Exercises: Sine, Cosine and Tangent on the unit circle

The cosine is positive in the first quadrant and in the fourth quadrant.

1st angle

$\cos \left(\gamma \right)=\frac{1}{2}\sqrt{2}\backslash cos\backslash left(\backslash gamma\backslash right)=\backslash frac\{1\}\{2\}\backslash sqrt\{2\}$

${\gamma }_1={45}^{\circ }\backslash gamma\_1=45^\{\backslash circ\}$

2nd angle

$\cos ({360}^{\circ }-{\gamma }_1)=\frac{1}{2}\sqrt{2}\backslash cos\backslash left(360^\{\backslash circ\}-\backslash gamma\_1\backslash right)=\backslash frac\{1\}\{2\}\backslash sqrt\{2\}$

$\cos ({360}^{\circ }-{45}^{\circ })=\frac{1}{2}\sqrt{2}\backslash cos\backslash left(360^\{\backslash circ\}-45^\{\backslash circ\}\backslash right)=\backslash frac\{1\}\{2\}\backslash sqrt\{2\}$

${\gamma }_2={315}^{\circ }\backslash gamma\_2=315^\{\backslash circ\}$

3rd angle

$\cos ({360}^{\circ }+{\gamma }_1)=\frac{1}{2}\sqrt{2}\backslash cos\backslash left(360^\{\backslash circ\}+\backslash gamma\_1\backslash right)=\backslash frac\{1\}\{2\}\backslash sqrt\{2\}$

$\cos ({360}^{\circ }+{45}^{\circ })=\frac{1}{2}\sqrt{2}\backslash cos\backslash left(360^\{\backslash circ\}+45^\{\backslash circ\}\backslash right)=\backslash frac\{1\}\{2\}\backslash sqrt\{2\}$

${\gamma }_3={405}^{\circ }\backslash gamma\_3=405^\{\backslash circ\}$

4th angle

$\cos ({720}^{\circ }-{\gamma }_1)=\frac{1}{2}\sqrt{2}\backslash cos\backslash left(720^\{\backslash circ\}-\backslash gamma\_1\backslash right)=\backslash frac\{1\}\{2\}\backslash sqrt\{2\}$

$\cos ({720}^{\circ }-{45}^{\circ })=\frac{1}{2}\sqrt{2}\backslash cos\backslash left(720^\{\backslash circ\}-45^\{\backslash circ\}\backslash right)=\backslash frac\{1\}\{2\}\backslash sqrt\{2\}$

${\gamma }_3={675}^{\circ }\backslash gamma\_3=675^\{\backslash circ\}$

5th angle

$\cos (0^{\circ }-{\gamma }_1)=\frac{1}{2}\sqrt{2}\backslash cos\backslash left(0^\{\backslash circ\}-\backslash gamma\_1\backslash right)=\backslash frac\{1\}\{2\}\backslash sqrt\{2\}$

$\cos (0^{\circ }-{45}^{\circ })=\frac{1}{2}\sqrt{2}\backslash cos\backslash left(0^\{\backslash circ\}-45^\{\backslash circ\}\backslash right)=\backslash frac\{1\}\{2\}\backslash sqrt\{2\}$

${\gamma }_4=-{45}^{\circ }\backslash gamma\_4=-45^\{\backslash circ\}$

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

The sine is positive in the first and second quadrant.

$y\in [-\pi ;3\pi ]y\backslash in\backslash left[-\backslash mathrm\backslash pi;\backslash ;3\backslash mathrm\backslash pi\backslash right]$

1st angle in radiant

$\frac{x}{2}=y\backslash frac\{x\}\{2\}=y$

$\sin \left(y\right)=1\backslash sin\backslash left(y\backslash right)=1$

$y_1=\frac{1}{2}\pi y\_1=\backslash frac\{1\}\{2\}\backslash mathrm\{\backslash pi\}$

$x_1=\pi x\_1=\backslash mathrm\{\backslash pi\}$

2nd angle in radiant

$\sin (2\pi +y_1)=1\backslash sin\backslash left(2\backslash mathrm\{\backslash pi\}+y\_1\backslash right)=1$

