Trigonometric functions

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1. Radian measure and arc length

Discover

Wrap a piece of string as long as the radius around the edge of a circle. The angle it covers at the center is the same for every circle. How many of these angles make a half turn?

  1. About 3.14, that is π\pi
  2. 180

Key idea

One radian is the angle that cuts off an arc as long as the radius.

180∘=π180^\circ = \pi radians. Degrees to radians: multiply by π180\dfrac{\pi}{180}. Radians to degrees: multiply by 180π\dfrac{180}{\pi}. Arc length: s=rθs = r\theta, with θ\theta in radians.

Worked example

Convert 135∘135^\circ to radians, then find the arc it cuts off on a circle of radius 4 cm.

135⋅π180=3π4135 \cdot \dfrac{\pi}{180} = \dfrac{3\pi}{4}
Multiply by π/180.
s=4⋅3π4=3πs = 4 \cdot \dfrac{3\pi}{4} = 3\pi cm
s = rθ, θ in radians.

Your turn, guided

Your turn: convert 5π6\dfrac{5\pi}{6} radians to degrees.

5⋅1806\dfrac{5 \cdot 180}{6}
Replace π by 180°.
=150∘= 150^\circ
900 ÷ 6.

Practice ★

  1. Convert 225∘225^\circ to radians. Give the exact value.
  2. Convert 330∘330^\circ to radians. Give the exact value.
  3. Convert 300∘300^\circ to radians. Give the exact value.
  4. Convert 30∘30^\circ to radians. Give the exact value.
  5. Convert 180∘180^\circ to radians. Give the exact value.

Watch out!

Practice ★★

  1. Convert π\pi radians to degrees.
  2. Convert 5π4\frac{5 \pi}{4} radians to degrees.
  3. Convert 3π3 \pi radians to degrees.
  1. A circle has radius 4 cm. What is the length of the arc cut off by a central angle of 3π4\frac{3 \pi}{4}? Give the exact value, in cm.
  2. A circle has radius 8 cm. What is the length of the arc cut off by a central angle of 2π3\frac{2 \pi}{3}? Give the exact value, in cm.
  3. A circle has radius 12 cm. What is the length of the arc cut off by a central angle of 5π6\frac{5 \pi}{6}? Give the exact value, in cm.

Exit ticket

Three questions to check you have got it.

  1. A circle has radius 4 cm. What is the length of the arc cut off by a central angle of 3π4\frac{3 \pi}{4}? Give the exact value, in cm.
  2. A circle has radius 8 cm. What is the length of the arc cut off by a central angle of 2π3\frac{2 \pi}{3}? Give the exact value, in cm.
  3. A circle has radius 12 cm. What is the length of the arc cut off by a central angle of 5π6\frac{5 \pi}{6}? Give the exact value, in cm.

Summary card

Radians

  • 180∘=π180^\circ = \pi rad.
  • Degrees → radians: ×π180\times \dfrac{\pi}{180}. Radians → degrees: ×180π\times \dfrac{180}{\pi}.
  • Arc length: s=rθs = r\theta, θ\theta in radians.

Watch out: never put degrees into s=rθs = r\theta.

2. Sine and cosine on the unit circle: special angles

Warm-up

Three quick questions from earlier lessons.

Discover

On the circle of radius 1 centered at the origin, start at (1,0)(1, 0) and turn by an angle θ\theta. You reach a point. At θ=π2\theta = \dfrac{\pi}{2}, which point is it?

  1. (0,1)(0, 1)
  2. (1,0)(1, 0)

Key idea

The unit circle has radius 1. Turning by θ\theta from (1,0)(1, 0) lands on a point we name with sine and cosine.

The point at angle θ\theta is (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta): cosine is the xx-coordinate, sine is the yy-coordinate. Special angles: cos⁡π6=32\cos\dfrac{\pi}{6} = \dfrac{\sqrt{3}}{2}, sin⁡π6=12\sin\dfrac{\pi}{6} = \dfrac{1}{2}; cos⁡π4=sin⁡π4=22\cos\dfrac{\pi}{4} = \sin\dfrac{\pi}{4} = \dfrac{\sqrt{2}}{2}; cos⁡π3=12\cos\dfrac{\pi}{3} = \dfrac{1}{2}, sin⁡π3=32\sin\dfrac{\pi}{3} = \dfrac{\sqrt{3}}{2}.

