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 the same sine or the same cosine as a first-quadrant angle, apart from the sign. This lets you calculate any ratio without memorizing a table, beyond the special angles $30º$, $45º$, and $60º$.
Reducing any angle to the first quadrant
Two angles that are symmetric about the $y$-axis (like $30º$ and $150º$) have points at the same height: the same sine. Two angles that are symmetric about the $x$-axis (like $30º$ and $330º$) have the same $x$-coordinate: the same cosine. This symmetry is what lets us "reduce" any angle to one in the first quadrant.
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:
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$\sin 210º$
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$\cos 225º$
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$\tan 240º$
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:
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$\sin 330º$
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$\cos 315º$
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$\tan 300º$
The fundamental identity, watching the sign
The fundamental identity $\sin^2\alpha + \cos^2\alpha = 1$ holds for any angle, not just acute ones: it comes from applying the Pythagorean theorem to the reference triangle, whose hypotenuse is always the radius $1$. What changes compared to an ordinary right triangle is that, when you solve for a square root, you have to choose the sign according to the quadrant, instead of always keeping the positive one.
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 $\tan \alpha$.
Given that $\cos \alpha = -\frac{1}{3}$ and that $\alpha$ belongs to the third quadrant, calculate the exact value of $\sin \alpha$ and $\tan \alpha$.
Given that $\cos \alpha = -\frac{1}{3}$ and that $\alpha$ belongs to the third quadrant, calculate the exact value of $\sin \alpha$ 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:
$\sin \alpha = \sqrt{1 - 0.64} = \sqrt{0.36} = -0.6$»
Find the mistake made, explain why it is a mistake, and calculate $\sin \alpha$ and $\tan \alpha$ correctly.
Trigonometric equations: 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.
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.
Find all angles $x$ in $[0º, 360º)$ that satisfy $\cos x = -\frac{\sqrt{2}}{2}$, stating which quadrants they belong to.
Find all angles $x$ in $[0º, 360º)$ that satisfy $\cos x = -\frac{\sqrt{2}}{2}$, stating which quadrants they belong to.
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.
Beyond 360º: infinitely many solutions
Since trigonometric ratios repeat exactly the same way every full turn, an equation like $\tan\ x = 1$ doesn't have just the solutions between $0º$ and $360º$: 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, like the blade of a wind turbine.
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 $[0º, 720º)$…
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Find all angles $x$ in $[0º, 360º)$ that satisfy $\tan x = 1$.
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Using the unit circle, reason how many solutions the same equation would have in the interval $[0º, 720º)$ and write them all down.
Find all angles $x$ in $[0º, 360º)$ that satisfy $\sin^2 x = \frac{1}{4}$, reasoning how many solutions there are and which quadrants they belong to.
Find all angles $x$ in $[0º, 360º)$ that satisfy $\sin^2 x = \frac{1}{4}$, reasoning how many solutions there are and which quadrants they belong to.
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 counterclockwise, the height of the tip relative to the center is $\sin \theta$ and its horizontal displacement is $\cos \theta$.
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Complete a table with the height and horizontal displacement of the tip for $\theta = 0º,\ 90º,\ 180º$ and $270º$.
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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 center. What values of $\theta$ between $0º$ and $360º$ correspond to this height?
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What about a height of $-0.5\ \text{m}$?
Proving identities by reducing step by step
The reduction formulas ($180º-\alpha$, $180º+\alpha$, $360º-\alpha$...) are not just for calculating a numerical value: they also serve to prove that an expression with several related angles is always equal to a simpler one, whatever $\alpha$ is.
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:
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Calculate the exact value of $\sin \alpha$ and $\cos \alpha$.
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Calculate $\sin(180º - \alpha)$.
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Calculate $\cos(180º + \alpha)$.
Prove that, for any angle $\alpha$, the following holds: $$\sin(180º - \alpha)\times\cos(360º - \alpha) + \cos(180º + \alpha)\times\sin(180º + \alpha) = 2\times\sin…
Prove that, for any angle $\alpha$, the following holds:
$$\sin(180º - \alpha)\times\cos(360º - \alpha) + \cos(180º + \alpha)\times\sin(180º + \alpha) = 2\times\sin \alpha\times\cos \alpha$$(Hint: first reduce each ratio to the first quadrant.)