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The sum and double-angle formulas and what they compute, and the inverse trigonometric functions with their restricted ranges and compositions.
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 get two formulas — one for $\sin(a + b)$ and one for $\cos(a + b)$ — and find that most of trigonometry beyond the unit circle falls out of them. Setting both angles equal gives the double-angle formulas; splitting an angle into two you already know gives exact values at angles like $15^\circ$ and $75^\circ$. Then you turn the machinery around: sine, cosine and tangent each take the same value at infinitely many angles, so their inverses are defined to return exactly one, and knowing which one is the whole skill.
You know the unit circle and the exact values at $30^\circ$, $45^\circ$ and $60^\circ$, and you know the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$. You can also read a right triangle: opposite over hypotenuse is the sine, adjacent over hypotenuse the cosine. Everything below is built from two formulas you are about to be given, plus that algebra.
Identity: an equation true for every value of the variable, not something to solve.
Sum formula: an expression for $\sin(a + b)$ or $\cos(a + b)$ in terms of the sines and cosines of $a$ and $b$ separately.
Double-angle formula: the sum formula with $a = b$.
Exact value: written with roots and fractions, like $\frac{\sqrt{6} + \sqrt{2}}{4}$, rather than as a rounded decimal.
Acute: between $0^\circ$ and $90^\circ$, where every trigonometric ratio is positive.
Everything here comes from $$\sin(a + b) = \sin a\cos b + \cos a\sin b, \qquad \cos(a + b) = \cos a\cos b - \sin a\sin b.$$ Replacing $b$ by $-b$ gives the difference formulas (the middle signs swap). Setting $a = b = \theta$ gives the double-angle formulas: $\sin 2\theta = 2\sin\theta\cos\theta$ and $\cos 2\theta = \cos^2\theta - \sin^2\theta$. Using $\sin^2\theta = 1 - \cos^2\theta$ rewrites the second as $2\cos^2\theta - 1$, which is the version to reach for when only the cosine is known. Dividing the two sum formulas gives $\tan 2\theta = \frac{2\tan\theta}{1 - \tan^2\theta}$. The formulas earn their keep twice over. They let you compute with ratios: from $\sin\theta = \frac{3}{5}$, $\cos\theta = \frac{4}{5}$ you get $\sin 2\theta = 2\cdot\frac{3}{5}\cdot\frac{4}{5} = \frac{24}{25}$. And they let you reach exact values at new angles by splitting: $75^\circ = 45^\circ + 30^\circ$, so $\sin 75^\circ = \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2}\cdot\frac{1}{2} = \frac{\sqrt{6} + \sqrt{2}}{4}$.
Another way: picture
The unit circle with an angle $\theta$ drawn from the positive $x$-axis and a second copy of the same angle stacked on top of it, so the terminal ray sits at $2\theta$: doubling the angle is stacking, and the coordinates of the far ray are what the formulas compute.
Another way: steps
"$\sin 2\theta = 2\sin\theta$." Never. Test it at $\theta = 30^\circ$: $\sin 60^\circ \approx 0.866$, but $2\sin 30^\circ = 1$. The sine of a doubled angle is $2\sin\theta\cos\theta$.
Dropping the $-1$ in $\cos 2\theta = 2\cos^2\theta - 1$. With $\cos\theta = \frac{1}{3}$ the answer is $\frac{2}{9} - 1 = -\frac{7}{9}$, which is negative — and that minus sign is exactly the part that gets lost.
"$\cos(a + b) = \cos a + \cos b$." No formula distributes like that. The cosine sum formula even flips the sign in the middle: $\cos a\cos b - \sin a\sin b$.
Only the cosine is given, so use $\cos 2\theta = 2\cos^2\theta - 1$.
Choose the version that needs what you have.
$2 \times \tfrac{1}{9} - 1 = \tfrac{2}{9} - \tfrac{9}{9} = -\tfrac{7}{9}$.
