Geometry and Trigonometry · Trigonometry of general angles
Reduction to the first quadrant and the fundamental identity
You already know how to place an angle in its quadrant and work out the sign of its ratios. The next step is to take advantage of the symmetry of the unit circle: different angles can share, apart from the sign, the same sine or the same cosine as an angle in the first quadrant. This allows you to calculate any ratio without memorising any table, beyond the special angles $30º$, $45º$ and $60º$.
Before you start, make sure you know: Symmetries (3rd ESO)
Reducing any angle to the first quadrant
Two angles that are symmetrical about the $Y$ axis (such as $30º$ and $150º$) have points at the same height: the same sine. Two angles that are symmetrical about the $X$ axis (such as $30º$ and $330º$) have the same $x$-coordinate: the same cosine. This symmetry is what allows you to "reduce" any angle to one in the first quadrant.
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The figure shows the angles $30º$ and $150º$ on the unit circle. Looking at the symmetry of the figure, express $\sin 150º$ and $\cos 150º$ in terms of $\sin 30º$ and…
The figure shows the angles $30º$ and $150º$ on the unit circle.
Looking at the symmetry of the figure, express $\sin 150º$ and $\cos 150º$ in terms of $\sin 30º$ and $\cos 30º$, stating the correct sign of each.
Express each of these trigonometric ratios in terms of the ratio of a first-quadrant angle, taking care with the sign: $\sin 210º$ $\cos 225º$ $\tan 240º$
Express each of these trigonometric ratios in terms of the ratio of a first-quadrant angle, taking care with the sign:
Express each of these trigonometric ratios in terms of the ratio of a first-quadrant angle, taking care with the sign: $\sin 330º$ $\cos 315º$ $\tan 300º$
Express each of these trigonometric ratios in terms of the ratio of a first-quadrant angle, taking care with the sign:
The fundamental identity, taking care with the sign
The fundamental identity $\sin^2\alpha + \cos^2\alpha = 1$ is still valid for any angle, not just acute ones: it comes from applying Pythagoras to the reference triangle, whose hypotenuse is always the radius $1$. What changes compared with an ordinary right-angled triangle is that, when you solve for a square root, you have to choose the sign according to the quadrant, instead of always taking the positive one.
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Given that $\sin \alpha = \frac{3}{5}$ and that $\alpha$ belongs to the second quadrant, calculate the exact value of $\cos \alpha$ and $\tan \alpha$.
Given that $\sin \alpha = \frac{3}{5}$ and that $\alpha$ belongs to the second quadrant, calculate the exact value of $\cos \alpha$ and $\tan \alpha$.
A student wrote this working: > «If $\cos \alpha = -0,8$ and $\alpha$ is in the second quadrant, then: > $\sin \alpha = \sqrt{1 - 0,64} = \sqrt{0,36} = -0,6$» Find the…
A student wrote this working:
«If $\cos \alpha = -0,8$ and $\alpha$ is in the second quadrant, then:
Trigonometric equations: finding all the solutions
If you are told that $\sin\ x = \dfrac12$, more than one angle between $0º$ and $360º$ satisfies it: the horizontal line $y=\dfrac12$ cuts the unit circle at two points, one in each of two different quadrants.
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The figure shows the unit circle and the horizontal line $\sin x = \frac{1}{2}$, which intersects it at two points. Find all angles $x$ in $[0º, 360º)$ that satisfy…
The figure shows the unit circle and the horizontal line $\sin x = \frac{1}{2}$, which intersects it at two points.
Find all angles $x$ in $[0º, 360º)$ that satisfy $\sin x = \frac{1}{2}$, reasoning from the figure.
The figure shows the angles $60º$ and $120º$, which are supplementary. Given that $\sin 60º = \frac{\sqrt{3}}{2} \approx 0,866$ and $\cos 60º = \frac{1}{2}$, calculate…
The figure shows the angles $60º$ and $120º$, which are supplementary.
Given that $\sin 60º = \frac{\sqrt{3}}{2} \approx 0,866$ and $\cos 60º = \frac{1}{2}$, calculate $\sin 120º$ and $\cos 120º$ reasoning from the symmetry of the figure.
Since trigonometric ratios repeat in exactly the same way every complete turn, an equation such as $\tan\ x = 1$ does not have only the solutions in $[0º,360º)$: it has infinitely many, one every $180º$ (for sine and cosine, every $360º$). This is called periodicity, and it is key in any phenomenon that rotates or repeats, such as the blade of a wind turbine.
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Find all angles $x$ in $[0º, 360º)$ that satisfy $\tan x = 1$. Using the unit circle, reason how many solutions the same equation would have in the interval $[0º, 720º)$…
Find all angles $x$ in $[0º, 360º)$ that satisfy $\tan x = 1$.
Using the unit circle, reason how many solutions the same equation would have in the interval $[0º, 720º)$ and write them all down.
The tip of a wind turbine blade describes a circle of radius $1\ \text{m}$ (in relative units) around the rotor axis. If $\theta$ is the angle that the blade makes with…
The tip of a wind turbine blade describes a circle of radius $1\ \text{m}$ (in relative units) around the rotor axis. If $\theta$ is the angle that the blade makes with the horizontal, measured anticlockwise, the height of the tip relative to the centre is $\sin \theta$ and its horizontal displacement is $\cos \theta$.
Complete a table with the height and horizontal displacement of the tip for $\theta = 0º,\ 90º,\ 180º$ and $270º$.
The blade rotates $40$ times per minute. After a period of time, the tip is at a height of $0,5\ \text{m}$ above the centre. What values of $\theta$ between $0º$ and $360º$ correspond to this height?
The reduction formulae ($180º-\alpha$, $180º+\alpha$, $360º-\alpha$...) are not only for calculating a numerical value: they are also used to prove that an expression with several related angles is always equal to a simpler one, whatever $\alpha$ is.
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Given that $\tan \alpha = -\frac{3}{4}$ and that $\alpha$ belongs to the fourth quadrant: Calculate the exact value of $\sin \alpha$ and $\cos \alpha$. Calculate…
Given that $\tan \alpha = -\frac{3}{4}$ and that $\alpha$ belongs to the fourth quadrant:
Calculate the exact value of $\sin \alpha$ and $\cos \alpha$.