Worked example

Find sin⁡π3\sin\dfrac{\pi}{3} and cos⁡π3\cos\dfrac{\pi}{3}.

The point is (12,32)\left(\dfrac{1}{2}, \dfrac{\sqrt{3}}{2}\right)
π/3 = 60°: the point is high and close to the y-axis.
cos⁡π3=12\cos\dfrac{\pi}{3} = \dfrac{1}{2}
Cosine is the x-coordinate.
sin⁡π3=32\sin\dfrac{\pi}{3} = \dfrac{\sqrt{3}}{2}
Sine is the y-coordinate.

Your turn, guided

Your turn: θ=π6\theta = \dfrac{\pi}{6}.

The point is (32,12)\left(\dfrac{\sqrt{3}}{2}, \dfrac{1}{2}\right)
π/6 = 30°: the point is low and far to the right.
sin⁡π6=12\sin\dfrac{\pi}{6} = \dfrac{1}{2}
The y-coordinate.

Practice ★

  1. Find the exact value of cos⁡(π6)\cos\left(\frac{\pi}{6}\right).
  2. Find the exact value of cos⁡(π3)\cos\left(\frac{\pi}{3}\right).
  3. Find the exact value of cos⁡(π2)\cos\left(\frac{\pi}{2}\right).
  4. Find the exact value of cos⁡(0)\cos\left(0\right).

Watch out!

Practice ★★

  1. Find the exact value of sin⁡(π2)\sin\left(\frac{\pi}{2}\right).
  2. Find the exact value of sin⁡(π6)\sin\left(\frac{\pi}{6}\right).
  3. Find the exact value of sin⁡(π3)\sin\left(\frac{\pi}{3}\right).
  4. Find the exact value of sin⁡(0)\sin\left(0\right).
  1. Match each angle with its point (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta) on the unit circle.
    1. π6\dfrac{\pi}{6}
    2. π4\dfrac{\pi}{4}
    3. π3\dfrac{\pi}{3}
    4. π2\dfrac{\pi}{2}
    1. (32,12)\left(\dfrac{\sqrt{3}}{2}, \dfrac{1}{2}\right)
    2. (22,22)\left(\dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{2}}{2}\right)
    3. (12,32)\left(\dfrac{1}{2}, \dfrac{\sqrt{3}}{2}\right)
    4. (0,1)(0, 1)

Exit ticket

Three questions to check you have got it.

  1. Find the exact value of sin⁡(π2)\sin\left(\frac{\pi}{2}\right).
  2. Find the exact value of sin⁡(π6)\sin\left(\frac{\pi}{6}\right).
  3. Find the exact value of sin⁡(π3)\sin\left(\frac{\pi}{3}\right).

Summary card

The unit circle

  • The point at angle θ\theta is (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta).
  • π6\dfrac{\pi}{6}: (32,12)\left(\dfrac{\sqrt{3}}{2}, \dfrac{1}{2}\right); π4\dfrac{\pi}{4}: (22,22)\left(\dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{2}}{2}\right); π3\dfrac{\pi}{3}: (12,32)\left(\dfrac{1}{2}, \dfrac{\sqrt{3}}{2}\right).

Watch out: cosine is xx, sine is yy.

3. Reference angles and signs in each quadrant

Warm-up

Three quick questions from earlier lessons.

Discover

The point at 150∘150^\circ is the mirror image of the point at 30∘30^\circ across the yy-axis. What is cos⁡150∘\cos 150^\circ?

  1. −32-\dfrac{\sqrt{3}}{2}
  2. 32\dfrac{\sqrt{3}}{2}

Key idea

Every angle has a reference angle: the acute angle between its side and the xx-axis.

sin⁡\sin and cos⁡\cos of an angle have the same size as for its reference angle; the quadrant gives the sign. Quadrant I: both positive. II: sin⁡>0\sin > 0, cos⁡<0\cos < 0. III: both negative. IV: sin⁡<0\sin < 0, cos⁡>0\cos > 0.

Worked example

Find sin⁡225∘\sin 225^\circ.

