Trigonometric Identities 1 page 1. Sample Problems. Prove each of the ... 10. 1 2
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Trigonometric Identities 1
Lecture Notes
page 1
Sample Problems Prove each of the following identities.
1. tan x sin x + cos x = sec x 2.
1 1 + tan x = tan x sin x cos x
3. sin x 4.
5.
sin x cos2 x = sin3 x
cos 1 + sin
+
cos x 1 sin x
6. cos2 x =
1 + sin cos
csc x cos x tan x + cot x
sin4 x sin2 x
8.
tan2 x = sin2 x tan2 x + 1
9.
1
cos4 x =1 cos2 x
sin x cos x = cos x 1 + sin x
c copyright Hidegkuti, Powell, 2009
2 cos2 x =
tan2 x 1 tan2 x + 1
11. tan2 = csc2 tan2 12. sec x + tan x =
= 2 sec
cos x = 2 tan x 1 + sin x
7.
10. 1
13.
csc sin
cot tan
1
cos x 1 sin x
=1
14. sin4 x
cos4 x = 1
15. (sin x
cos x)2 + (sin x + cos x)2 = 2
2 cos2 x
sin2 x + 4 sin x + 3 3 + sin x 16. = 2 cos x 1 sin x 17.
cos x 1 sin x
tan x = sec x
18. tan2 x + 1 + tan x sec x =
1 + sin x cos2 x
Last revised: May 8, 2013
Trigonometric Identities 1
Lecture Notes
page 2
Practice Problems Prove each of the following identities.
1. tan x +
cos x 1 = 1 + sin x cos x
11.
12. (sin x + cos x) (tan x + cot x) = sec x + csc x
2. tan2 x + 1 = sec2 x
3.
1
1 sin x
sin3 x + cos3 x 13. =1 sin x + cos x
1 = 2 tan x sec x 1 + sin x
4. tan x + cot x = sec x csc x
5.
1 + tan2 x 1 = 2 1 tan x cos2 x sin2 x
6. tan2 x
7.
8.
1
cot x 1 1 tan x = cot x + 1 1 + tan x
sin2 x = tan2 x sin2 x
14.
cos x + 1 csc x = 3 1 cos x sin x
15.
1 + sin x 1 sin x
16. csc4 x
cos x sin x + = 2 csc x sin x 1 cos x
sin x cos x
1 sin x = 4 tan x sec x 1 + sin x
cot4 x = csc2 x + cot2 x
sin2 x 1 cos x 17. = 2 cos x + 3 cos x + 2 2 + cos x
sec x 1 1 cos x = sec x + 1 1 + cos x
18.
tan x + tan y = tan x tan y cot x + cot y
19.
1 + tan x cos x + sin x = 1 tan x cos x sin x
9. 1 + cot2 x = csc2 x
10.
csc2 x 1 = cos2 x csc2 x
20. (sin x
tan x) (cos x
cot x) = (sin x
c copyright Hidegkuti, Powell, 2009
1) (cos x
1)
Last revised: May 8, 2013
Trigonometric Identities 1
Lecture Notes
page 3
Sample Problems - Solutions 1. tan x sin x + cos x = sec x Solution: We will only use the fact that sin2 x + cos2 x = 1 for all values of x. sin2 x sin2 x cos2 x sin x LHS = tan x sin x + cos x = sin x + cos x = + cos x = + cos x cos x cos x cos x sin2 x + cos2 x 1 = = = RHS cos x cos x 2.
1 1 + tan x = tan x sin x cos x Solution: We will only use the fact that sin2 x + cos2 x = 1 for all values of x. LHS =
1 cos x sin x cos2 x + sin2 x 1 + tan x = + = = = RHS tan x sin x cos x sin x cos x sin x cos x
sin x cos2 x = sin3 x
3. sin x
Solution: We will only use the fact that sin2 x + cos2 x = 1 for all values of x. LHS = sin x 4.
cos 1 + sin
+
1 + sin cos
sin x cos2 x = sin x 1
cos2 x = sin x sin2 x = RHS
= 2 sec
Solution: We will only use the fact that sin2 x + cos2 x = 1 for all values of x. cos 1 + sin cos2 (1 + sin )2 cos2 + (1 + sin )2 + = + = 1 + sin cos (1 + sin ) cos (1 + sin ) cos (1 + sin ) cos 2 2 cos2 + 1 + 2 sin + sin cos2 + sin + 1 + 2 sin 2 + 2 sin = = = (1 + sin ) cos (1 + sin ) cos (1 + sin ) cos 2 (1 + sin ) 2 1 = = =2 = 2 sec = RHS (1 + sin ) cos cos cos
LHS =
5.
cos x 1 sin x
cos x = 2 tan x 1 + sin x
Solution: We will start with the left-hand side. First we bring the fractions to the common denominator. Recall that sin2 x + cos2 x = 1 for all values of x. cos x cos x cos x (1 + sin x) cos x (1 sin x) = 1 sin x 1 + sin x (1 sin x) (1 + sin x) (1 sin x) (1 + sin x) cos x (1 + sin x) cos x (1 sin x) cos x + cos x sin x cos x + cos x sin x 2 sin x cos x = = = 2 (1 sin x) (1 + sin x) cos2 x 1 sin x 2 sin x = = 2 tan x = RHS cos x
LHS =
c copyright Hidegkuti, Powell, 2009
Last revised: May 8, 2013
Trigonometric Identities 1
Lecture Notes
page 4
csc x cos x tan x + cot x
6. cos2 x =
Solution: We will start with the right-hand side. We will re-write everything in terms of sin x and cos x and simplify. We will again run into the Pythagorean identity, sin2 x + cos2 x = 1. cos x 1 1 cos x cos x cos x csc x cos x sin x sin x sin x 1 = = RHS = = sin x = 1 sin x cos x tan x + cot x sin2 x + cos2 x sin2 x cos2 x + + sin x cos x cos x sin x sin x cos x sin x cos x sin x cos x 2 cos x cos x sin x cos x = = = cos2 x = LHS sin x 1 1 7.
sin4 x sin2 x
cos4 x =1 cos2 x
Solution: We can factor the numerator via the di¤erence of squares theorem. 2
sin2 x sin4 x cos4 x LHS = = sin2 x cos2 x sin2 x = sin2 x + cos2 x = 1 = RHS 8.
