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Use $\sin^2\theta + \cos^2\theta = 1$ to find one ratio from another, and let the quadrant decide the sign.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
In this lesson you meet the one identity that connects sine and cosine: $\sin^2\theta + \cos^2\theta = 1$, which is nothing but the equation of the unit circle written with the coordinates called by their names. You use it to get either ratio from the other, and the tangent from both — but the identity only ever fixes a size, so every answer needs the quadrant to settle its sign, and getting that habit is the point of the lesson.
You know Pythagoras: in a right triangle the two legs squared add to the hypotenuse squared. You know that the point at angle $\theta$ on the unit circle is $(\cos\theta, \sin\theta)$, and that its two coordinates change sign from quadrant to quadrant. You can square a fraction and take the square root of a perfect square. That is the whole tool kit.
Identity: an equation true for every value of its letter, not just for some.
$\sin^2\theta$: shorthand for $(\sin\theta)^2$ — square the ratio, not the angle.
Tangent: $\tan\theta = \frac{\sin\theta}{\cos\theta}$.
Quadrant signs: in quadrant 1 both ratios are positive; in quadrant 2 only sine; in quadrant 3 neither; in quadrant 4 only cosine.
Pythagorean triple: three whole numbers with $a^2 + b^2 = c^2$, like $3, 4, 5$ or $5, 12, 13$.
The point at angle $\theta$ on the unit circle is $(\cos\theta, \sin\theta)$, and every point $(x, y)$ of that circle satisfies $x^2 + y^2 = 1$. Substituting the coordinates gives the Pythagorean identity: $$\sin^2\theta + \cos^2\theta = 1.$$ It is not a new fact. It is Pythagoras on the triangle whose legs are $|\cos\theta|$ and $|\sin\theta|$ and whose hypotenuse is the radius $1$ — which is why it holds for every angle, including the ones with no triangle.
Rearranged, it turns one ratio into the other: $$\cos^2\theta = 1 - \sin^2\theta, \qquad \sin^2\theta = 1 - \cos^2\theta.$$ Use it in three moves. Square the ratio you were given. Subtract it from $1$ and take the square root, which gives the size of the ratio you want. Choose the sign from the quadrant — the identity deals only in squares, so it cannot know whether the answer is positive or negative, and an answer without the quadrant is only half done.
So if $\sin\theta = \frac{5}{13}$ and $\theta$ is in quadrant 2: $\cos^2\theta = 1 - \frac{25}{169} = \frac{144}{169}$, so $\cos\theta = \pm\frac{12}{13}$, and quadrant 2 is to the left of the axis, so $\cos\theta = -\frac{12}{13}$.
Once both ratios are known the tangent is free: $\tan\theta = \frac{\sin\theta}{\cos\theta}$, and since the two share a denominator it cancels, leaving one numerator over the other.
Another way: picture
The unit circle with the radius to a point drawn in, and the point's height and horizontal offset completing a right triangle. The legs are $|\sin\theta|$ and $|\cos\theta|$, the hypotenuse is $1$, and Pythagoras on it is the identity itself.
Another way: story
The identity is a budget. The two squares must always add to exactly $1$: spend more on one and the other must shrink. Which side of zero each one falls on is not in the budget — that is the quadrant's decision.
Subtracting the ratio instead of its square. From $\sin\theta = \frac{3}{5}$, $\cos\theta$ is not $1 - \frac{3}{5} = \frac{2}{5}$. Square first: $1 - \frac{9}{25} = \frac{16}{25}$, and then the root is $\frac{4}{5}$.
Forgetting that a square root has two signs. $\cos^2\theta = \frac{16}{25}$ leaves $\cos\theta = \pm\frac{4}{5}$. The identity cannot choose between them; only the quadrant can.
Reading $\sin^2\theta$ as $\sin(\theta^2)$. The exponent belongs to the whole ratio. $\sin^2 30^\circ = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$, which is nothing like $\sin 900^\circ$.
Square: $\cos^2\theta = \frac{64}{289}$; the minus sign disappears.
Squares are never negative.
$\sin^2\theta = 1 - \frac{64}{289} = \frac{225}{289}$, so $\sin\theta = \pm\frac{15}{17}$.
$225$ and $289$ are both perfect squares.
Quadrant 3 is below the axis, so the $y$ is negative: $\sin\theta = -\frac{15}{17}$.
The quadrant, not the identity, picks the sign.
$\cos^2\theta = 1 - \frac{400}{841} = \frac{441}{841}$, so $\cos\theta = \frac{21}{29}$, positive in quadrant 1.
$\tan\theta = \frac{20}{29} \div \frac{21}{29} = \frac{20}{21}$.
The shared denominator cancels.
$\cos^2\theta = 1 - \frac{49}{625} = \frac{576}{625}$, so $\cos\theta = \pm\frac{24}{25}$.
Quadrant 4 is to the right of the axis, so $\cos\theta = \frac{24}{25}$.
$\sin\theta = \frac{28}{53}$ and $\theta$ is in quadrant 1. Find $\cos\theta$.
Answer:
$\cos\theta = -\frac{45}{53}$ and $\theta$ is in quadrant 2. Find $\sin\theta$.
Answer:
$\sin\theta = \frac{24}{25}$ and $\cos\theta = \frac{7}{25}$. Find $\tan\theta$.
Answer:
With $\sin\theta = \frac{40}{41}$ and $\cos\theta = \frac{9}{41}$ the identity checks out: $1600 + 81 = 1681$. Why must $\sin^2\theta + \cos^2\theta = 1$ for every angle, and not only for this one?
A game stores a facing direction as a unit vector $(u, v)$ with $u^2 + v^2 = 1$. A character has $v = \frac{24}{25}$ and is facing left, so $u < 0$. What is $u$?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
$\sin\theta = \frac{40}{41}$ and $\theta$ is in quadrant 1. Find $\cos\theta$.
Answer:
You can find one trigonometric ratio from another and give it the right sign. If $\cos\theta = -\frac{3}{5}$ and $\theta$ is in quadrant 3, find $\sin\theta$ and $\tan\theta$, and say which step needed the quadrant.
8. Your turn: $\sin\theta = -\frac{7}{25}$ in quadrant 4. Find $\cos\theta$, step 2