Quadrant III
225° is between 180° and 270°.
Reference angle: 225∘−180∘=45∘225^\circ - 180^\circ = 45^\circ
Measure from 180°.
sin⁡225∘=−22\sin 225^\circ = -\dfrac{\sqrt{2}}{2}
sin 45° = √2/2, and sine is negative in quadrant III.

Your turn, guided

Your turn: cos⁡300∘\cos 300^\circ.

Reference angle: 360∘−300∘=60∘360^\circ - 300^\circ = 60^\circ
300° is between 270° and 360°: quadrant IV.
cos⁡300∘=12\cos 300^\circ = \dfrac{1}{2}
cos 60° = 1/2, and cosine is positive in quadrant IV.

Practice ★

  1. What is the reference angle of 200∘200^\circ, in degrees?
  2. What is the reference angle of 240∘240^\circ, in degrees?
  3. What is the reference angle of 250∘250^\circ, in degrees?

Watch out!

Practice ★★

  1. Find the exact value of cos⁡(5π4)\cos\left(\frac{5 \pi}{4}\right).
  2. Find the exact value of cos⁡(4π3)\cos\left(\frac{4 \pi}{3}\right).
  3. Find the exact value of cos⁡(2π3)\cos\left(\frac{2 \pi}{3}\right).
  4. Find the exact value of cos⁡(7π4)\cos\left(\frac{7 \pi}{4}\right).
  1. In which quadrant is sin⁡θ<0\sin\theta < 0 and cos⁡θ>0\cos\theta > 0?
    1. Quadrant II
    2. Quadrant III
    3. Quadrant IV

Exit ticket

Three questions to check you have got it.

  1. Find the exact value of cos⁡(5π4)\cos\left(\frac{5 \pi}{4}\right).
  2. Find the exact value of cos⁡(4π3)\cos\left(\frac{4 \pi}{3}\right).
  3. Find the exact value of cos⁡(2π3)\cos\left(\frac{2 \pi}{3}\right).

Summary card

Reference angles

  1. Find the quadrant.
  2. Reference angle: acute angle to the xx-axis (180∘−θ180^\circ - \theta, θ−180∘\theta - 180^\circ or 360∘−θ360^\circ - \theta).
  3. Value of the reference angle, with the quadrant's sign.

Watch out: cosine is negative on the left (II, III), sine below (III, IV).

4. The Pythagorean identity and simplifying

Warm-up

Three quick questions from earlier lessons.

Discover

The point (cos⁡x,sin⁡x)(\cos x, \sin x) is on the unit circle, whose equation is X2+Y2=1X^2 + Y^2 = 1. What does that tell you?

  1. cos⁡2x+sin⁡2x=1\cos^2 x + \sin^2 x = 1
  2. cos⁡x+sin⁡x=1\cos x + \sin x = 1

Key idea

An identity is an equation that is true for every value of the variable.

sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1, so cos⁡2x=1−sin⁡2x\cos^2 x = 1 - \sin^2 x and sin⁡2x=1−cos⁡2x\sin^2 x = 1 - \cos^2 x. Also tan⁡x=sin⁡xcos⁡x\tan x = \dfrac{\sin x}{\cos x}. Use them to find one value from another (the quadrant gives the sign) and to simplify.

Worked example

cos⁡θ=−35\cos\theta = -\dfrac{3}{5} and θ\theta is in quadrant III. Find sin⁡θ\sin\theta.

sin⁡2θ=1−925=1625\sin^2\theta = 1 - \dfrac{9}{25} = \dfrac{16}{25}
The Pythagorean identity.
sin⁡θ=±45\sin\theta = \pm\dfrac{4}{5}
Two square roots are possible.
sin⁡θ=−45\sin\theta = -\dfrac{4}{5}
Sine is negative in quadrant III.

Your turn, guided

Your turn: sin⁡θ=513\sin\theta = \dfrac{5}{13} and θ\theta is in quadrant I.

cos⁡2θ=144169\cos^2\theta = \dfrac{144}{169}
1 − 25/169.
cos⁡θ=1213\cos\theta = \dfrac{12}{13}
Positive in quadrant I.

