Pre-Class Problems 7 for Monday, February 19

Illustration of the definition of the cosine function, sine function, tangent function, secant function, cosecant function, and cotangent function for the acute angle  using right triangle trigonometry. Second illustration of the cosine function, sine function, tangent function, secant function, cosecant function, and cotangent function.

Illustration of the definition of all the six trigonometric functions for the acute angle  using right triangle trigonometry. Second illustration of all the six trigonometric functions.

These are the type of problems that you will be working on in class. These problems are from Lesson 6.

Solution to Problems on the Pre-Exam.

You can go to the solution for each problem by clicking on the problem letter.

Objective of the following problems: To use right triangle trigonometry to find the exact value of the six trigonometric functions for an acute angle. You will also need to apply the Pythagorean Theorem.

1. Find the exact value of cosine, sine, and tangent for the following angles.

a. b. 11 

7 5 6  c.

Objective of the following problems: To use right triangle trigonometry to find the exact value of the other five trigonometric functions for an acute angle if given the value of one of the trigonometric function of the angle.

2. Use a right triangle to find the exact value of the other five trigonometric functions of the angle if given the following information.

4 a. cos   and  is an acute angle 7

3 b. tan   and  is an acute angle 5

8 c. csc   and  is an acute angle 3

Objective of the following problems: To determine what quadrant the terminal side of an angle is in if given the sign of two trigonometric functions of the angle.

3. Determine what quadrant the following angles are in if given the following information.

a. cos   0 and sin   0 b. sin   0 and sec   0

c. cos   0 and csc   0 d. sin   0 and tan   0 e. cos   0 and cot   0 f. tan   0 and sec   0

g. cot   0 and csc   0 h. csc   0 and sec   0

Additional problems available in the textbook: Page 492 … 13 – 18, 25 - 30. Page 483 … Examples1 and 2. Page 505 … 9 – 14.

Solutions:

NOTE: You use the Pythagorean Theorem to find the value of 24 .

adj 24 opp 5 opp 5 cos    , sin    , tan    hyp 7 hyp 7 adj 24

24 5 5 Answer: cos   , sin   , tan   7 7 24

Back to Problem 1.

1b. 11 

NOTE: You use the Pythagorean Theorem to find the value of 5.

adj 11 opp 5 opp 5 cos    , sin    , tan    hyp 6 hyp 6 adj 11

11 5 5 Answer: cos   , sin   , tan   6 6 11

Back to Problem 1.

 80  4 5 8

adj 8 2 opp 4 1 cos     , sin     , hyp 4 5 5 hyp 4 5 5 opp 4 1 tan     adj 8 2

2 1 1 Answer: cos   , sin   , tan   5 5 2

Back to Problem 1.

4 2a. cos   and  is an acute angle 7

4 7 cos    sec   NOTE: We do not need to use right triangle 7 4 trigonometry to find this value.

4 adj NOTE: cos    7 hyp

opp 33 hyp 7 sin    csc    hyp 7 opp 33 opp 33 adj 4 tan    cot    adj 4 opp 33

33 33 7 7 Answer: sin   , tan   , sec   , csc   , 7 4 4 33

4 cot   33

Back to Problem 2.

3 2b. tan   and  is an acute angle 5

3 5 tan    cot   NOTE: We do not need to use right 5 3 triangle trigonometry to find this value.

3 opp NOTE: tan    5 adj

adj 5 hyp 28 cos    sec    hyp 28 adj 5 opp 3 hyp 28 sin    csc    hyp 28 opp 3

5 3 28 Answer: cos   , sin   , sec   , 28 28 5

28 5 csc   , cot   3 3

Back to Problem 2.

8 2c. csc   and  is an acute angle 3

8 3 csc    sin   NOTE: We do not need to use right triangle 3 8 trigonometry to find this value.

3 opp NOTE: sin    8 hyp

NOTE: You use the Pythagorean Theorem to find the value of 55 . adj 55 hyp 8 cos    sec    hyp 8 adj 55

opp 3 adj 55 tan    cot    adj 55 opp 3

55 3 3 Answer: cos   , sin   , tan   , 8 8 55

8 55 sec   , cot   55 3

Back to Problem 2.

3a. cos   0 and sin   0

Using Method 1, which is Unit Circle Trigonometry, from Lesson 6:

Recall Unit Circle Trigonometry: P ( )  (cos  , sin  )

cos   0  x  0

sin   0  y  0

Answer: IV

NOTE: The quadrant where x-coordinates are positive and y-coordinates are negative is the fourth quadrant.

