Trigonometric Function Flashcards

1
Q

the questions ask for

sin θ = ??

A

sin θ = y/r

r = sqrt(x^2 + y^2)

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2
Q

cos θ = ??

A

cos θ = x/r

r = sqrt(x^2 + y^2)

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3
Q

tan θ = ???

A

tan θ = y/x

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4
Q

csc θ = ??

A

csc θ = r/y

r = sqrt(x^2 + y^2)

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5
Q

sec θ = ??

A

sec θ = r/x

r = sqrt(x^2 + y^2)

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6
Q

cot θ = ??

A

cot θ = x/y

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7
Q

What if the question asks you to express the given trigonometric functions in terms of the same function of a positive acute angle??

sin 160 = ??

cos 250 = ??

A

for sin 160

180 - 160 = 20

is in the Q2 (starts from the left of the Q1 where -x,y is)

so sin(20)

for cos 250

it must be -cos(70)

it is not 270 - 250 = 20

is

250 - 180 = 70

why? Because cosine is measured from the x-axis

why negative? because the x is negative in Q2.

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8
Q

when getting the reference angle for sine and cosine, what should you know?

A

SINE is correlate with Y axis

COSINE is correlate with X axis

so ….

if the y is negative in the quadrant, it become a negative.

and when minusing, you must minus from the nearest vertical (y) axis for SINE and horizontal for COSINE (x)

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9
Q

how do you find the radians?

A

radian = degree * pi/180

to find degree

degree = radian ^ 180/pi

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10
Q

What does the complement refer to?

A

it refers to the angle that adds up to the quadrant degree like 90

so if angle = 58 degrees

complement is 90-58

32 degrees is the complement angle.

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11
Q

What does the supplement refer to?

A

add up to 180

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12
Q

how do you find rad or degree?

A

rad = degree * pi/180

degree = rad * 180/pi

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13
Q

cot theta = 1/9, how would you find sec and sin theta?

A

cot theta = x/y

so x = 1

y = 9

so find r = sqrt(82)

now sub it in in sec and sin.

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14
Q

how about tan theta = 1.073, how can you find sin theta and sec theta?

A

ez

make 1.073 to fraction

1073/1000

now you have x and y. 1073,1000.

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15
Q

SOH, CAH, TOA

A

sin = opposite of a / hypotenuse

cos = adjacent of a / hypotenuse

tan = opposite/adjacent

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16
Q

word question: When the question says the bottom of the face is 31 degrees, the top is 34 degrees, and the mountain’s base is 270, what does it look like?

A

ez

There’s a triangle for 31 degrees with the base as 270

Another one is for 34 degrees with the base at 270.

always draw it visually and act accordingly to the question.

17
Q

What is the sine rule?

A

sin A / a = sin B / b (for angle)

a / sin A = b / sin B (for side)

18
Q

What is the cosine rule?

A

In the cheat sheet, too much.

19
Q

How do you solve the distance between things using bearing?

A

Make a huge compass, then label it.

Then make angles

20
Q

determine the signs of the given function

cos 331

csc 260

A

refer to the quadrants

Cos is relate to X so follow the angle and see what x position is, is positive.

csc is relate to sin

and sin relate to y

so follow and see where the y lands, is negative

so cos 331 is positive

and csc 260 is negative.

21
Q

determine the quadrants in which the terminal side of the angle lies

cos theta = 0.9500

A

cos theta = -0.9500

this makes x/r

x is negative and r is positive

when x is negative, it is in QII and QIII, that the answer.

22
Q

where is the negative quadrant and positive quadrant for tan?

A

for tan

QI = positive

QII = negative

QIII = positive

QIV = negative

23
Q

Find the common quad for

csc theta < 0, tan theta < 0

A

csc theta relates to sin

sin is y

is says less than 0, meaning negative

that means QIII and QIV is negative

for tan, also negative, QII and QIV

the common quad is

QIV

24
Q

decide whether the given ratio is positive and negative

II, x/y

A

visit the quad

QII is x negative and y positive

so when negative and positive combined or - and +, it becomes negative, so negative.

in case of r/x , r is always positive.

25
express the given trigonometric functions in terms of the same function of a positive acute angle. sin 130, cos 210
follow sin 130, it lands at QII for QII: 180 - 130 = 50 so the answer is sin 50 follow cos 210, it lands at QIII for QIII: 210 + 180 = 390 wait but is ask for positive acute? so do 390 -360 or 210 - 180 = 30 and because cos relates to x and QIII is a negative x overall -cos30 so overall answer sin 50, - cos 30.
26
What if it goes beyond 360?
minus 360 till it reaches between 0-360
27
express the given trigonometric functions in terms of the same function of a positive acute angle. cos 1028 tan - 1028
cos 1028 - 360 = 668 668 - 360 = 308 cos 308 lands at QIV QIV: 360 - 308 = 52 x is positive so cos 52 tan -1028 = -308 go the opposite way QI QI is positive for tan so tan 52 so cos 52, tan 52.
28