Sample Problems cos 2 x = tan2 x tan 2 = csc 2 tan sec x + tan x = cos x sin 4 x cos 4 x = 1 2 cos 2 x

Sample Problems

Prove each of the following identities.

1. tan x sin x + cos x = sec x

2. 1

tan x + tan x = 1 sin x cos x 3. sin x sin x cos2_{x = sin}3_x

4. cos 1 + sin + 1 + sin cos = 2 sec 5. cos x 1 sin x cos x 1 + sin x = 2 tan x 6. cos2x = csc x cos x tan x + cot x 7. sin 4_{x cos}4_x sin2x cos2_x = 1 8. tan 2_x tan2_{x + 1} = sin 2_x 9. 1 sin x cos x = cos x 1 + sin x 10. 1 2 cos2_{x =} tan 2_{x 1} tan2_{x + 1} 11. tan2 _{= csc}2 _tan2 ₁

12. sec x + tan x = cos x

1 sin x

13. csc sin

cot

tan = 1

14. sin4x cos4_{x = 1 2 cos}2_x

15. (sin x cos x)2+ (sin x + cos x)2 = 2

16. sin 2_{x + 4 sin x + 3} cos2_x = 3 + sin x 1 sin x 17. cos x

1 sin x tan x = sec x

18. tan2x + 1 + tan x sec x = 1 + sin x cos2_x

Practice Problems

Prove each of the following identities.

1. tan x + cos x 1 + sin x = 1 cos x 2. tan2x + 1 = sec2x 3. 1 1 sin x 1

1 + sin x = 2 tan x sec x 4. tan x + cot x = sec x csc x

5. 1 + tan

2_x

1 tan2_x =

1 cos2_{x sin}2_x

6. tan2_{x sin}2_{x = tan}2_{x sin}2_x

7. 1 cos x sin x + sin x 1 cos x = 2 csc x 8. sec x 1 sec x + 1 = 1 cos x 1 + cos x 9. 1 + cot2_{x = csc}2_x 10. csc 2_{x 1} csc2_x = cos 2_x 11. cot x 1 cot x + 1 = 1 tan x 1 + tan x

12. (sin x + cos x) (tan x + cot x) = sec x + csc x

13. sin

3_{x + cos}3_x

sin x + cos x = 1 sin x cos x

14. cos x + 1 sin3x = csc x 1 cos x 15. 1 + sin x 1 sin x 1 sin x

1 + sin x = 4 tan x sec x 16. csc4x cot4x = csc2x + cot2x 17. sin 2_x cos2_{x + 3 cos x + 2} = 1 cos x 2 + cos x 18. tan x + tan y

cot x + cot y = tan x tan y

19. 1 + tan x

1 tan x =

cos x + sin x cos x sin x

Sample Problems - Solutions

1. tan x sin x + cos x = sec x

Solution: We will only use the fact that sin2x + cos2_{x = 1 for all values of x.}

LHS = tan x sin x + cos x = sin x

cos x sin x + cos x = sin2x cos x + cos x = sin2x cos x + cos2_x cos x = sin 2_{x + cos}2_x cos x = 1 cos x =RHS 2. 1 tan x + tan x = 1 sin x cos x

Solution: We will only use the fact that sin2x + cos2x = 1 for all values of x.

LHS = 1 tan x + tan x = cos x sin x + sin x cos x = cos2x + sin2x sin x cos x = 1 sin x cos x =RHS 3. sin x sin x cos2_{x = sin}3_x

Solution: We will only use the fact that sin2x + cos2x = 1 for all values of x. LHS = sin x sin x cos2x = sin x 1 cos2x = sin x sin2x =RHS

4. cos

1 + sin +

1 + sin

cos = 2 sec

Solution: We will only use the fact that sin2x + cos2_{x = 1 for all values of x.}

LHS = cos 1 + sin + 1 + sin cos = cos2 (1 + sin ) cos + (1 + sin )2 (1 + sin ) cos = cos2 + (1 + sin )2 (1 + sin ) cos = cos 2 _{+ 1 + 2 sin + sin}2 (1 + sin ) cos =

cos2 _{+ sin}2 _{+ 1 + 2 sin}

(1 + sin ) cos = 2 + 2 sin (1 + sin ) cos = 2 (1 + sin ) (1 + sin ) cos = 2 cos = 2 1 cos = 2 sec =RHS 5. cos x 1 sin x cos x 1 + sin x = 2 tan x

Solution: We will start with the left-hand side. First we bring the fractions to the common denominator. Recall that sin2x + cos2_{x = 1 for all values of x.}

LHS = cos x 1 sin x cos x 1 + sin x = cos x (1 + sin x) (1 sin x) (1 + sin x) cos x (1 sin x) (1 sin x) (1 + sin x)

6. cos2x = csc x cos x tan x + cot 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, sin2x + cos2_{x = 1.}

RHS = csc x cos x tan x + cot x = 1 sin x cos x sin x cos x + cos x sin x = 1 sin x cos x 1 sin2x sin x cos x + cos2_x sin x cos x = cos x sin x sin2x + cos2_x sin x cos x = cos x sin x 1 sin x cos x = cos x sin x cos x sin x 1 = cos2x 1 = cos 2_{x =LHS} 7. sin 4_{x cos}4_x sin2x cos2_x = 1

Solution: We can factor the numerator via the di¤erence of squares theorem.

