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CHAPTER
CHAPTER
TABLE OF CONTENTS
13-1 First-Degree Trigonometric Equations
13-2 Using Factoring to Solve Trigonometric Equations
13-3 Using the Quadratic Formula to Solve Trigonometric Equations
13-4 Using Substitution to Solve Trigonometric Equations Involving More Than One Function
13-5 Using Substitution to Solve Trigonometric Equations Involving Different Angle Measures Chapter Summary Vocabulary Review Exercises Cumulative Review
TRIGONOMETRIC
EQUATIONS
The triangle is a rigid figure, that is, its shape cannot
be changed without changing the lengths of its sides.
This fact makes the triangle a basic shape in construc-
tion. The theorems of geometry give us relationships
among the measures of the sides and angles of triangle.
Precisely calibrated instruments enable surveyors to
obtain needed measurements. The identities and for-
mulas of trigonometry enable architects and builders to
formulate plans needed to construct the roads, bridges,
and buildings that are an essential part of modern life.
A trigonometric equation is an equation whose variable is expressed in terms of a trigonometric function value. To solve a trigonometric equation, we use the same procedures that we used to solve algebraic equations. For example, in the equation 4 sin u 1 5 5 7, sin u is multiplied by 4 and then 5 is added. Thus, to solve for sin u, first add the opposite of 5 and then divide by 4. 4 sin u 1 5 5 7 2 5 5 2 5 4 sin u 5 2
5 sin u 5
We know that sin 30° 5 , so one value of u is 30°; u 1 5 30. We also know that since sin u is positive in the second quadrant, there is a second- quadrant angle, u 2 , whose sine is. Recall the relationship between an angle in any quadrant to the acute angle called the reference angle. The following table compares the degree measures of u from 2 90° to 360°, the radian measures of u from 2 to 2p, and the measure of its reference angle.
The reference angle for the second-quadrant angle whose sine is has a degree measure of 30° and u 2 5 180° 2 30° or 150°. Therefore, sin u 2 5. For 0° u , 360°, the solution set of 4 sin u 1 5 5 7 is {30°, 150°}. In radian mea- sure, the solution set is. In the example given above, it was possible to give the exact value of u that makes the equation true. Often it is necessary to use a calculator to find an approximate value. Consider the solution of the following equation. 5 cos u 1 7 5 3 5 cos u 5 2 4 cos u 5 2 u 5 arccos (^) A 245 B
4 5
Up 6 , 56 p V
1 2
1 2
p 2
1 2
1 2
1 2
2 4
4 sin u 4
13-1 FIRST-DEGREE TRIGONOMETRIC EQUATIONS
First-Degree Trigonometric Equations 519
Fourth First Second Third Fourth Quadrant Quadrant Quadrant Quadrant Quadrant
Angle 2 90° , u , 0° 0° , u , 90° 90° , u , 180° 180° , u , 270° 270° , u , 360° , u , 0 , u , p p , u , , u , 2 p Reference 2u u 180° 2 u u 2 180° 360° 2 u Angle 2u u p 2 u u 2 p 2 p 2 u
3 p 2
3 p 2
p 0 , u , (^2) p (^22) p 2
The following table will help you find the locations of the angles that satisfy trigonometric equations. The values in the table follow from the definitions of the trigonometric functions on the unit circle.
EXAMPLE 1
Find the solution set of the equation 7 tan u 5 1 tan u in the interval 0 ° u , 360 °.
Solution How to Proceed
(1) Solve the equation for tan u: 7 tan u 5 1 tan u 6 tan u 5
tan u 5
(2) Since tan u is positive, u 1 can be a u 1 5 30 ° first-quadrant angle:
(3) Since u is a first-quadrant angle, R 5 30 ° R 5 u:
(4) Tangent is also positive in the third u 2 5 180 ° 1 R quadrant. Therefore, there is a u 2 5 180 ° 1 30 ° 5 210 ° third-quadrant angle such that tan u 5. In the third quadrant, u 2 5 180 ° 1 R :
Answer The solution set is {30°, 210°}.