Negative, so $2\theta$ is obtuse — sensible, since $\cos\theta = \frac13$ means $\theta$ is over $70^\circ$.
$15^\circ = 45^\circ - 30^\circ$, so use $\cos(a - b) = \cos a\cos b + \sin a\sin b$.
The difference formula has the $+$ in the middle.
$\tfrac{\sqrt{2}}{2}\cdot\tfrac{\sqrt{3}}{2} + \tfrac{\sqrt{2}}{2}\cdot\tfrac{1}{2} = \tfrac{\sqrt{6}}{4} + \tfrac{\sqrt{2}}{4} = \tfrac{\sqrt{6} + \sqrt{2}}{4} \approx 0.9659$.
Pythagoras gives $\cos\theta = \tfrac{12}{13}$, positive because $\theta$ is acute.
$\sin 2\theta = 2 \cdot \tfrac{5}{13} \cdot \tfrac{12}{13} = \tfrac{120}{169}$.
$\sin\theta = \dfrac{5}{13}$ and $\cos\theta = \dfrac{12}{13}$, with $\theta$ acute. Find $\sin 2\theta$ as a fraction.
Answer:
$\cos\theta = \dfrac{4}{9}$. Find $\cos 2\theta$ as a fraction.
Answer:
$\tan\theta = \dfrac{1}{8}$. Find $\tan 2\theta$ as a fraction.
Answer:
Match each expression to what it equals.
| $\sin 2\theta$ | $\cos 2\theta$ | $1$ | |
|---|---|---|---|
| $2\sin\theta\cos\theta$ | |||
| $\cos^2\theta - \sin^2\theta$ | |||
| $\sin^2\theta + \cos^2\theta$ |
You know what an inverse function is — it undoes the original, and only a one-to-one function has one — and you can read the unit circle in both directions. You have also solved right triangles: given two sides you can find the third with Pythagoras. This lesson is about the trouble that arises because sine, cosine and tangent are emphatically not one-to-one.
$\arcsin x$ (also written $\sin^{-1} x$): the angle in $[-90^\circ, 90^\circ]$ whose sine is $x$. The $-1$ is not a reciprocal.
$\arccos x$: the angle in $[0^\circ, 180^\circ]$ whose cosine is $x$.
$\arctan x$: the angle strictly between $-90^\circ$ and $90^\circ$ whose tangent is $x$.
Principal value: the one angle the inverse is defined to return.
Restricted domain: the stretch of the original function that was kept so an inverse could exist.
Composition: applying one function to the output of another, as in $\cos(\arcsin x)$.
Sine takes the value $\tfrac{1}{2}$ at $30^\circ$, $150^\circ$, $390^\circ$ and endlessly on, so it has no inverse as it stands. The fix is to keep a stretch on which it is one-to-one and throw the rest away. For sine that stretch is $[-90^\circ, 90^\circ]$, for cosine it is $[0^\circ, 180^\circ]$, and for tangent it is the open interval $(-90^\circ, 90^\circ)$. The inverse then returns exactly one angle, the principal value: $\arcsin\tfrac{1}{2} = 30^\circ$, $\arctan 1 = 45^\circ$, $\arccos(-1) = 180^\circ$. Note that arcsine gives negative answers for negative inputs while arccosine gives obtuse ones — the ranges are different on purpose, so that each covers every possible value exactly once. Compositions like $\cos(\arcsin\tfrac{3}{5})$ are answered with a picture, not a calculator: the inner function names an angle in a right triangle with opposite $3$ and hypotenuse $5$, Pythagoras gives the adjacent side $4$, and the cosine is $\tfrac{4}{5}$ — positive, because arcsine only ever answers where the cosine is positive.
