The Lumenary · Grades 9–12

Trigonometry

Name: ______________________

Date: ______________________

You will learn to

Define sine, cosine and tangent for acute angles via similar right triangles, and for all angles via the unit circle
Convert between degrees and radians, and sketch the graphs of $\sin$ , $\cos$ and $\tan$
Prove $\sin^2\theta + \cos^2\theta = 1$ and the addition formulas, and derive the double angle formulas
Prove and apply the law of cosines and the law of sines
Recognise common errors, such as reading $\sin\theta$ as a product or using degrees where radians are required

1.
A ladder $10$ m long leans against a vertical wall. The angle $\theta$ between the ladder and the level ground has $\cos\theta = 0.6$ . How many metres is the foot of the ladder from the wall?
2.
Convert $\dfrac{3\pi}{4}$ radians to degrees.
3.
What is $\sin 120^\circ$ ?
- A. $\dfrac{\sqrt3}{2}$
- B. $-\dfrac{\sqrt3}{2}$
- C. $-\dfrac12$
- D. $\dfrac12$
4.
Which expression equals $\sin 75^\circ$ ?
- A. $\dfrac{\sqrt6 + \sqrt2}{4}$
- B. $\sin 45^\circ + \sin 30^\circ = \dfrac{\sqrt2 + 1}{2}$
- C. $\sin 45^\circ \cdot \sin 30^\circ = \dfrac{\sqrt2}{4}$
- D. $\dfrac{5}{2}\sin 30^\circ = \dfrac54$
5.
In triangle $ABC$ , $a = 5$ , $b = 8$ and the angle $C$ between them is $60^\circ$ . Find $c$ .
6.
Our proof of the law of cosines puts $C$ at the origin, $B = (a, 0)$ and $A = (b\cos C, b\sin C)$ . Why does the proof also work when the angle $C$ is obtuse?
- A.Because the unit-circle definition gives the point at angle $C$ as $(\cos C, \sin C)$ for every angle; for obtuse $C$ , $\cos C$ is simply negative
- B.It does not: the law of cosines only holds for acute triangles
- C.Because an obtuse triangle can be split into two right triangles
- D.Because $\sin C$ is negative for obtuse $C$
7.
How many angles $x$ with $0^\circ \le x < 720^\circ$ satisfy $\sin x = 0.3$ ?
8.
Compute $8 \sin 10^\circ \sin 50^\circ \sin 70^\circ$ exactly. (Hint below if you need it.)
9.
Which Lean statement says correctly that $\sin(\alpha + \beta) = \sin\alpha + \sin\beta$ is not an identity?
- A.¬ ∀ α β : ℝ, sin (α + β) = sin α + sin β
- B.∀ α β : ℝ, sin (α + β) ≠ sin α + sin β
- C.∃ α β : ℝ, sin (α + β) = sin α + sin β
- D.sin (π / 2 + π / 2) = 2
10.
Write third_side(a, b, angle_c_degrees) that returns the side $c$ of a triangle with sides $a$ , $b$ and the angle $C$ between them given in degrees, using the law of cosines. Remember that math.cos expects radians.
Computational — Practice

Answer key — Trigonometry

1.
The ladder is the hypotenuse ( $10$ m) and the distance from the wall is the side adjacent to $\theta$ . So the distance is $10\cos\theta = 10 \times 0.6 = 6$ m. (By Pythagoras the ladder reaches $\sqrt{100 - 36} = 8$ m up the wall.)
2.
$\dfrac{3\pi}{4} \times \dfrac{180^\circ}{\pi} = 3 \times 45^\circ = 135^\circ$ .
3.
The point at $120^\circ$ is the reflection in the $y$ -axis of the point at $60^\circ$ , which is $\left(\tfrac12, \tfrac{\sqrt3}{2}\right)$ . Reflection changes the sign of $x$ only, so the point is $\left(-\tfrac12, \tfrac{\sqrt3}{2}\right)$ and $\sin 120^\circ = \tfrac{\sqrt3}{2}$ .
4.
$\sin 75^\circ = \sin 45^\circ\cos 30^\circ + \cos 45^\circ \sin 30^\circ = \tfrac{\sqrt2}{2}\cdot\tfrac{\sqrt3}{2} + \tfrac{\sqrt2}{2}\cdot\tfrac12 = \tfrac{\sqrt6 + \sqrt2}{4} \approx 0.966$ .
5.
$c^2 = 5^2 + 8^2 - 2\cdot 5\cdot 8 \cdot \tfrac12 = 25 + 64 - 40 = 49$ , so $c = 7$ .
6.
The unit-circle definition places $A$ at $(b\cos C, b\sin C)$ for every angle $C$ . For obtuse $C$ the point is in the second quadrant, with $\cos C < 0$ and $\sin C > 0$ . The computation $c^2 = (b\cos C - a)^2 + (b\sin C)^2 = a^2 + b^2 - 2ab\cos C$ uses only algebra and $\sin^2 + \cos^2 = 1$ , so it is valid in all cases.
7.
In $[0^\circ, 360^\circ)$ the solutions are $x_1 = \arcsin 0.3 \approx 17.46^\circ$ and $x_2 = 180^\circ - x_1 \approx 162.54^\circ$ (the line $y = 0.3$ meets the unit circle twice). Adding $360^\circ$ gives two more, $\approx 377.46^\circ$ and $\approx 522.54^\circ$ . So there are $4$ .
8.
By the addition formulas, $\sin(60^\circ - \theta)\sin(60^\circ + \theta) = \sin^2 60^\circ\cos^2\theta - \cos^2 60^\circ\sin^2\theta = \tfrac34(1 - \sin^2\theta) - \tfrac14\sin^2\theta = \tfrac34 - \sin^2\theta$ . So $\sin\theta\sin(60^\circ - \theta)\sin(60^\circ + \theta) = \tfrac34\sin\theta - \sin^3\theta = \tfrac14(3\sin\theta - 4\sin^3\theta) = \tfrac14\sin 3\theta$ , using the triple angle formula (from $\sin(2\theta + \theta)$ ). With $\theta = 10^\circ$ : $\sin 10^\circ\sin 50^\circ\sin 70^\circ = \tfrac14\sin 30^\circ = \tfrac18$ . Hence the answer is $8 \cdot \tfrac18 = 1$ .
9.
An identity is a statement of the form ∀ α β, …. Its negation is ¬ ∀ α β, …, which is equivalent to ∃ α β, sin (α + β) ≠ sin α + sin β: one counterexample suffices. The statement with ≠ inside the ∀ would claim the equation never holds, which is false.
10.
```
import math

def third_side(a, b, angle_c_degrees):
    c_radians = math.radians(angle_c_degrees)
    return math.sqrt(a**2 + b**2 - 2*a*b*math.cos(c_radians))
```
The starter computes $\cos 60$ in radians ( $\approx -0.952$ ) instead of $\cos 60^\circ = 0.5$ , which is the degree/radian misconception in code. With the conversion, $(3, 4, 90^\circ)$ gives $5$ (Pythagoras), $(5, 8, 60^\circ)$ gives $7$ , and $(1, 1, 120^\circ)$ gives $\sqrt3$ .