! 3 3
! 3 3
First-Degree Trigonometric Equations 521
Sign of a and b (0 * | a | * 1, b 0) 1 2 sin u 5 a Quadrants I and II Quadrants III and IV cos u 5 a Quadrants I and IV Quadrants II and III tan u 5 b Quadrants I and III Quadrants II and IV
EXAMPLE 2
Find, to the nearest hundredth, all possible solutions of the following equation in radians: 3(sin A 1 2) 5 3 2 sin A
Solution How to Proceed (1) Solve the equation for sin A : 3(sin A 1 2) 5 3 2 sin A 3 sin A 1 6 5 3 2 sin A 4 sin A 5 2 3 sin A 5
(2) Use a calculator to find one value of A (be sure that the calculator is in RADIAN mode):
(3) Find the reference angle: R 5 2 A 5 2( 2 0.848) 5 0. (4) Sine is negative in quadrants In quadrant III: A 1 5 p 1 0.848 3. III and IV. Use the In quadrant IV: A 2 5 2 p 2 0.848 5. reference angle to find a value of A in each of these quadrants: (5) Write the solution set: {3.99 1 2 p n , 5.44 1 2 p n } Answer
Note: When sin^21 was entered, the calculator returned the value in the interval , the range of the inverse of the sine function. This is the measure of a fourth-quadrant angle that is a solution of the equation. How- ever, solutions are usually given as angle measures in radians between 0 and 2 p plus multiples of 2p. Note that the value returned by the calculator is 5.43 1 2 p ( 2 1) 2 0.85.
EXAMPLE 3
Find all possible solutions to the following equation in degrees: 1 2 (^ sec^ u 1^3 )^5 sec^ u 1^
5 2
2 p 2 # u # p 2
A 2 3 4 B
522 Trigonometric Equations
ENTER : 3 4
DISPLAY:
One value of A is 2 0.848.
-. 8 4 8 0 6 2 0 7 9
s i n -^1 ( - 3 / 4 )
ENTER
2nd SIN ^1 (-) )
STEP 3. The solutions are the x -coordinates of the intersection points of the graphs. We can find the intersection points by using the intersect function. Press 5 to select both curves. When the calculator asks you for a guess, move the cursor near one of the intersection points using the arrow keys and then press. Repeat this process to find the other intersection point.
As before, the solutions in the interval 0 u , 2 p are approximately 3.99 and 5.44. The solution set is {3.99 1 2 p n , 5.44 1 2 p n }.
Writing About Mathematics
1. Explain why the solution set of the equation 2 x 1 4 5 8 is {2} but the solution set of the equation 2 sin x 1 4 5 8 is { }, the empty set. 2. Explain why 2 x 1 4 5 8 has only one solution in the set of real numbers but the equation 2 tan x 1 4 5 8 has infinitely many solutions in the set of real numbers.
Developing Skills
In 3–8, find the exact solution set of each equation if 0° u , 360 °.
3. 2 cos u 2 1 5 0 4. 3 tan u 1 5 0 5. 4 sin u 2 1 5 2 sin u 1 1 6. 5(cos u 1 1) 5 5 7. 3(tan u 2 2) 5 2 tan u 2 7 8. sec u 1 5
In 9–14, find the exact values for u in the interval 0 u , 2 p.
9. 3 sin u 2 5 sin u 10. 5 cos u 1 3 5 3 cos u 1 5 11. tan u 1 12 5 2 tan u 1 11 12. sin u 1 5 13. 3 csc u 1 5 5 csc u 1 9 14. 4(cot u 1 1) 5 2(cot u 1 2)
! 2 ! (^22)
Exercises
I n t e r s e c t i o n X = 5. 4 3 5 1 2 3 2 Y = 3. 7 5
I n t e r s e c t i o n X = 3. 9 8 9 6 5 4 7 Y = 3. 7 5
ENTER
2nd CALC ENTER ENTER
524 Trigonometric Equations
In 15–20, find, to the nearest degree, the measure of an acute angle for which the given equation is true.
15. sin u 1 3 5 5 sin u 16. 3 tan u 2 1 5 tan u 1 9 17. 5 cos u 1 1 5 8 cos u 18. 4(sin u 1 1) 5 6 2 sin u 19. csc u 2 1 5 3 csc u 2 11 20. cot u 1 8 5 3 cot u 1 2
In 21–24, find, to the nearest tenth, the degree measures of all u in the interval 0° u , 360 ° that make the equation true.
21. 8 cos u 5 3 2 4 cos u 22. 5 sin u 2 1 5 1 2 2 sin u 23. tan u 2 4 5 3 tan u 1 4 24. 2 2 sec u 5 5 1 sec u
In 25–28, find, to the nearest hundredth, the radian measures of all u in the interval 0 u , 2 p that make the equation true.
