Công thức và phương trình lượng giác tiếng anh

Trigonometric Identities and Equations 
 I. Fundamental Trigonometric Identities 
 A. Reciprocal identities 
1. θθ cos
1sec = 2. θθ sin
1csc = 3. θθ tan
1cot = 
B.Quotient identities 
 1. θ
θθ
cos
sintan = 2. . θ
θθ
sin
coscot = 
C. Pythagorean identities 
 1. 2. 3. 1cossin 22 =+ θθ θθ 22 sec1tan =+ θθ 22 csccot1 =+
D. Sum and difference identities 
 1. φθφθφθ sincoscossin)sin( ±=± 
 2. φθφθφθ sinsincoscos)cos( m=± 
 3. φθ
φθφθ
tantan1
tantan)tan( m
±=± 
E. Double angle identities 
 1. θθθ cossin22sin = 
 2. 1cos2sin21sincos2cos 2222 −=−=−= θθθθθ
 3. θ
θθ 2tan1
tan22tan −= 
F. Half angle identities 
 1. 
2
cos1
2
sin 2 θθ −=⎟⎠
⎞⎜⎝
⎛ 2.
2
cos1
2
cos2 θθ +=⎟⎠
⎞⎜⎝
⎛ 3. θ
θθ
cos1
cos1
2
tan2 +
−=⎟⎠
⎞⎜⎝
⎛ 
1 
G. Miscellaneous identities 
 1. θθ sin)sin( −=− 
 2. θθ cos)cos( =− 
 3. θθ tan)tan( −=− 
 4. ⎟⎠
⎞⎜⎝
⎛⎟⎠
⎞⎜⎝
⎛ ±=±
2
cos
2
sin2sinsin ϕθϕθϕθ m 
 5. ⎟⎠
⎞⎜⎝
⎛ −⎟⎠
⎞⎜⎝
⎛ +=+
2
cos
2
cos2coscos ϕθϕθϕθ 
 6. ⎟⎠
⎞⎜⎝
⎛ −⎟⎠
⎞⎜⎝
⎛ +−=−
2
sin
2
sin2coscos ϕθϕθϕθ 
 7. )sin(
2
1)sin(
2
1cossin ϕθϕθϕθ −++= 
 8. )sin(
2
1)sin(
2
1sincos ϕθϕθϕθ −−+= 
 9. )cos(
2
1)cos(
2
1coscos ϕθϕθϕθ −++= 
 10. )cos(
2
1)cos(
2
1sinsin ϕθϕθϕθ +−−= 
 11. θ
θ
θ
θθ
sin
cos1
cos1
sin
2
tan −=+=⎟⎠
⎞⎜⎝
⎛ 
H. Useful suggestions for proving trigonometric identities 
 1. Avoid aimless transformations. Any transformation that is made in one of the 
 members should lead in some way to the form of the other. 
 2. Start with the more complicated member of the identity and transform it into 
 the form of the simpler member. 
 3. Where possible, express different functions in terms of the same function. 
 4. It is often useful to express all functions in terms of sines and cosines, or in 
 terms of tangents and secants. 
2 
 5. As a rule, trigonometric functions of a double angle, a half angle, or the sums 
 and differences of angles should be expressed in terms of functions of the 
 single angle. 
 6. Simplify expressions by utilizing basic identities and combining like terms. 
 7. Simplify fractions. For example, transform complex fractions into simple 
 fractions or divide the terms of a fraction by the common factors. 
