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.

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