Trigonometry Flashcards

1
Q

1 βˆ’ 𝑠𝑖𝑛^2 πœƒ =

A

π‘π‘œπ‘ ^2 πœƒ

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2
Q

1 + π‘‘π‘Žπ‘›^2 πœƒ =

A

𝑠𝑒𝑐^2 πœƒ

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3
Q

1 + π‘π‘œπ‘‘^2 πœƒ =

A

π‘π‘œπ‘ π‘’π‘^2 πœƒ

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4
Q

𝑠𝑖𝑛^4 πœƒ + π‘π‘œπ‘ ^4 πœƒ =

A

1 βˆ’ 2 𝑠𝑖𝑛^2 πœƒ π‘π‘œπ‘ ^2 πœƒ

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5
Q

𝑠𝑖𝑛^6 πœƒ + π‘π‘œπ‘ ^6 πœƒ =

A

1 βˆ’ 3 𝑠𝑖𝑛^2 πœƒ π‘π‘œπ‘ ^2 πœƒ

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6
Q

If π‘₯ = π‘Žπ‘ π‘–π‘› πœƒ + 𝑏 π‘π‘œπ‘  πœƒ & 𝑦 = a cos πœƒ βˆ’ 𝑏 sin πœƒ OR π‘₯ = π‘Ž 𝑠𝑖𝑛 πœƒ + 𝑏 π‘π‘œπ‘  πœƒ & 𝑦 = 𝑏 𝑠𝑖𝑛 πœƒ βˆ’ π‘Ž π‘π‘œπ‘  πœƒ
then

A

π‘₯^2 + 𝑦^2 = π‘Ž^2 + 𝑏^2

Approach: square & add the equations

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7
Q

𝑠𝑖𝑛(𝐴 + 𝐡) =

A

𝑠𝑖𝑛 𝐴 π‘π‘œπ‘  𝐡 + π‘π‘œπ‘  𝐴 𝑠𝑖𝑛 B

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8
Q

𝑠𝑖𝑛(𝐴 βˆ’ 𝐡) =

A

𝑠𝑖𝑛 𝐴 π‘π‘œπ‘  𝐡 βˆ’ π‘π‘œπ‘  𝐴 𝑠𝑖𝑛 B

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9
Q

π‘π‘œπ‘ (𝐴 + 𝐡) =

A

π‘π‘œπ‘  𝐴 π‘π‘œπ‘  𝐡 βˆ’ 𝑠𝑖𝑛 𝐴 𝑠𝑖𝑛 B

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10
Q

π‘π‘œπ‘ (𝐴 βˆ’ 𝐡) =

A

π‘π‘œπ‘  𝐴 π‘π‘œπ‘  𝐡 + 𝑠𝑖𝑛 𝐴 𝑠𝑖𝑛 B

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11
Q

π‘‘π‘Žπ‘›(𝐴 + 𝐡) =

A

π‘‘π‘Žπ‘› 𝐴 + π‘‘π‘Žπ‘› 𝐡 / 1 βˆ’ π‘‘π‘Žπ‘› 𝐴 π‘‘π‘Žπ‘› B

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12
Q

π‘‘π‘Žπ‘›(𝐴 βˆ’ 𝐡) =

A

π‘‘π‘Žπ‘› 𝐴 βˆ’ π‘‘π‘Žπ‘› 𝐡 / 1 + π‘‘π‘Žπ‘› 𝐴 π‘‘π‘Žπ‘› 𝐡

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13
Q

π‘π‘œπ‘‘(𝐴 + 𝐡) =

A

π‘π‘œπ‘‘ 𝐴 π‘π‘œπ‘‘ 𝐡 βˆ’ 1 / π‘π‘œπ‘‘ 𝐴 + π‘π‘œπ‘‘ 𝐡

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14
Q

π‘‘π‘Žπ‘› (πœ‹/4+ πœƒ) =

A

= 1 + π‘‘π‘Žπ‘› πœƒ / 1 βˆ’ π‘‘π‘Žπ‘› πœƒ

= cos πœƒ + sin πœƒ / cos πœƒ βˆ’ sin πœƒ

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15
Q

𝑠𝑖𝑛 πœƒ + π‘π‘œπ‘  πœƒ =

A

√2 𝑠𝑖𝑛 (πœƒ +πœ‹/4)