$\sin (2\pi +\frac{\pi }{2})=1\backslash sin\backslash left(2\backslash mathrm\{\backslash pi\}+\backslash frac\{\backslash mathrm\{\backslash pi\}\}\{2\}\backslash right)=1$

${\gamma }_2=2.5\pi \backslash gamma\_2=2.5\backslash mathrm\{\backslash pi\}$

$x=5\pi x=5\backslash mathrm\{\backslash pi\}$

3rd angle in radiant

$\sin (0-y_1)=1\backslash sin\backslash left(0-y\_1\backslash right)=1$

$\sin (-\frac{\pi }{2})=1\backslash sin\backslash left(-\backslash frac\{\backslash mathrm\{\backslash pi\}\}\{2\}\backslash right)=1$

$y_3=-\frac{1}{2}\pi y\_3=-\backslash frac\{1\}\{2\}\backslash mathrm\{\backslash pi\}$

$x=-\pi x=-\backslash mathrm\{\backslash pi\}$

$\sin \left(x\right)=-2\backslash sin\backslash left(x\backslash right)=-2$

Can never hold, since $-1\le \sin \left(x\right)\le 1-1\backslash le\backslash sin\backslash left(x\backslash right)\backslash le1$

Draw the coordinate system.

Construct the unit circle with your compass.

Use a triangle to draw a $60\mathrm{°}60°$ angle with respect to the $xx$-axis. Mark the intersection of the resulting straight line with the unit circle.

Draw the perpendicular to the $xx$-axis that passes through the marked point.

Now you can measure the length of the sine with your ruler.

$\sin (60\mathrm{°})\approx 0.87\backslash sin(60°)\backslash approx0.87$

In a right triangle with a $60\mathrm{°}60°$ angle, the third interior angle has $30\mathrm{°}30°$. In the unit circle the hypotenuse has length $11$.

Extend the triangle to an equilateral triangle by mirroring the given triangle at the edge with $\sin (60\mathrm{°})\backslash sin(60°)$. The triangle must be equilateral because it has three $60\mathrm{°}60°$ angles.

In an equilateral triangle all sides have the same length, i.e. length $11$.

$\cos (60\mathrm{°})\backslash cos(60°)$ is exactly half the side length of the triangle, i.e. $\frac{1}{2}\backslash frac\{1\}\{2\}$.

Now use the Pythagorean theorem in the right triangle to calculate $\sin (60\mathrm{°})\backslash sin(60°)$.

Plug in the value for the cosine. ${(\cos (60\mathrm{°}))}^2+{(\sin (60\mathrm{°}))}^2=1^2\backslash left(\backslash cos(60°)\backslash right)^2+\backslash left(\backslash sin(60°)\backslash right)^2=1^2$

${\left(\frac{1}{2}\right)}^2+{(\sin (60\mathrm{°}))}^2=1\backslash left(\backslash frac\{1\}\{2\}\backslash right)^2+\backslash left(\backslash sin(60°)\backslash right)^2=1$

$\frac{1}{4}+{(\sin (60\mathrm{°}))}^2=1\backslash frac\{1\}\{4\}+\backslash left(\backslash sin(60°)\backslash right)^2=1$ $\mid -\frac{1}{4}\backslash mid-\backslash frac\{1\}\{4\}$

${(\sin (60\mathrm{°}))}^2=\frac{3}{4}\backslash left(\backslash sin(60°)\backslash right)^2=\backslash frac\{3\}\{4\}$ $\mid \sqrt{}\backslash mid\backslash sqrt\{\; \}$

$\sin (60\mathrm{°})=\pm \sqrt{\frac{3}{4}}\backslash sin(60°)=\backslash pm\backslash sqrt\{\backslash frac\{3\}\{4\}\}$ since $\sin (60\mathrm{°})\backslash sin(60°)$ was applied above the $xx$-axis, only the positive result comes into question.

$\sin (60\mathrm{°})=\sqrt{\frac{3}{4}}\backslash sin(60°)=\backslash sqrt\{\backslash frac\{3\}\{4\}\}$

$\sin (60\mathrm{°})=\frac{1}{2}\sqrt{3}\backslash sin(60°)=\backslash frac\{1\}\{2\}\backslash sqrt\{3\}$