Practice ★

  1. sin⁡θ=15\sin\theta = \frac{1}{5} and θ\theta is in quadrant II. Find the exact value of cos⁡θ\cos\theta.
  2. sin⁡θ=38\sin\theta = \frac{3}{8} and θ\theta is in quadrant II. Find the exact value of cos⁡θ\cos\theta.
  3. sin⁡θ=16\sin\theta = \frac{1}{6} and θ\theta is in quadrant II. Find the exact value of cos⁡θ\cos\theta.
  4. sin⁡θ=13\sin\theta = \frac{1}{3} and θ\theta is in quadrant II. Find the exact value of cos⁡θ\cos\theta.

Watch out!

Practice ★★

  1. Which expression is equal to 1−cos⁡2xsin⁡x\dfrac{1 - \cos^2 x}{\sin x} (where sin⁡x≠0\sin x \ne 0)?
    1. sin⁡x\sin x
    2. cos⁡x\cos x
    3. 1−cos⁡x1 - \cos x
  1. Which equation is true for every xx?
    1. sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1
    2. sin⁡x+cos⁡x=1\sin x + \cos x = 1
    3. sin⁡2x−cos⁡2x=1\sin^2 x - \cos^2 x = 1
    4. tan⁡x=cos⁡xsin⁡x\tan x = \dfrac{\cos x}{\sin x}

Exit ticket

Three questions to check you have got it.

  1. sin⁡θ=15\sin\theta = \frac{1}{5} and θ\theta is in quadrant II. Find the exact value of cos⁡θ\cos\theta.
  2. sin⁡θ=38\sin\theta = \frac{3}{8} and θ\theta is in quadrant II. Find the exact value of cos⁡θ\cos\theta.
  3. sin⁡θ=16\sin\theta = \frac{1}{6} and θ\theta is in quadrant II. Find the exact value of cos⁡θ\cos\theta.

Summary card

The Pythagorean identity

  • sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1 for every xx.
  • From one value: square, subtract from 1, take the root, choose the sign by quadrant.

Watch out: the identity needs the squares.

5. Sinusoidal graphs: amplitude, period, midline

Warm-up

Three quick questions from earlier lessons.

Discover

Move the sliders and watch the graph of y=asin⁡(bx)+ky = a\sin(bx) + k.

The graph of y = a sin(bx) + k: a wave around the line y = k, a units high, that repeats faster as b grows.

What happens to the waves when bb goes from 1 to 2?

  1. They get twice as narrow
  2. They get twice as high
  3. They get twice as wide

Key idea

y=Asin⁡(Bx)+Dy = A\sin(Bx) + D and y=Acos⁡(Bx)+Dy = A\cos(Bx) + D are waves.

Amplitude ∣A∣|A|: the height above the midline. Midline y=Dy = D. Period 2πB\dfrac{2\pi}{B} (for B>0B > 0): the length of one cycle. Maximum D+∣A∣D + |A|, minimum D−∣A∣D - |A|.

Worked example

Describe y=3sin⁡(2x)+1y = 3\sin(2x) + 1.

Amplitude: 33
The number in front of sin.
Midline: y=1y = 1
The number added.
Period: 2π2=π\dfrac{2\pi}{2} = \pi
2π divided by B = 2.
Range: from −2-2 to 44
Midline ± amplitude.

Your turn, guided

Your turn: y=5cos⁡(4x)−2y = 5\cos(4x) - 2.

Amplitude: 55, midline y=−2y = -2
In front of cos.
Period: π2\dfrac{\pi}{2}
2π ÷ 4.

Practice ★

  1. What is the amplitude of y=2sin⁡(5x)+1y = 2\sin(5x) + 1?
  2. What is the amplitude of y=4sin⁡(5x)+3y = 4\sin(5x) + 3?
  3. What is the amplitude of y=4sin⁡(2x)+7y = 4\sin(2x) + 7?
  4. What is the amplitude of y=6sin⁡(3x)+8y = 6\sin(3x) + 8?

Watch out!

Practice ★★

  1. What is the period of y=sin⁡(6x)y = \sin(6x)?
  2. What is the period of y=sin⁡(5x)y = \sin(5x)?
  3. What is the period of y=sin⁡(3x)y = \sin(3x)?
  1. What is the midline of y=9cos⁡x−4y = 9\cos x - 4?
    1. y=−4y = -4
    2. y=4y = 4
    3. y=9y = 9
  2. What is the midline of y=8cos⁡x−9y = 8\cos x - 9?
    1. y=−9y = -9
    2. y=9y = 9
    3. y=8y = 8

Exit ticket

Three questions to check you have got it.

