obtain the sign of the ratio by noting the quadrant,.In summary, to find the trigonometric ratio of an angle between 0° and 360° we In general, if lies in the fourth quadrant, the acute angle is called the related angle for θ. The angle POQ is 30° and is called the related angle for 330°.Ĭos 330° = cos 30° = and sin 330° = −sin 30° = −. The coordinates of P are (cos 330°, sin 330°). To find the sine and cosine of 330° we locate the corresponding point P in the fourth quadrant. In this quadrant, we can see that the sine and tangent ratios are negative and the cosine ratio is positive. In general, if lies in the third quadrant, the acute angle θ − 180° is called the related angle for θ.įinally, if θ lies between 270° and 360° the corresponding point P is in the fourth quadrant. The angle POQ is 30° and is called the related angle for 210°.Ĭos 210° = −cos 30° = − and sin 210° = −sin 30° = −.
The coordinates of P are (cos 210°, sin 210°). To find the sine and cosine of 210° we locate the corresponding point P in the third quadrant. In this quadrant, we can see that the sine and cosine ratios are negative and the tangent ratio positive.
Will place the corresponding point P in the third quadrant. We introduced this idea in the module, Further Trigonometry. In general, if θ lies in the second quadrant, the acute angle 180° − θ is called the related angle for θ. From POQ we can see that OQ = cos 30° and PQ = sin 30°, so the coordinates of P are −cos 30°, sin 30°). The angle POQ is 30° and is called the related angle for 150°. To the point P in the second quadrant with coordinates Suppose we wish to find the exact°and sin 150°. We will examine this quadrant by quadrant.
To find the trigonometric ratio of angles beyond 90°, we introduce the notion of the related angle. Ī 150° lies in the second quadrant so cos 150° is negative.ī 300° lies in the fourth quadrant so sin 300° is negative.Ĭ 235° lies in the third quadrant so tan 235° is positive. Note that this is the same as saying that tan θ equals the. You will see in the following exercise why this is the case. Between 0° and 360°, this will happen when θ = 90°, or θ = 270°. Unless cos In this case, we say that the tangent ratio is undefined. For angles that are greater than 90° we define the tangent of θ by
the sine of θ to be the y-coordinate of the point P.įor acute angles, we know that tan θ =. the cosine of θ to be the x-coordinate of the point P. Since each angle θ determines a point P on the unit circle, we will define For the time being we will concentrate on positive angles between 0° and 360°. Angles measured clockwise from OA are called negative angles. We measure angles anticlockwise from OA and call these positive angles. Since the length OQ = cos θ is the x-coordinate of P, and PQ = sin θ is the y-coordinate of P, we see that the point P has coordinates From the point P on the circle in the first quadrant we can construct a right-angled triangle POQ with O at the origin and Q on the x-axis. Recall also the isosceles right triangle and find the described ratios in itįor the rest of the first quarter just memorize that the common 30-60-90 degree triangle has its shortest to longest edge ratio of $1:2$ (see the image linked above) so sine reaches $1/2$ at one-third of the right angle.We begin by taking the circle of radius 1, centre the origin, in the plane. As it's symmetric it reduces to really only the three cases above.įirst of all, you know the square diagonal is $\sqrt 2$ of the side, so sine and cosine of $45^\circ$ is $1/\sqrt 2 = \sqrt 2/2$ and tangent is $1$. As the circular picture is perfectly symmetric it's easy to memorize even though it does, technically, have 48 values. Which values depend upon whether $|x| > |y|$ or $|y| > |x|$ and which quadrant $(x,y) lie in.Īnd yes, that circular picture helps. Unfortunately, you have to memorize the definitions of the sine and tangent: $\sin=\text$ The triangle has two sides of length $1$ if you have memorized the theorem of Pythagoras, you can figure out that the length of the third side is $\sqrt2.$ Either one of these triangles has angles of $90^\circ,45^\circ,45^\circ$ no need to memorize $45$, just divide $90$ by $2$. Next, cut the square along the diagonal, making two triangles. The square has four sides of equal length, which we take to be $1$. The trig functions for $30^\circ,45^\circ,60^\circ$ are based on two simple geometric figures: the square and the equilateral triangle. (Never memorized them myself.) I would expect a student to have enough understanding to be able to figure them out in seconds. I would not expect a student to memorize trig functions of easy angles.
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