25. 10 sin u 1 1 5 3 2 2 sin u 26. 9 2 2 cos u 5 8 2 4 cos u 27. 15 tan u 2 7 5 5 tan u 2 3 28. cot u 2 6 5 2 cot u 1 2
Applying Skills
29. The voltage E (in volts) in an electrical circuit is given by the function E 5 20 cos (p t ) where t is time in seconds. a. Graph the voltage E in the interval 0 t 2. b. What is the voltage of the electrical circuit when t 5 1? c. How many times does the voltage equal 12 volts in the first two seconds? d. Find, to the nearest hundredth of a second, the times in the first two seconds when the voltage is equal to 12 volts. (1) Let u 5 p t. Solve the equation 20 cos u 5 12 in the interval 0 u , 2 p. (2) Use the formula u 5 p t and your answers to part (1) to find t when 0 u , 2 p and the voltage is equal to 12 volts. 30. A water balloon leaves the air cannon at an angle of u with the ground and an initial velocity of 40 feet per second. The water bal- loon lands 30 feet from the can- non. The distance d traveled by the water balloon is given by the formula
where v is the initial velocity.
d 5 321 v^2 sin 2 u
First-Degree Trigonometric Equations 525
30 ft
θ
ENTER : 2 DISPLAY:
To the nearest tenth of a degree, the measure of u in the first quadrant is 63.4. This is also the reference angle for the third-quadrant angle. In quadrant I: u 1 5 63.4° In quadrant III: u 2 5 180 1 R 5 180 1 63.4 5 243.4°
There are two values of u in the interval 0° u , 360 ° for which tan u 5 , one in the second quadrant and one in the fourth quadrant.
ENTER : 21 3 DISPLAY:
The calculator will display the measure of a fourth-quadrant angle, which is negative. To the nearest tenth of a degree, one measure of u in the fourth quad- rant is 2 18.4. The opposite of this measure, 18.4, is the measure of the reference angle for the second- and fourth-quadrant angles. In quadrant II: u 3 5 180 2 18.4 5 161.6° In quadrant IV: u 4 5 360 2 18.4 5 341.6°
The solution set of 3 tan 2 u 2 5 tan u 2 2 5 0 is {63.4°, 161.6°, 243.4°, 341.6°} when 0° u , 360 °.
EXAMPLE 1
Find all values of u in the interval 0 u , 2 p for which 2 sin u 2 1 5.
Solution How to Proceed (1) Multiply both sides of the equation by sin u: (2) Write an equivalent equation with 0 as the right side: (3) Factor the left side: (4) Set each factor equal to 0 and solve for sin u:
(5) Find all possible values of u:
Answer u 5 32 p
3 sin u
ENTER - 1 8. 4 3 4 9 4 8 8 2
t a n -^1 ( - 1 / 3 )
2nd TAN ^1 )
6 3. 4 3 4 9 4 8 8 2
t a n -^1 ( 2 )
2nd TAN ^1 ) ENTER
Using Factoring to Solve Trigonometric Equations 527
2 sin u 2 1 5 2 sin^2 u 2 sin u 5 3 2 sin 2 u 2 sin u 2 3 5 0
(2 sin u 2 3)(sin u 1 1) 5 0 2 sin u 2 3 5 0 sin u 1 1 5 0 2 sin u 5 3 sin u 5 2 1 sin u 5 There is no value of u such that sin u. 1. For sin u 5 21, u 5 32 p.
3 2
3 sin u
EXAMPLE 2
Find the solution set of 4 sin 2 A 2 1 5 0 for the degree measures of A in the interval 0° A , 360 °.
Solution METHOD 1 METHOD 2 Factor the left side. 4 sin 2 A 2 1 5 0 (2 sin A 2 1)(2 sin A 1 1) 5 0 2 sin A 2 1 5 0 2 sin A 1 1 5 0 2 sin A 5 1 2 sin A 5 2 1 sin A 5 sin A 5
If sin A 5 , A 5 30 ° or A 5 150 °. If sin A 5 , A 5 210 ° or A 5 330 °.