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16
Q

𝑠𝑖𝑛(𝐴 + 𝐡) 𝑠𝑖𝑛(𝐴 βˆ’ 𝐡) =

A

𝑠𝑖𝑛^2 𝐴 βˆ’ 𝑠𝑖𝑛^2 𝐡

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17
Q

π‘π‘œπ‘ (𝐴 + 𝐡) π‘π‘œπ‘ (𝐴 βˆ’ 𝐡) =

A

π‘π‘œπ‘ 2 𝐴 βˆ’ 𝑠𝑖𝑛2 B

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18
Q

(1 + π‘‘π‘Žπ‘› πœƒ)(1 + π‘‘π‘Žπ‘›(45Β° βˆ’ πœƒ)) =

A

2

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19
Q

2 𝑠𝑖𝑛 𝐴 π‘π‘œπ‘  𝐡 =

A

𝑠𝑖𝑛(𝐴 + 𝐡) + 𝑠𝑖𝑛(𝐴 βˆ’ 𝐡)

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20
Q

2 π‘π‘œπ‘  𝐴 𝑠𝑖𝑛 𝐡 =

A

𝑠𝑖𝑛(𝐴 + 𝐡) βˆ’ 𝑠𝑖𝑛(𝐴 βˆ’ 𝐡)

21
Q

2 π‘π‘œπ‘  𝐴 π‘π‘œπ‘  𝐡 =

A

π‘π‘œπ‘ (𝐴 + 𝐡) + π‘π‘œπ‘ (𝐴 βˆ’ 𝐡)

22
Q

2 𝑠𝑖𝑛 𝐴 𝑠𝑖𝑛 𝐡 =

A

π‘π‘œπ‘ (𝐴 βˆ’ 𝐡) βˆ’ π‘π‘œπ‘ (𝐴 + 𝐡)

