$
$\frac{1}{4} + {(\sin (60 °))}^{2} = 1$ $| - \frac{1}{4}$
${(\sin (60 °))}^{2} = \frac{3}{4}$ $| \sqrt{}$
$\sin (60 °) = \pm \sqrt{\frac{3}{4}}$ since $\sin (60 °)$ was applied above the $x$ -axis, only the positive result comes into question.
$\sin (60 °) = \sqrt{\frac{3}{4}}$
$\sin (60 °) = \frac{1}{2} \sqrt{3}$

Exercises: Sine, Cosine and Tangent on the unit circle

1st angle

2nd angle

3rd angle

4th angle

5th angle

1st angle in radiant

2nd angle in radiant

3rd angle in radiant

1st angle

2nd angle