Answer {30°, 150°, 210°, 330°}
Note: The graphing calculator does not use the notation sin 2 A , so we must enter the square of the trig function as (sin A )^2 or enter sin ( A )^2. For example, to check the solution A 5 30 ° for Example 2:
ENTER : 4 30 1
DISPLAY:
Factoring Equations with Two Trigonometric Functions
To solve an equation such as 2 sin u cos u 1 sin u 5 0, it is convenient to rewrite the left side so that we can solve the equation with just one trigonometric func- tion value. In this equation, we can rewrite the left side as the product of two factors. Each factor contains one function.
0
4 ( s i n ( 3 0 ) ) 2 - 1
0
4 s i n ( 3 0 ) 2 - 1
SIN ) x 2 ENTER
( SIN ) ) x 2 ENTER
528 Trigonometric Equations
Solve for sin 2 A and take the square root of each side of the equation. 4 sin 2 A 2 1 5 0 4 sin 2 A 5 1 sin 2 A 5 sin A 5 6^12
1 4
Developing Skills
In 3–8, find the exact solution set of each equation if 0° u , 360 °.
3. 2 sin 2 u 1 sin u 2 1 5 0 4. 3 tan 2 u 5 1 5. tan 2 u 2 3 5 0 6. 2 sin 2 u 2 1 5 0 7. 6 cos 2 u 1 5 cos u 2 4 5 0 8. 2 sin u cos u 1 cos u 5 0
In 9–14, find, to the nearest tenth of a degree, the values of u in the interval 0° u , 360 ° that sat- isfy each equation.
9. tan 2 u 2 3 tan u 1 2 5 0 10. 3 cos 2 u 2 4 cos u 1 1 5 0 11. 9 sin 2 u 2 9 sin u 1 2 5 0 12. 25 cos 2 u 2 4 5 0 13. tan 2 u 1 4 tan u 2 12 5 0 14. sec^2 u 2 7 sec u 1 12 5 0
In 15–20, find, to the nearest hundredth of a radian, the values of u in the interval 0 u , 2 p that satisfy the equation.
15. tan^2 u 2 5 tan u 1 6 5 0 16. 4 cos 2 u 2 3 cos u 5 1 17. 5 sin 2 u 1 2 sin u 5 0 18. 3 sin 2 u 1 7 sin u 1 2 5 0 19. csc^2 u 2 6 csc u 1 8 5 0 20. 2 cot 2 u 2 13 cot u 1 6 5 0 21. Find the smallest positive value of u such that 4 sin 2 u 2 1 5 0. 22. Find, to the nearest hundredth of a radian, the value of u such that sec u 5 and , u , p. 23. Find two values of A such that (sin A )(csc A ) 5 2sin A.
Not all quadratic equations can be solved by factoring. It is often useful or nec- essary to use the quadratic formula to solve a second-degree trigonometric equation. The trigonometric equation 2 cos 2 u 2 4 cos u 1 1 5 0 is similar in form to the algebraic equation 2 x^2 2 4 x 1 1 5 0. Both are quadratic equations that can- not be solved by factoring over the set of integers but can be solved by using the quadratic formula with a 5 2, b 5 24, and c 5 1.
13-3 USING THE QUADRATIC FORMULA TO SOLVE
TRIGONOMETRIC EQUATIONS
p 2
5 sec u
530 Trigonometric Equations
Algebraic equation : Trigonometric equation : 2 x^2 2 4 x 1 1 5 0 2 cos^2 u 2 4 cos u + 1 5 0
x 5 cos u 5
x 5 cos u 5
x 5 cos u 5
x 5 cos u 5
x 5 cos u 5
There are no differences between the two solutions up to this point.
However, for the algebraic equation, the solution is complete. There are two val-
ues of x that make the equation true: x 5 or x 5.
For the trigonometric equation, there appear to be two values of cos u. Can we find values of u for each of these two values of cos u?
cos u 5
There is no value of u such that cos u. 1.
cos u 5
There are values of u in the first quadrant and in the fourth quadrant such that cos u is a positive number less than 1. Use a calculator to approximate these values to the nearest degree.
ENTER : 2 DISPLAY:
To the nearest degree, the value of u in the first quadrant is 73°. This is also the value of the reference angle. Therefore, in the fourth quadrant, u 5 360 ° 2 73 ° or 287°. In the interval 0° u , 360 °, the solution set of 2 cos^2 u 2 4 cos u 1 1 5 0 is: {73°, 287°}
) ENTER
2nd ¯ ) )
7 2. 9 6 8 7 5 1 5 4
COS ( c o s - (^1) ( ( 2 - √ ( 2 ) ) / 2 ) 2nd ^1
2 2 1. 2 5 0.