Using Method 2 from Lesson 6:

cos   0 sin   0 y y

x x X X X

Answer: IV

Back to Problem 3.

3b. sin   0 and sec   0

Recall Unit Circle Trigonometry: P ( )  (cos  , sin  )

sin   0  y  0

sec   0  cos   0  x  0

Answer: II

NOTE: The quadrant where x-coordinates are negative and y-coordinates are positive is the second quadrant.

NOTE: Cosine and secant are reciprocals of each other. The sign of reciprocals is the same.

sin   0 sec   0 or cos   0 y y X X X

Answer: II

Back to Problem 3.

3c. cos   0 and csc   0

Recall Unit Circle Trigonometry: P (  )  (cos  , sin  )

cos   0  x  0

csc   0  sin   0  y  0

Answer: III

NOTE: The quadrant where x-coordinates are negative and y-coordinates are negative is the third quadrant.

NOTE: Sine and cosecant are reciprocals of each other. The sign of reciprocals is the same.

cos   0 csc   0 or sin   0 y y X

x x X X X

Answer: III

Back to Problem 3.

3d. sin   0 and tan   0

Recall Unit Circle Trigonometry: P ( )  (cos  , sin  )

sin   0  y  0

y (  ) tan   0 : (  )  tan     x  0 x ?

Answer: IV

NOTE: A negative divided by a positive will produce a negative result. Division with unlike signs will produce a negative.

sin   0 tan   0 y y X

x x X X X

Answer: IV

Back to Problem 3.

3e. cos   0 and cot   0

cos   0  x  0

y ? cot   0  tan   0 : (  )  tan     y  0 x (  )

Answer: III

NOTE: The quadrant where x-coordinates are negative and y-coordinates are negative is the third quadrant.

NOTE: Tangent and cotangent are reciprocals of each other. The sign of reciprocals is the same.

NOTE: A negative divided by a negative will produce a positive result. Division with like signs will produce a positive.

cos   0 cot   0 or tan   0 y y

x x X X

Answer: III

Back to Problem 3.

3f. tan   0 and sec   0

Recall Unit Circle Trigonometry: P ( )  (cos  , sin  )

sec   0  cos   0  x  0

y ? tan   0 : (  )  tan     y  0 x (  )

Answer: II

NOTE: A positive divided by a negative will produce a negative result. Division with unlike signs will produce a negative. Using Method 2 from Lesson 6:

tan   0 sec   0 or cos   0 y y

x x X X

Answer: II

Back to Problem 3.

3g. cot   0 and csc   0

Recall Unit Circle Trigonometry: P (  )  (cos  , sin  )

csc   0  sin   0  y  0 y (  ) cot   0  tan   0 : (  )  tan     x  0 x ?

Answer: II

NOTE: Sine and cosecant are reciprocals of each other. The sign of reciprocals is the same.

NOTE: Tangent and cotangent are reciprocals of each other. The sign of reciprocals is the same. NOTE: A positive divided by a negative will produce a negative result. Division with unlike signs will produce a negative.

cot   0 or tan   0 csc   0 or sin   0 y y

Answer: II Back to Problem 3.

3h. csc   0 and sec   0

csc   0  sin   0  y  0

sec   0  cos   0  x  0

Answer: IV

NOTE: Sine and cosecant are reciprocals of each other. The sign of reciprocals is the same. NOTE: Cosine and secant are reciprocals of each other. The sign of reciprocals is the same.

csc   0 or sin   0 sec   0 or cos   0 y y

x x X X X

Answer: IV

Back to Problem 3.

Solution to Problems on the Pre-Exam: Back to Page 1.

11. Given: 7 Find the exact value of

3 a. cos   4 3 4 4  b. csc   7

8. If tan   0 and sec   0 , then  lies in which quadrant?

Using Method 1, which is Unit Circle Trigonometry, from Lesson 6: Recall Unit Circle Trigonometry: P ( )  (cos  , sin  ) sec   0  cos   0  x  0

y ? tan   0 : (  )  tan     y  0 x (  )

Answer: II

NOTE: A positive divided by a negative will produce a negative result. Division with unlike signs will produce a negative.

tan   0 sec   0 or cos   0 y y

x x X X

Answer: II