23
Q

𝑠𝑖𝑛 𝐢 + 𝑠𝑖𝑛 𝐷 =

A

2 𝑠𝑖𝑛 𝐢+𝐷 /2 π‘π‘œπ‘  πΆβˆ’π· /2

24
Q

𝑠𝑖𝑛 𝐢 βˆ’ 𝑠𝑖𝑛 𝐷 =

A

2 π‘π‘œπ‘  𝐢 + 𝐷 / 2 𝑠𝑖𝑛 𝐢 βˆ’ 𝐷 /2

25
Q

π‘π‘œπ‘  𝐢 + π‘π‘œπ‘  𝐷 =

A

2 π‘π‘œπ‘  𝐢 + 𝐷 /2 π‘π‘œπ‘  𝐢 βˆ’ 𝐷 /2

26
Q

π‘π‘œπ‘  𝐢 βˆ’ π‘π‘œπ‘  𝐷 =

A

βˆ’2 𝑠𝑖𝑛 𝐢 + 𝐷 /2 𝑠𝑖𝑛 𝐢 βˆ’ 𝐷 /2

27
Q

𝑠𝑖𝑛 2πœƒ =

A

2 𝑠𝑖𝑛 πœƒ π‘π‘œπ‘  πœƒ

28
Q

π‘π‘œπ‘  2πœƒ =

A

=π‘π‘œπ‘ ^2 πœƒ βˆ’ 𝑠𝑖𝑛^2 πœƒ

= 2 π‘π‘œπ‘ ^2 πœƒ βˆ’ 1

= 1 βˆ’ 2 𝑠𝑖𝑛^2 πœƒ

29
Q

1 βˆ’ π‘π‘œπ‘  2πœƒ =

A

2 𝑠𝑖𝑛^2 πœƒ

30
Q

1 + π‘π‘œπ‘  2πœƒ =

A

2 π‘π‘œπ‘ ^2 πœƒ

31
Q

1 βˆ’ 𝑠𝑖𝑛 2πœƒ =

A

(𝑠𝑖𝑛 πœƒ βˆ’ π‘π‘œπ‘  πœƒ)^2

32
Q

1 + 𝑠𝑖𝑛 2πœƒ =

A

(𝑠𝑖𝑛 πœƒ + π‘π‘œπ‘  πœƒ)^2

33
Q

1 βˆ’ π‘π‘œπ‘  2πœƒ / 1 + π‘π‘œπ‘  2πœƒ =

A

π‘‘π‘Žπ‘›^2 πœƒ

34
Q

2 𝑠𝑖𝑛^2 πœƒ =

A

1 βˆ’ π‘π‘œπ‘  2πœƒ

35
Q

2 π‘π‘œπ‘ ^2 πœƒ =

A

1 + π‘π‘œπ‘  2πœƒ

36
Q

𝑠𝑖𝑛 πœƒ π‘π‘œπ‘  πœƒ =

A

1/2 𝑠𝑖𝑛 2πœƒ

37
Q

𝑠𝑖𝑛 2πœƒ =
(T)

A

2 π‘‘π‘Žπ‘› πœƒ / 1 + π‘‘π‘Žπ‘›^2 πœƒ

38
Q

π‘π‘œπ‘  2πœƒ =
(T)

A

1 βˆ’ π‘‘π‘Žπ‘›^2 πœƒ / 1 + π‘‘π‘Žπ‘›^2 πœƒ

39
Q

π‘‘π‘Žπ‘› 2πœƒ =
(T)

A

2 π‘‘π‘Žπ‘› πœƒ / 1 + π‘‘π‘Žπ‘›^2 πœƒ

40
Q

√1 + 𝑠𝑖𝑛 2πœƒ =

A

|𝑠𝑖𝑛 πœƒ + π‘π‘œπ‘  πœƒ|

41
Q

√1 βˆ’ 𝑠𝑖𝑛 2πœƒ =

A

|𝑠𝑖𝑛 πœƒ βˆ’ π‘π‘œπ‘  πœƒ|

42
Q

𝑠𝑖𝑛 3πœƒ =

A

3 𝑠𝑖𝑛 πœƒ βˆ’ 4 𝑠𝑖𝑛^3 πœƒ

43
Q

π‘π‘œπ‘  3πœƒ =

A

4 π‘π‘œπ‘ ^3 πœƒ βˆ’ 3 π‘π‘œπ‘  πœƒ

44
Q

π‘‘π‘Žπ‘› 3πœƒ =

A

3 π‘‘π‘Žπ‘› πœƒ βˆ’ π‘‘π‘Žπ‘›^3 πœƒ / 1 βˆ’ 3 π‘‘π‘Žπ‘›^2 πœƒ

45
Q

𝑠𝑖𝑛^3 πœƒ =

A

1/4 (3 𝑠𝑖𝑛 πœƒ βˆ’ 𝑠𝑖𝑛 3πœƒ)

46
Q

π‘π‘œπ‘ ^3πœƒ =

A

1/4 (3 π‘π‘œπ‘  𝐴 + π‘π‘œπ‘  3𝐴)

47
Q

𝑠𝑖𝑛(60 βˆ’ 𝐴) 𝑠𝑖𝑛 𝐴 𝑠𝑖𝑛(60 + 𝐴) =

A

1/4 𝑠𝑖𝑛 3𝐴

48
Q

π‘π‘œπ‘ (600 βˆ’ 𝐴) π‘π‘œπ‘  𝐴 π‘π‘œπ‘ (600 + 𝐴)

A

1/4 π‘π‘œπ‘  3𝐴

49
Q

π‘‘π‘Žπ‘›(600 βˆ’ 𝐴) π‘‘π‘Žπ‘› 𝐴 π‘‘π‘Žπ‘›(600 + 𝐴) =

A

π‘‘π‘Žπ‘› 3A