2 2 " 2 CASE 2 2
1 1 1. 2 5 1.
2 1! 2 CASE 1 2
2 2! 2 2
2 1! 2 2
2 6! 2 2
2 6! 2 2
4 6! 8 4
4 6! 8 4
4 6! 16 2 8 4
4 6! 16 2 8 4
2 ( 24 ) 6 "( 24 )^2 2 4 ( 2 )( 1 ) 2 ( 2 )
2 ( 24 ) 6 "( 24 )^2 2 4 ( 2 )( 1 ) 2 ( 2 )
2 b 6 "b 2 2 4ac 2a
2 b 6 "b^2 2 4ac 2a
Using the Quadratic Formula to Solve Trigonometric Equations 531
ENTER : 23 DISPLAY:
The sine function is positive in the first and second quadrants. In quadrant I: B 5 27 ° In quadrant II: B 5 180 2 27 5 153 °
Let sin B 5.
Since 2 1.46 is not a number between 2 1 and 1, it is not in the range of the sine function. There are no values of B such that sin B 5. ✘
Enter Y 1 5 3 sin 2 X 1 3 sin X 2 2 into the menu.
ENTER : 3 3 2
With the calculator set to DEGREE mode, graph the function in the following viewing window: Xmin 5 0, Xmax 5 360, Xscl 5 30, Ymin 5 25, Ymax 5 5 The solutions are the x -coordinates of the x - intercepts, that is, the roots of Y 1. Use the zero function of your graphing calculator to find the roots. Press. Use the arrows to enter a left bound to the left of one of the zero values, a right bound to the right of the zero value, and a guess near the zero value. The calculator will dis- play the coordinates of the point at which the graph intersects the x -axis. Repeat to find the other root.
As before, we find that the solutions in the interval 0° u , 360 ° are approxi- mately 27° and 153°.
Answer B 5 27 ° 1 360 n or B 5 153 ° 1 360 n for integral values of n.
Z e r o X = 1 5 2. 7 7 8 7 9 Y = 0
Z e r o X = 2 7. 2 2 1 2 0 8 Y = 0
2nd CALC 2
Y SIN X,T, ,n ) x 2 SIN X,T, ,n )^
Calculator^ Y Solution
23 2! 33 6
23 2! 33 6 <
23 2! 33 CASE 2 6
) ENTER
2nd (^) ¯ ) ) 2 7. 2 2 1 2 0 7 6 8
s i n -^1 ( ( - 3 + √ ( 3 3 ) ) / 6 )
2nd SIN ^1 (
Using the Quadratic Formula to Solve Trigonometric Equations 533
Writing About Mathematics
1. The discriminant of the quadratic equation tan 2 u 1 4 tan u 1 5 5 0 is 2 4. Explain why the solution set of this equation is the empty set. 2. Explain why the solution set of 2 csc 2 u 2 csc u 5 0 is the empty set.
Developing Skills
In 3–14, use the quadratic formula to find, to the nearest degree, all values of u in the interval 0 ° u , 360 ° that satisfy each equation.
3. 3 sin 2 u 2 7 sin u 2 3 5 0 4. tan 2 u 2 2 tan u 2 5 5 0 5. 7 cos 2 u 2 1 5 5 cos u 6. 9 sin 2 u 1 6 sin u 5 2 7. tan 2 u 1 3 tan u 1 1 5 0 8. 8 cos 2 u 2 7 cos u 1 1 5 0 9. 2 cot^2 u 1 3 cot u 2 4 5 0 10. sec 2 u 2 2 sec u 2 4 5 0 11. 3 csc 2 u 2 2 csc u 5 2 12. 2 tan u (tan u 1 1) 5 3 13. 3 cos u 1 1 5 14. 15. Find all radian values of u in the interval 0 u , 2 p for which. 16. Find, to the nearest hundredth of a radian, all values of u in the interval 0 u , 2 p for which.