  1. What is the period of y=sin⁡(6x)y = \sin(6x)?
  2. What is the period of y=sin⁡(5x)y = \sin(5x)?
  3. What is the period of y=sin⁡(3x)y = \sin(3x)?

Summary card

Sinusoidal graphs y=Asin⁡(Bx)+Dy = A\sin(Bx) + D

  • Amplitude ∣A∣|A|, midline y=Dy = D.
  • Period 2πB\dfrac{2\pi}{B}.
  • Maximum D+∣A∣D + |A|, minimum D−∣A∣D - |A|.

Watch out: a bigger BB means a shorter period.

6. Phase shift and sinusoidal models

Warm-up

Three quick questions from earlier lessons.

Discover

y=sin⁡(x−π2)y = \sin\left(x - \dfrac{\pi}{2}\right) uses the same wave as y=sin⁡xy = \sin x. Where does it cross its midline going up for the first time with x≥0x \ge 0?

  1. At x=π2x = \dfrac{\pi}{2}
  2. At x=−π2x = -\dfrac{\pi}{2}

Key idea

A sinusoid can also be shifted sideways.

y=Asin⁡(B(x−h))+Dy = A\sin\big(B(x - h)\big) + D is shifted hh to the right (the phase shift). In the form sin⁡(Bx−C)\sin(Bx - C), factor out BB first: the shift is CB\dfrac{C}{B}.

To model a situation, read AA, DD and the period from the data, then B=2πperiodB = \dfrac{2\pi}{\text{period}}.

Worked example

Find the phase shift of y=cos⁡(3x−π)y = \cos(3x - \pi).

y=cos⁡(3(x−π3))y = \cos\left(3\left(x - \dfrac{\pi}{3}\right)\right)
Factor out B = 3.
Shift: π3\dfrac{\pi}{3} to the right
The form B(x − h).

Your turn, guided

Your turn: y=sin⁡(2x−6)y = \sin(2x - 6).

y=sin⁡(2(x−3))y = \sin\big(2(x - 3)\big)
Factor out 2.
Shift: 33 to the right
h in B(x − h).

Practice ★

  1. What is the largest value of y=5sin⁡x+4y = 5\sin x + 4?
  2. What is the largest value of y=2sin⁡x+9y = 2\sin x + 9?
  3. What is the largest value of y=6sin⁡x+2y = 6\sin x + 2?

Watch out!

Practice ★★

  1. The graph of y=cos⁡(2x−4)y = \cos(2x - 4) is the graph of y=cos⁡(2x)y = \cos(2x) shifted to the right. By how much?
  2. The graph of y=cos⁡(3x−15)y = \cos(3x - 15) is the graph of y=cos⁡(3x)y = \cos(3x) shifted to the right. By how much?
  3. The graph of y=cos⁡(4x−16)y = \cos(4x - 16) is the graph of y=cos⁡(4x)y = \cos(4x) shifted to the right. By how much?
  1. Write y=Asin⁡(Bx)+Dy = A\sin(Bx) + D, with A,B>0A, B > 0, for a sine curve with amplitude 2, period π2\frac{\pi}{2} and midline y=9y = 9.
  2. Write y=Asin⁡(Bx)+Dy = A\sin(Bx) + D, with A,B>0A, B > 0, for a sine curve with amplitude 2, period 2π3\frac{2 \pi}{3} and midline y=6y = 6.
  3. Write y=Asin⁡(Bx)+Dy = A\sin(Bx) + D, with A,B>0A, B > 0, for a sine curve with amplitude 4, period 2π3\frac{2 \pi}{3} and midline y=1y = 1.

Exit ticket

Three questions to check you have got it.

  1. Write y=Asin⁡(Bx)+Dy = A\sin(Bx) + D, with A,B>0A, B > 0, for a sine curve with amplitude 2, period π2\frac{\pi}{2} and midline y=9y = 9.
  2. Write y=Asin⁡(Bx)+Dy = A\sin(Bx) + D, with A,B>0A, B > 0, for a sine curve with amplitude 2, period 2π3\frac{2 \pi}{3} and midline y=6y = 6.
  3. Write y=Asin⁡(Bx)+Dy = A\sin(Bx) + D, with A,B>0A, B > 0, for a sine curve with amplitude 4, period 2π3\frac{2 \pi}{3} and midline y=1y = 1.