LHS = sin

4_{x cos}4_x

sin2x cos2_x =

sin2x 2 (cos2_x)2

sin2x cos2_x =

sin2x + cos2_{x sin}2_{x cos}2_x

sin2x cos2_x = sin2x + cos2x = 1 =RHS 8. tan 2_x tan2_{x + 1} = sin 2_x Solution: LHS = tan 2_x tan2_{x + 1} = sin x cos x 2 sin x cos x 2 + 1 = sin2x cos2_x sin2x cos2_x+ 1 = sin2x cos2_x sin2x cos2_x + cos2x cos2_x = sin2x cos2_x sin2x + cos2_x cos2_x = sin2x cos2_x 1 cos2_x = sin 2_x cos2_x cos2_x 1 = sin 2_{x =RHS} 9. 1 sin x cos x = cos x 1 + sin x Solution: LHS = 1 sin x cos x = 1 sin x cos x 1 = 1 sin x cos x 1 + sin x 1 + sin x = (1 sin x) (1 + sin x) cos x (1 + sin x) = 1 sin2x cos x (1 + sin x) = cos 2_x cos x (1 + sin x) = cos x 1 + sin x =RHS

10. 1 2 cos2_{x =} tan 2_{x 1} tan2x + 1 Solution: RHS = tan 2_{x 1} tan2x + 1 = sin2x cos2_x 1 sin2x cos2_x + 1 = sin2x cos2_x cos2x cos2_x sin2x cos2_x + cos2_x cos2_x = sin2x cos2x cos2_x sin2x + cos2_x cos2_x = sin 2_{x cos}2_x cos2_x cos2_x sin2x + cos2_x = sin2x cos2_x sin2x + cos2_x = sin2x cos2_x 1 = sin 2 x cos2x

= 1 cos2x cos2x = 1 2 cos2x =LHS 11. tan2 = csc2 tan2 1 RHS = csc2 tan2 1 = 1 sin2 sin cos 2 1 = 1 sin2 sin2 cos2 1 = 1 cos2 1 = 1 cos2 cos2 cos2 = 1 cos2 cos2 = sin2 cos2 = sin cos 2 = tan2 =LHS

12. sec x + tan x = cos x 1 sin x Solution: RHS = cos x 1 sin x = cos x 1 sin x 1 = cos x 1 sin x 1 + sin x 1 + sin x = cos x (1 + sin x) (1 sin x) (1 + sin x) = cos x (1 + sin x) 1 sin2x = cos x (1 + sin x) cos2_x = 1 + sin x cos x = 1 cos x + sin x cos x =LHS 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 + cos}2_{x = 1 for all angles}

x. LHS = csc sin cot tan = 1 sin sin 1 cos sin sin cos = 1 sin 1 sin cos sin cos sin = 1 sin2 cos2 sin2 = 1 cos 2 sin2 =

sin2 + cos2 cos2

sin2 =

sin2

sin2 = 1 =RHS 14. sin4x cos4_{x = 1 2 cos}2_x

15. (sin x cos x)2+ (sin x + cos x)2 = 2 Solution:

LHS = (sin x cos x)2 + (sin x + cos x)2

= sin2x + cos2x 2 sin x cos x + sin2x + cos2x + 2 sin x cos x = 2 sin2x + 2 cos2x = 2 sin2x + cos2x = 2 1 = 2 =RHS 16. sin 2_{x + 4 sin x + 3} cos2_x = 3 + sin x 1 sin x Solution: LHS = sin 2_{x + 4 sin x + 3} cos2_x = (sin x + 1) (sin x + 3) 1 sin2x = (sin x + 1) (sin x + 3) (1 + sin x) (1 sin x) = sin x + 3 1 sin x =RHS 17. cos x

1 sin x tan x = sec x Solution: LHS = cos x 1 sin x tan x = cos x 1 sin x sin x cos x =

cos2_{x sin x (1 sin x)}

cos x (1 sin x) =

cos2_{x sin x + sin}2_x

cos x (1 sin x) = cos 2_{x + sin}2_{x sin x} cos x (1 sin x) = 1 sin x cos x (1 sin x) = 1 cos x =RHS

18. tan2x + 1 + tan x sec x = 1 + sin x cos2_x

Solution:

LHS = tan2x + 1 + tan x sec x = sin

2_x cos2_x + 1 + sin x cos x 1 cos x = sin2_x cos2_x + cos2_x cos2_x + sin x cos2_x = sin 2_{x + cos}2_{x + sin x} cos2_x = 1 + sin x cos2_x =RHS

For more documents like this, visit our page at http://www.teaching.martahidegkuti.com and click on Lecture Notes. E-mail questions or comments to mhidegkuti@ccc.edu.