When an equation contains two different functions, it may be possible to factor in order to write two equations, each with a different function. We can also use identities to write an equivalent equation with one function. The equation cos^2 u 1 sin u 5 1 cannot be solved by factoring. We can use the identity cos 2 u 1 sin 2 u 5 1 to change the equation to an equivalent equa- tion in sin u by replacing cos 2 u with 1 2 sin 2 u. cos 2 u 1 sin u 5 1 1 2 sin 2 u 1 sin u 5 1 2 sin^2 u 1 sin u 5 0 sin u ( 2 sin u 1 1) 5 0 sin u 5 0 2 sin u 1 1 5 0 u 15 0 ° 1 5 sin u u 2 5 180 ° u 3 5 90 °
13-4 USING SUBSTITUTION TO SOLVE TRIGONOMETRIC
EQUATIONS INVOLVING MORE THAN ONE FUNCTION
cos u 3 5
1 3 cos u 1 1
sin u 1 5
1 2 sin u
sin u 2 5
3 sin u 1 2
1 cos u
Exercises
534 Trigonometric Equations
Any of the eight basic identities or the related identities can be substituted in a given equation.
EXAMPLE 1
Find all values of A in the interval 0° A , 360 ° such that 2 sin A 1 1 5 csc A.
Solution How to Proceed (1) Write the equation: 2 sin A 1 1 5 csc A
(2) Replace csc A with : 2 sin A 1 1 5 (3) Multiply both sides of the 2 sin^2 A 1 sin A 5 1 equation by sin A : (4) Write an equivalent equation 2 sin 2 A 1 sin A 2 1 5 0 with 0 as the right side: (5) Factor the left side: (2 sin A 2 1)(sin A 1 1) 5 0 (6) Set each factor equal to 0 2 sin A 2 1 5 0 sin A 1 1 5 0 and solve for sin A : (^) 2 sin A 5 1 sin A 5 2 1
sin A 5 A 5 270 ° A 5 30 ° or A 5 150 °
Answer A 5 30 ° or A 5 150 ° or A 5 270 °
Often, more than one substitution is necessary to solve an equation.
EXAMPLE 2
If 0 u , 2 p, find the solution set of the equation 2 sin u 5 3 cot u.
Solution How to Proceed (1) Write the equation: 2 sin u 5 3 cot u
(2) Replace cot u with : 2 sin u 5 3 (3) Multiply both sides of the 2 sin 2 u 5 3 cos u equation by sin u: (4) Replace sin 2 u with 1 2 cos 2 u: 2(1 2 cos 2 u) 5 3 cos u (5) Write an equivalent equation 2 2 2 cos^2 u 5 3 cos u in standard form:
cos sin uu A cos sin uu B
1 2
1 sin A
1 sin A
536 Trigonometric Equations
2 2 2 cos 2 u 2 3 cos u 5 0 2 cos 2 u 1 3 cos u 2 2 5 0
(6) Factor and solve for cos u: (2 cos u 2 1)(cos u 1 2) 5 0 (7) Find all values of u in the 2 cos u 2 1 5 0 cos u 1 2 5 0 given interval: (^) 2 cos u 5 1 cos u 5 2 2 ✘
cos u 5 No solution
u 5 or u 5
Answer
The following identities from Chapter 10 will be useful in solving trigono- metric equations:
Up 3 , 53 p V
5 p 3
p 3
1 2
Using Substitution to Solve Trigonometric Equations Involving More than One Function 537
Reciprocal Identities Quotient Identities Pythagorean Identities sec u 5 tan u 5 cos^2 u 1 sin^2 u 5 1
csc u 5 cot u 5 1 1 tan^2 u 5 sec^2 u
cot u 5 (^) tan^1 u cot^2 u 1 1 5 csc^2 u
cos u sin u
1 sin u
sin u cos u
1 cos u
Writing About Mathematics
1. Sasha said that sin u 1 cos u 5 2 has no solution. Do you agree with Sasha? Explain why or why not. 2. For what values of u is sin u 5 true?
Developing Skills
In 3–14, find the exact values of u in the interval 0° u , 360 ° that satisfy each equation.
3. 2 cos 2 u 2 3 sin u 5 0 4. 4 cos 2 u 1 4 sin u 2 5 5 0 5. csc 2 u 2 cot u 2 1 5 0 6. 2 sin u 1 1 5 csc u 7. 2 sin 2 u 1 3 cos u 2 3 5 0 8. 3 tan u 5 cot u 9. 2 cos u 5 sec u 10. sin u 5 csc u 11. tan u 5 cot u 12. 2 cos 2 u 5 sin u 1 2 13. cot 2 u 5 csc u 1 1 14. 2 sin 2 u 2 tan u cot u 5 0
" 1 2 cos 2 u
Exercises