Summary card

Phase shift and models

  • Asin⁡(B(x−h))+DA\sin\big(B(x - h)\big) + D: shift hh right.
  • sin⁡(Bx−C)\sin(Bx - C): shift CB\dfrac{C}{B}, after factoring out BB.
  • Model: AA = half the range, DD = middle value, B=2πperiodB = \dfrac{2\pi}{\text{period}}.

Watch out: factor out BB before reading the shift.

7. Area of a triangle: ½ab sin C

Warm-up

Three quick questions from earlier lessons.

Discover

A triangle has sides 6 and 8 with a right angle between them. What is its area?

  1. 24
  2. 48

Key idea

Area = ½ab sin C.

If two sides aa and bb make an angle CC, the height on side aa is bsin⁡Cb \sin C, so the area is 12absin⁡C\tfrac12 ab \sin C. Use the angle between the two sides. For C=90°C = 90° it gives 12ab\tfrac12 ab.

Worked example

Sides 10 and 7 with a 30° angle between them.

7sin⁡30°=3.57 \sin 30° = 3.5
Height.
12⋅10⋅7⋅12=17.5\tfrac12 \cdot 10 \cdot 7 \cdot \tfrac12 = 17.5
Area.

Your turn, guided

Your turn: sides 4 and 6, angle 30°. Area?

66
½ · 4 · 6 · ½.

Practice ★

  1. A triangle has sides 5 cm and 12 cm with an angle of 30° between them. What is its area, in cm²?
  2. A triangle has sides 12 cm and 6 cm with an angle of 30° between them. What is its area, in cm²?
  1. In triangle PQR, PQ = 9, PR = 11. To find the area with ½ · PQ · PR · sin(angle), which angle do you use?
    1. The angle at P, between the two sides
    2. The angle at Q
    3. Any angle of the triangle

Watch out!

Practice ★★

  1. A triangle has sides 5 m and 3 m with an angle of 60° between them. What is its exact area, in m²?
  2. A triangle has sides 8 m and 11 m with an angle of 60° between them. What is its exact area, in m²?
  1. A triangle has sides 5 cm and 12 cm with an angle of 30° between them. What is its area, in cm²?

Exit ticket

Three questions to check you have got it.

  1. A triangle has sides 5 cm and 12 cm with an angle of 30° between them. What is its area, in cm²?
  2. A triangle has sides 12 cm and 6 cm with an angle of 30° between them. What is its area, in cm²?
  3. A triangle has sides 3 cm and 11 cm with an angle of 30° between them. What is its area, in cm²?

Summary card

Area of any triangle

½ab sin C, with C between a and b.

Watch out: use the included angle.

8. The law of sines

Warm-up

Three quick questions from earlier lessons.

Discover

In a triangle, the largest angle is opposite which side?

  1. The longest side
  2. The shortest side

Key idea

Sides are proportional to the sines of their opposite angles.

In any triangle, asin⁡A=bsin⁡B=csin⁡C\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}. Use it when you know a side and its opposite angle plus one more part: two angles and a side, or two sides and an angle that is not between them (check for a second possible triangle).

Worked example

A = 30°, B = 45°, a = 6.

bsin⁡45°=6sin⁡30°\dfrac{b}{\sin 45°} = \dfrac{6}{\sin 30°}
Set up.
b=6⋅2/21/2=62b = 6 \cdot \dfrac{\sqrt2/2}{1/2} = 6\sqrt2
Solve.

Your turn, guided

Your turn: A = 30°, a = 5. What is a ÷ sin A?

1010
5 ÷ ½.

Practice ★

  1. You know some parts of a triangle that has no right angle. When is the law of sines the right tool?
    1. When you know two angles and a side, such as A = 40°, B = 65° and a = 7
    2. When you know the three sides
    3. When you know two sides and the angle between them
  2. You know some parts of a triangle that has no right angle. When is the law of sines the right tool?
    1. When you know two angles and a side, such as A = 40°, B = 65° and a = 3
    2. When you know the three sides
    3. When you know two sides and the angle between them
  1. In triangle ABC, angle A = 30°, angle B = 45° and side a = 2 (opposite A). Use the law of sines to find side b (opposite B), exactly.