I. Examples 
 1. Using trigonometric identities and fundamental trigonometric function values, 
 find each of the following: 
 (a) 
2
32
4
32
2
2
31
2
60cos1
2
30sin15sin −=−=−=−=⎟⎟⎠
⎞
⎜⎜⎝
⎛=
ooo 
 (b) −⎟⎟⎠
⎞
⎜⎜⎝
⎛
⎟⎟⎠
⎞
⎜⎜⎝
⎛=−=+=
2
3
2
230sin45sin30cos45cos)3045cos(75cos ooooooo 
4
26
2
1
2
2 −=⎟⎠
⎞⎜⎝
⎛
⎟⎟⎠
⎞
⎜⎜⎝
⎛
 (c) 
2
130sin)]15(2sin[15cos15sin2 === oooo 
 (d) 32
32
1
2
31
2
1
30cos1
30sin
2
30tan15tan −=+=+
=+=⎟
⎟
⎠
⎞
⎜⎜⎝
⎛= o
ooo 
 (e) =−++= )5.75.37cos(
2
1)5.75.37cos(
2
15.7cos5.37cos oooooo 
4
32
2
3
2
1
2
2
2
130cos
2
145cos
2
1 +=⎟⎟⎠
⎞
⎜⎜⎝
⎛⎟⎠
⎞⎜⎝
⎛+⎟⎟⎠
⎞
⎜⎜⎝
⎛⎟⎠
⎞⎜⎝
⎛=+ oo 
3 
 2. Prove: xx
x
x tansin
csc
sec1 +=+ 
 ( )( ) +=+=+=+=+ xxxx
x
xx
x
x
xx
x sin1
sin
cos
1sin
sin
1
cos
1
sin
csc
sec
csc
1
csc
sec1 
 xx
x
x tansin
cos
sin += 
3. Prove: 
xx
x
x
x
2sin
2
cos
sin
sin
cos =+ 
 ==+=+=+
xxxx
xx
xx
xx
xx
xx
x
x
x
x
cossin
1
cossin
sincos
cossin
)(sinsin
cossin
)(coscos
cos
sin
sin
cos 22 
xxx 2sin
2
cossin2
2 = 
 4. Prove: 
yx
yx
yx
yx
tantan
tantan
)sin(
)sin(
−
+=−
+ 
 =
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −
⎟⎟⎠
⎞
⎜⎜⎝
⎛ +
=
−
+
=−
+
))(cos(cos
cos
sin
cos
sin
))(cos(cos
cos
sin
cos
sin
cos
sin
cos
sin
cos
sin
cos
sin
tantan
tantan
yx
y
y
x
x
yx
y
y
x
x
y
y
x
x
y
y
x
x
yx
yx 
)sin(
)sin(
cossincossin
cossincossin
yx
yx
xyyx
xyyx
−
+=−
+ 
II. Solution of Trigonometric Equations 
 A. Useful suggestions for solving trigonometric equations 
 1. Simplify the equation by clearing fractions, removing parentheses, combining 
 like terms, and removing radicals. 
 2. Express functions of a double angle, a half angle, or the sums and differences 
 of angles in terms of functions of the single angle; then express the different 
 functions of the single angle in terms of a single function of that angle. 
4 
 3. Solve the resulting equation, whether it be linear or quadratic in nature, for all 
 the values of the angle in the given domain. 
 4. Checks the results by substituting into the original equation. 
 B. Examples 
 1. Solve for x in the interval :)2,0[ π 03sin2 =+x 
2
3sin3sin203sin2 −=⇒−=⇒=+ xxx . Get the reference angle 
 ).60(
32
3sin 1 oπ=⎟⎟⎠
⎞
⎜⎜⎝
⎛− Since is negative, x lies in the 3xsin rd and 4th 
 quadrants. Thus, )240(
3
4)60(
3
)180( ooo πππ =+=x or −= )360(2 oπx
 )300(
3
5)60(
3
oo ππ = . Both of these values do check. 
 2. Solve for x in the interval :)2,0[ π xxx tancoscos = 
 ⇒=−⇒=−⇒= 0)tan1(cos0tancoscostancoscos xxxxxxxx 
 or 0cos =x 0cos0tan1 =⇒=− xx or 1tan =x . 
2
0cos π=⇒= xx ( )90o
 or ).270(
2
3 oπ=x ⇒=1tan x reference angle = )45(
4
)1(tan 1 oπ=− . Since 
 xtan is positive, x lies in the 1st and 3rd quadrants. Thus, )45(
4
oπ=x or x = 
 ).225(
4
5)45(
4
)180( ooo πππ =+ Thus, the solutions are 
2
3,
4
5,
2
,
4
ππππ orx = 
 and they all check. 
 3. Solve for x in the interval :)2,0[ π 13cos2 =x 
2
13cos13cos2 =⇒= xx . Since π20 <≤ x , π60 <≤ x . The reference angle 
 for 3x is )60(
32
1cos 1 oπ=⎟⎠
⎞⎜⎝
⎛− and is positive in the 1xcos st and 4th quadrants. 
 Thus, )60(
3
3 oπ=x , )300(
3
5)60(
3
)360(23 ooo πππ =−=x , += )360(23 oπx
5 
 )420(
3
7)60(
3
oo ππ = , )660(
3
11)60(
3
)720(43 ooo πππ =−=x , += )720(43 oπx
 )780(
3
13)60(
3
oo ππ = , and ⇒=−= )1020(
3
17)60(
3
)1080(63 ooo πππx 
9
17,
9
13,
9
11,
9
7,
9
5,
9
ππππππ orx = and they all check. 
 4. Solve for x in the interval :)2,0[ π xx cos2cos = 
 −⇒=−−⇒=−⇒= xxxxxxx 22 cos20cos)1cos2(0cos2coscos2cos
 01cos20)1)(cos1cos2(01cos =+⇒=−+⇒=− xxxx or ⇒=− 01cos x
2
1cos −=x or . 1cos =x ⇒−=
2
1cos x reference angle is 
32
1cos 1 π=⎟⎠
⎞⎜⎝
⎛− and 
 x lies in the 2nd or 3rd quadrants since is negative xcos
3
2
3
πππ =−=⇒ x or 
3
4
3
πππ =+=x . 01cos =⇒= xx . Thus, 
3
4,
3
2,0 ππ orx = and they all 
check. 