2
(cos2 x) sin2 x + cos2 x = cos2 x sin2 x
sin2 x cos2 x
cos2 x
tan2 x = sin2 x tan2 x + 1 Solution: 2
sin x sin2 x sin2 x 2 2x tan x cos x cos2 x LHS = = = cos = 2 2 2 2 tan x + 1 sin x sin x cos2 x sin x +1 + +1 cos2 x cos2 x cos2 x cos x sin2 x sin2 x 2 2 cos2 x cos2 x = sin x cos x = sin2 x = RHS = = 1 cos2 x 1 sin2 x + cos2 x 2 cos x cos2 x 9.
1
sin x cos x = cos x 1 + sin x
Solution: sin x 1 sin x 1 sin x 1 + sin x (1 sin x) (1 + sin x) 1 sin2 x = 1= = = cos x cos x cos x 1 + sin x cos x (1 + sin x) cos x (1 + sin x) 2 cos x cos x = = = RHS cos x (1 + sin x) 1 + sin x
LHS =
1
c copyright Hidegkuti, Powell, 2009
Last revised: May 8, 2013
Trigonometric Identities 1
Lecture Notes
10. 1
2 cos2 x =
page 5
tan2 x 1 tan2 x + 1
Solution: sin2 x cos2 x sin2 x sin2 x cos2 x 1 2 2 2 tan x 1 cos2 x = cos2 x cos2 x = RHS = = cos2 x 2 2 tan x + 1 sin x + cos2 x sin x cos x sin x + + 1 cos2 x cos2 x cos2 x cos2 x 2 2 2 2 2 cos x sin x cos x sin x cos x sin2 x cos2 x = = = = sin2 x cos2 x 1 sin2 x + cos2 x sin2 x + cos2 x = 1 cos2 x cos2 x = 1 2 cos2 x = LHS 2
11. tan2 = csc2 tan2
1
RHS = csc2 tan2 1 = cos2 12. sec x + tan x =
cos2 x
cos2 cos2
1= =
1 sin2 1
sin cos
cos2 cos2
2
sin2 = cos2
=
sin2 cos2
1 sin2
1=
sin cos
1=
1 cos2
1
2
= tan2 = LHS
cos x 1 sin x
Solution: cos x cos x cos x 1 + sin x cos x (1 + sin x) = 1= = 1 sin x 1 sin x 1 sin x 1 + sin x (1 sin x) (1 + sin x) cos x (1 + sin x) cos x (1 + sin x) 1 + sin x 1 sin x = = = = + = LHS 2 2 cos x cos x cos x cos x 1 sin x
RHS =
13.
csc sin
cot tan
=1
Solution: We will start with the left-hand side. We will re-write everything in terms of sin and cos and simplify. We will again run into the Pythagorean identity, sin2 x + cos2 x = 1 for all angles x. cos 1 csc cot sin sin LHS = = sin sin sin tan 1 cos 2 2 2 sin + cos 1 cos = = 2 sin sin2 14. sin4 x
cos4 x = 1
= cos2
1 sin
1 sin
=
sin2 sin2
cos sin
cos sin
=
1 sin2
cos2 sin2
= 1 = RHS
2 cos2 x
Solution: LHS = sin4 x cos4 x = sin2 x = 1 sin2 x cos2 x = 1 c copyright Hidegkuti, Powell, 2009
2
2
cos2 x = sin2 x + cos2 x sin2 x cos2 x cos2 x cos2 x = 1 2 cos2 x = RHS Last revised: May 8, 2013
Trigonometric Identities 1
Lecture Notes 15. (sin x
page 6
cos x)2 + (sin x + cos x)2 = 2
Solution: LHS = (sin x cos x)2 + (sin x + cos x)2 = sin2 x + cos2 x 2 sin x cos x + sin2 x + cos2 x + 2 sin x cos x = 2 sin2 x + 2 cos2 x = 2 sin2 x + cos2 x = 2 1 = 2 = RHS 3 + sin x sin2 x + 4 sin x + 3 16. = 2 cos x 1 sin x Solution: LHS =
17.
cos x 1 sin x
(sin x + 1) (sin x + 3) sin x + 3 sin2 x + 4 sin x + 3 (sin x + 1) (sin x + 3) = = = = RHS 2 cos2 x (1 + sin x) (1 sin x) 1 sin x 1 sin x tan x = sec x
Solution: cos x sin x cos2 x sin x (1 sin x) cos2 x sin x + sin2 x cos x tan x = = = 1 sin x 1 sin x cos x cos x (1 sin x) cos x (1 sin x) 2 2 cos x + sin x sin x 1 sin x 1 = = = = RHS cos x (1 sin x) cos x (1 sin x) cos x
LHS =
18. tan2 x + 1 + tan x sec x =
1 + sin x cos2 x
Solution: sin2 x sin x 1 sin2 x cos2 x sin x LHS = tan x + 1 + tan x sec x = +1+ = + + 2 2 2 cos x cos x cos x cos x cos x cos2 x 2 2 sin x + cos x + sin x 1 + sin x = = = RHS 2 cos x cos2 x 2
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Last revised: May 8, 2013