Watch out!

Practice ★★

  1. In triangle ABC, angle A = 30°, side a = 7 and side b = 11. Use the law of sines to find sin B.
  2. In triangle ABC, angle A = 30°, side a = 11 and side b = 10. Use the law of sines to find sin B.
  1. In triangle ABC, angle A = 30°, angle B = 45° and side a = 2 (opposite A). Use the law of sines to find side b (opposite B), exactly.

Exit ticket

Three questions to check you have got it.

  1. In triangle ABC, angle A = 30°, angle B = 45° and side a = 2 (opposite A). Use the law of sines to find side b (opposite B), exactly.
  2. In triangle ABC, angle A = 30°, angle B = 45° and side a = 6 (opposite A). Use the law of sines to find side b (opposite B), exactly.
  3. In triangle ABC, angle A = 30°, angle B = 45° and side a = 4 (opposite A). Use the law of sines to find side b (opposite B), exactly.

Summary card

Law of sines

a / sin A = b / sin B = c / sin C.

Watch out: pair each side with its opposite angle.

9. The law of cosines

Warm-up

Three quick questions from earlier lessons.

Discover

If the angle between sides 3 and 4 is 90°, the third side is 5. If the angle is bigger than 90°, the third side is…

  1. Longer than 5
  2. Still 5

√(5² + 8²) ≈ 9.43. Set C above 90°. How long is c?

  1. Longer than 9.43
  2. Still 9.43

Key idea

The law of cosines corrects the Pythagorean theorem.

In any triangle, c2=a2+b2−2abcos⁡Cc^2 = a^2 + b^2 - 2ab \cos C, where CC is the angle opposite cc. Use it with two sides and the angle between them (to find the third side) or three sides (to find an angle: cos⁡C=a2+b2−c22ab\cos C = \dfrac{a^2 + b^2 - c^2}{2ab}). For C=90°C = 90° it is the Pythagorean theorem; for an obtuse CC, cos⁡C<0\cos C < 0 and cc gets longer.

Worked example

a = 5, b = 8, C = 60°.

c2=25+64−2⋅5⋅8⋅12c^2 = 25 + 64 - 2 \cdot 5 \cdot 8 \cdot \tfrac12
Formula.
c2=49c^2 = 49, c=7c = 7
Result.

Your turn, guided

Your turn: a = 3, b = 5, C = 120°. What is c²?

4949
9 + 25 + 15.

Practice ★

  1. In triangle ABC, a = 12, b = 8 and the angle between them is C = 60°. Use the law of cosines to find c².
  2. In triangle ABC, a = 3, b = 3 and the angle between them is C = 60°. Use the law of cosines to find c².
  1. In triangle ABC, a = 12, b = 5 and C = 120°. Use the law of cosines to find c².

Watch out!

Practice ★★

  1. A triangle has sides a = 5, b = 6 and c = 8. Find cos C, the cosine of the angle opposite c.
  2. A triangle has sides a = 4, b = 8 and c = 6. Find cos C, the cosine of the angle opposite c.
  1. Boss problem. Two hikers leave camp along straight paths that make a 60° angle. One walks 6 km, the other 8 km.

Exit ticket

Three questions to check you have got it.

  1. In triangle ABC, a = 12, b = 8 and the angle between them is C = 60°. Use the law of cosines to find c².
  2. In triangle ABC, a = 3, b = 3 and the angle between them is C = 60°. Use the law of cosines to find c².
  3. In triangle ABC, a = 8, b = 7 and the angle between them is C = 60°. Use the law of cosines to find c².

Summary card

Law of cosines

c² = a² + b² − 2ab cos C.

Watch out: Pythagoras alone works only for a right angle.

