UK Secondary (7-11)

Angles of any magnitude

Lesson

If we restrict our attention to right-angled triangles, then we can only think of the trigonometric functions of angles whose size is between $0^\circ$0° and $90^\circ$90°. We redefine our definitions to make it possible to apply the functions to angles of any magnitude. Observe that the definitions given below are consistent with the original right-angled triangle definitions while being more generally useful.

Consider the unit circle centred at the origin on the Cartesian plane. A radius to any point that is free to move on the circle makes an angle with the positive horizontal axis. By convention, the angle is measured anticlockwise from the positive horizontal axis. The angle can have any size, positive or negative, depending on how far the point has moved around the circle.

The *sine *of the angle is defined to be the coordinate of the point on the vertical axis. ($y$`y`-coordinate)

The *cosine *of the angle is defined to be the coordinate of the point on the horizontal axis. ($x$`x`-coordinate)

The *tangent *of the angle is defined to be the ratio of the angle in question.

Depending on where the point is on the unit circle, we say the angle is in one of four *quadrants*. From the diagram, if we imagine the point moving through the quadrants, it can be seen that the sine function is positive for angles in the first and second quadrants; cosine is positive for angles in the first and fourth quadrants; and tangent is positive for angles in the first and third quadrants.

The original right-angled triangle definitions of the trigonometric functions correspond to functions of angles in the first quadrant in the unit circle definitions.

Verify that the point with coordinates $\left(\frac{20}{29},-\frac{21}{29}\right)$(2029,−2129) is on the unit circle. Say in what quadrant the angle $\alpha$`α` lies that is made by the radius to the point and the positive horizontal axis. Write down the tangent of the angle.

By Pythagoras' theorem, the distance of the point from the origin is $d=\sqrt{\left(\frac{20}{29}\right)^2+\left(\frac{21}{29}\right)^2}=1$`d`=√(2029)2+(2129)2=1 . Therefore, the point is on the unit circle. The angle is in the fourth quadrant because the horizontal coordinate is positive and the vertical coordinate is negative. We have, $\sin\alpha=-\frac{21}{29}$`s``i``n``α`=−2129 and $\cos\alpha=\frac{20}{29}$`c``o``s``α`=2029. Using the definition of the tangent function, we have

$\tan\alpha=-\frac{\frac{21}{29}}{\frac{20}{29}}=-\frac{21}{20}$`t``a``n``α`=−21292029=−2120.

Write the following trigonometric ratio using an acute angle:

$\sin147^\circ$`s``i``n`147°

Solve $\sin\theta=0.9336$`s``i``n``θ`=0.9336

to the nearest degree for $0$0° ≤ `θ`

Find the acute angle $\theta$

`θ`that solves the equation.Now find

**ALL**solutions to $\theta$`θ`in the range 0 to 360°. (Write the solutions on the same line, separated by a comma.)

Evaluate $\cos387^\circ$`c``o``s`387° correct to 2 decimal places and make note of the sign of your answer.