 5. Solve for x in the interval :)2,0[ π xx cossin = 
 ⇒=⇒=⇒= 1tan1
cos
sincossin x
x
xxx reference angle is 
4
)1(tan 1 π=− and 
 x lies in the 1st or 3rd quadrants since xtan is positive 
4
π=⇒ x or += πx 
4
5
4
ππ = and they both check. 
6 
Practice Sheet – Trigonometric Identities and Equations 
 I. Verify the following identities: 
 (1) xx
x
x tansin
csc
sec1 +=+ (2) 
x
xx
2sin1
2cos
4
tan −=⎟⎠
⎞⎜⎝
⎛ +π 
 (3) 
x
xx
sin2
sec2csc = (4) 
x
xx
sec2
1sec
2
sin 2 −=⎟⎠
⎞⎜⎝
⎛ 
 (5) x
xx
xx 3tan
4sin2sin
2cos4cos =−
− 
 II. Solve the following equations for all values of x in the interval [0, π2 ) : 
 (1) (2) 3sin54sin3 −=− xx xxx cos3cossin2 = 
 (3) (4) 01cos4 2 =−x 3sin3cos2 2 =+ xx
 (5) 1cossin =− xx
Solution Key for Trigonometric Identities and Equations 
 I. (1) xxx
x
x
x
xx
x
x
xx
x tansin
1
sin
cos
1sin
sin
1
cos
1
sin
csc
sec
csc
1
csc
sec1 +=⎟⎠
⎞⎜⎝
⎛⎟⎠
⎞⎜⎝
⎛+=+=+=+ 
 (2) 
( )( ) =−+=−
+
=−
+=−
+=⎟⎠
⎞⎜⎝
⎛ +
xx
xx
x
x
x
x
x
x
x
x
x
sincos
sincos
cos
sin1
cos
sin1
tan1
tan1
tan4tan1
tan4tan
4
tan π
ππ 
x
x
xxxx
xx
xxxx
xxxx
2sin1
2cos
sinsincos2cos
sincos
)sin)(cossin(cos
)sin)(cossin(cos
22
22
−=+−
−=−−
−+ 
7 
 (3) 
x
xx
xxxxxx
x
sin2
secsec
sin2
1
cos
1
sin2
1
cossin2
1
2sin
12csc =⋅=⋅=== 
 (4) 
x
x
x
x
x
x
x
x
x
xx
sec2
1sec
sec2
cos
cos
cos
1
cos
2
cos
cos1
2
cos1
2
sin2 −=
−
=
−
=−=⎟⎠
⎞⎜⎝
⎛ 
 (5) =−
−=
⎟⎠
⎞⎜⎝
⎛ +⎟⎠
⎞⎜⎝
⎛ −
⎟⎠
⎞⎜⎝
⎛ −⎟⎠
⎞⎜⎝
⎛ +−
=−
−
xx
xx
xxxx
xxxx
xx
xx
3cos)sin(2
sin3sin2
2
42cos
2
42sin2
2
24sin
2
24sin2
4sin2sin
2cos4cos 
 x
x
x
xx
xx 3tan
3cos
3sin
3cossin2
sin3sin2 ==−
− 
II. (1) 
6
11,
6
7
2
1sin3sin54sin3 ππ=⇒−=⇒−=− xxxx and they both check. 
 (2) 
2
3sincos3cossin2 =⇒= xxxx or 
3
2,
3
0cos ππ=⇒= xx or 
2
3,
2
ππ=x 
 and they all check. 
 (3) 
3
5,
3
4,
3
2,
32
1cos01cos4 2 ππππ=⇒±=⇒=− xxx and they all check. 
 (4) ⇒=+−⇒=+−⇒=+ 01sin3sin23sin3)sin1(23sin3cos2 222 xxxxxx
2
1sin0)1)(sin1sin2( =⇒=−− xxx or 
6
5,
6
1sin ππ=⇒= xx or 
2
π=x and 
 they all check. 
 (5) ( ) ( ) +=⇒+=⇒+=⇒=− 1sincos1sincos1sin1cossin 222 xxxxxxx
 ⇒=+⇒++=−⇒+ 0cos2cos2coscos21cos1coscos2 2222 xxxxxxx
 0cos0)1(coscos2 =⇒=+ xxx or 
2
3,
2
1cos ππ=⇒−= xx or π=x . 
2
π=x or π=x are the solutions because they both check. However, 
2
3π=x 
 does not check in the original equation and thus is not a solution. [Note: 
2
3π=x is an extraneous root created by squaring both sides of the original 
equation.] 
8