Answers: Trigonometric functions

1. Radian measure and arc length

A

  1. 5π4\frac{5 \pi}{4}
  2. 11π6\frac{11 \pi}{6}
  3. 5π3\frac{5 \pi}{3}
  4. π6\frac{\pi}{6}
  5. π\pi

B

  1. 180180
  2. 225225
  3. 540540

C

  1. 3π3 \pi
  2. 16π3\frac{16 \pi}{3}
  3. 10π10 \pi

D

  1. 3π3 \pi
  2. 16π3\frac{16 \pi}{3}
  3. 10π10 \pi

2. Sine and cosine on the unit circle: special angles

A

  1. 32\frac{\sqrt{3}}{2}
  2. 12\frac{1}{2}
  3. 00
  4. 11

B

  1. 11
  2. 12\frac{1}{2}
  3. 32\frac{\sqrt{3}}{2}
  4. 00

C

  1. π6\dfrac{\pi}{6} → (32,12)\left(\dfrac{\sqrt{3}}{2}, \dfrac{1}{2}\right) · π4\dfrac{\pi}{4} → (22,22)\left(\dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{2}}{2}\right) · π3\dfrac{\pi}{3} → (12,32)\left(\dfrac{1}{2}, \dfrac{\sqrt{3}}{2}\right) · π2\dfrac{\pi}{2} → (0,1)(0, 1)

D

  1. 11
  2. 12\frac{1}{2}
  3. 32\frac{\sqrt{3}}{2}

3. Reference angles and signs in each quadrant

A

  1. 2020
  2. 6060
  3. 7070

B

  1. −22- \frac{\sqrt{2}}{2}
  2. −12- \frac{1}{2}
  3. −12- \frac{1}{2}
  4. 22\frac{\sqrt{2}}{2}

C

  1. Quadrant IV

D

  1. −22- \frac{\sqrt{2}}{2}
  2. −12- \frac{1}{2}
  3. −12- \frac{1}{2}

4. The Pythagorean identity and simplifying

A

  1. −265- \frac{2 \sqrt{6}}{5}
  2. −558- \frac{\sqrt{55}}{8}
  3. −356- \frac{\sqrt{35}}{6}
  4. −223- \frac{2 \sqrt{2}}{3}

B

  1. sin⁡x\sin x

C

  1. sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1

D

  1. −265- \frac{2 \sqrt{6}}{5}
  2. −558- \frac{\sqrt{55}}{8}
  3. −356- \frac{\sqrt{35}}{6}

5. Sinusoidal graphs: amplitude, period, midline

A

  1. 22
  2. 44
  3. 44
  4. 66

B

  1. π3\frac{\pi}{3}
  2. 2π5\frac{2 \pi}{5}
  3. 2π3\frac{2 \pi}{3}

C

  1. y=−4y = -4
  2. y=−9y = -9

D

  1. π3\frac{\pi}{3}
  2. 2π5\frac{2 \pi}{5}
  3. 2π3\frac{2 \pi}{3}

6. Phase shift and sinusoidal models

A

  1. 99
  2. 1111
  3. 88

B

  1. 22
  2. 55
  3. 44

C

  1. 2sin⁡(4x)+92 \sin{\left(4 x \right)} + 9
  2. 2sin⁡(3x)+62 \sin{\left(3 x \right)} + 6
  3. 4sin⁡(3x)+14 \sin{\left(3 x \right)} + 1

D

  1. 2sin⁡(4x)+92 \sin{\left(4 x \right)} + 9
  2. 2sin⁡(3x)+62 \sin{\left(3 x \right)} + 6
  3. 4sin⁡(3x)+14 \sin{\left(3 x \right)} + 1

7. Area of a triangle: ½ab sin C

A

  1. 1515
  2. 1818

B

  1. The angle at P, between the two sides

C

  1. 1534\frac{15 \sqrt{3}}{4}
  2. 22322 \sqrt{3}

D

  1. 1515

E

  1. 1515
  2. 1818
  3. 334\frac{33}{4}

8. The law of sines

A

  1. When you know two angles and a side, such as A = 40°, B = 65° and a = 7
  2. When you know two angles and a side, such as A = 40°, B = 65° and a = 3

B

  1. 222 \sqrt{2}

C

  1. 1114\frac{11}{14}
  2. 511\frac{5}{11}

D

  1. 222 \sqrt{2}

E

  1. 222 \sqrt{2}
  2. 626 \sqrt{2}
  3. 424 \sqrt{2}

9. The law of cosines

A

  1. 112112
  2. 99

B

  1. 229229

C

  1. −120- \frac{1}{20}
  2. 1116\frac{11}{16}

D

  1. 12312 \sqrt{3}

E

  1. 112112
  2. 99
  3. 5757