| $x$ | $\arcsin x$ | $\arccos x$ |
|---|---|---|
| $0$ | $0^\circ$ | $90^\circ$ |
| $\frac{1}{2}$ | $30^\circ$ | $60^\circ$ |
| $\frac{\sqrt{2}}{2}$ | $45^\circ$ | $45^\circ$ |
| $\frac{\sqrt{3}}{2}$ | $60^\circ$ | $30^\circ$ |
| $1$ | $90^\circ$ | $0^\circ$ |
| $-\frac{1}{2}$ | $-30^\circ$ | $120^\circ$ |
| $-\frac{\sqrt{2}}{2}$ | $-45^\circ$ | $135^\circ$ |
| $-\frac{\sqrt{3}}{2}$ | $-60^\circ$ | $150^\circ$ |
| $-1$ | $-90^\circ$ | $180^\circ$ |
Another way: picture
The sine curve with all of it faint except the arc from $-90^\circ$ to $90^\circ$, which is bold and rises once through every height from $-1$ to $1$; a horizontal line at height $\tfrac{1}{2}$ meets the faint curve many times and the bold arc exactly once.
Another way: story
Sine is a machine that turns many angles into one number, so it cannot be run backwards without a rule. The rule is: always hand back the angle nearest to zero that works — and for cosine, the one between zero and a straight angle.
"$\sin^{-1} x = \frac{1}{\sin x}$." No. $\sin^{-1}$ is the inverse function; the reciprocal is $\csc x$. The same superscript means two different things in $\sin^2 x$ and $\sin^{-1} x$, which is a genuinely bad piece of notation you simply have to know.
"$\arcsin(0.5)$ could be $150^\circ$." $\sin 150^\circ$ really is $0.5$, but $\arcsin$ returns only the angle in $[-90^\circ, 90^\circ]$, so the answer is $30^\circ$. If a problem needs the other angle, you must supply it yourself.
"$\arcsin(\sin 200^\circ) = 200^\circ$." It is $-20^\circ$. Composing the two only returns the input when the input was already inside the restricted range.
The cosine is negative, and arccosine answers in $[0^\circ, 180^\circ]$, so the angle is obtuse.
Fix the quadrant before the number.
$\cos 45^\circ = \tfrac{\sqrt{2}}{2}$, and the obtuse angle with the opposite cosine is $180^\circ - 45^\circ = 135^\circ$.
Let $\theta = \arcsin\tfrac{5}{13}$: a right triangle with opposite $5$ and hypotenuse $13$.
Draw the triangle the inner function names.
The adjacent side is $\sqrt{169 - 25} = 12$, so $\tan\theta = \tfrac{5}{12}$.
Arcsine answers in $[-90^\circ, 90^\circ]$, and a negative sine means a negative angle.
$\sin(-30^\circ) = -\tfrac{1}{2}$, so the answer is $-30^\circ$.
What is $\arcsin\left(-\frac{1}{2}\right)$, in degrees?
Answer:
What is $\arccos\left(-\frac{1}{2}\right)$, in degrees?
Answer:
What is $\arctan\left(1\right)$, in degrees?
Answer:
Find $\cos\left(\arcsin \dfrac{11}{61}\right)$ as a fraction.
Answer:
A loading ramp rises so that $\dfrac{\text{rise}}{\text{run}} = \sqrt{3}$. A regulation says the ramp must not be steeper than $65^\circ$. Does it pass?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Split $75^\circ$ into $45^\circ$ and $30^\circ$ to show that $\sin 75^\circ = \frac{\sqrt{6} + \sqrt{2}}{4}$. What is that value as a decimal, to four places?
Answer:
Why is $\arccos$ defined to give only angles between $0^\circ$ and $180^\circ$?
You can use the sum and double-angle formulas and read an inverse trigonometric value correctly. From memory: what is $\sin 2\theta$ in terms of $\sin\theta$ and $\cos\theta$, and why is $\arcsin(0.5)$ not $150^\circ$?
8. Your turn: $\sin 2\theta$ when $\sin\theta = \tfrac{5}{13}$ and $\theta$ is acute, step 2
19. Your turn: $\arcsin\left(-\tfrac{1}{2}\right)$, step 2