2nd lesson in trigonometric equations

8/3/2019 2nd Lesson in Trigonometric Equations

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To solve for trigonometric equationsmeans to find all the possible values

of angle , that will make the givenequation true.



For example, the possible values of from 0 to 360 , that will make

the trigonometric equation2Sin2 Cos 2 = 0 true are; 30 ,

150 , 210 , and 330 .



P roof:

If = 30 ;

2 Sin2 Cos 2 = 0

2 Sin2

30 Cos 2(30 ) = 02 Sin2 30 Cos 60 = 02(1/2) 2 1/2 = 0

2(1/4) ½ = 00 = 0



Furthermore, trying the valuesof = 210 and 330 will also make

the equation 2Sin 2 Cos 2 = 0true.



In the just concluded example, the

values of are restricted from 0 to360 .

Without restriction, the equation2Sin2 Cos 2 = 0, will yieldinfinite number of values for bothpositive (counter clockwise) andnegative(clockwise).



Tools for solving trigonometric

equations:Basic concepts in solving algebraic

equations (factoring, specialproducts, quadratic formula,completing the square, rationalizing,simplifying equations, etc.)



Basic theorems of special righttriangles (30 x 60 x 90 and45 x 45 x 90 )

Basic trigonometric identities



ILLUSTRATIVEP ROBLEMS:1.) Solve Sin = Cos , for 0 360

Solution:Sin = Cos

Sin /Cos = Cos / Cos Tan = 1 = 45 , 225

or, Sin = Cos Sin /Sin = Cos /Sin

1 = Cot Cot = 1 = 45 , 225



Check:

@ = 45 , Sin = Cos Sin 45 = Cos 45 2 /2 = 2 /2

@ = 225 , Sin = Cos

Sin 225 = Cos 225 2 /2 = 2 /2



2.) Solve 2Cos 2 Sin 2 = 2, for 0 360

Solution:2Cos2 Sin 2 = 2

2Cos2 ( 1 Cos 2 ) = 2

2Cos2 1 + Cos 2 = 23Cos 2 1 = 2

3Cos 2 = 3

Cos2 = 1Cos = ± 1

= 0 , 180 , 360



or,2Cos2 Sin 2 = 2

2(1 Sin 2 ) Sin 2 = 22 2Sin 2 Sin 2 = 2

2 3Sin 2 = 23Sin2 = 0Sin2 = 0Sin = 0

= 0 , 180 , 360



3.) Solve Tan + Cot = 2, for 0 360

Solution:Tan + Cot = 2Tan + ( 1 / Tan ) = 2

[Tan + (1 / Tan ) = 2] Tan Tan 2 + 1 = 2Tan

Tan 2 2tan + 1= 0(Tan 1) 2 = 0

Tan 1 = 0Tan = 1

= 45 , 225



or,

Tan + Cot = 21 /Cot + Cot = 2

[1 /Cot + Cot = 2] Cot

1 + Cot 2 = 2Cot Cot 2 2Cot + 1= 0

(Cot 1) 2 = 0

Cot 1 = 0Cot = 1

= 45 , 225



4.) Solve 2Sin 2 + Sin 2 2 = 2, for 0 360

Solution: 2Sin2 + Sin 2 2 = 22Sin2 + (Sin 2 ) 2 = 2

2Sin2 + (2Sin Cos ) 2 = 22Sin2 + 4Sin 2 Cos 2 = 2Sin2 + 2Sin 2 Cos 2 = 1

Sin2 (1 + 2Cos 2 ) = 1Sin2 = 1 1 + 2Cos 2 = 1Sin = ± 1 Cos = 0

= 90 , 270 = 90 , 270



or, 2Sin2 + Sin 2 2 = 22Sin2 + (Sin 2 ) 2 = 2

2Sin2 + ( 2Sin Cos )2 = 22Sin2 + 4Sin 2 Cos 2 = 2Sin2 + 2Sin 2 Cos 2 = 1

Sin2

+ 2Sin2

( 1 - Sin2

) = 1Sin2 + 2Sin 2 2Sin 4 = 1 2Sin 4 + 3Sin 2 = 1

2Sin4 3Sin 2 + 1 = 0(2Sin2 1)(Sin 2 1) = 0 factoring

2Sin2 1 = 0 Sin 2 1 = 0Sin = ± 2/2 Sin = ± 1

= 45 , 135 , 225 , 315 = 90 , 270



5.) Find the value of in the followingconditions

Tan (8 + 1) = Cot 17Tan 4 = Cot 6Sin 2 = Cos



Tan (8 + 1) = Cot 17

Tan (8 + 1) = Cot 17Tan (8 + 1) = Tan (90 - 17)

Cot A = Tan (90 A)Tan (8 + 1) = Tan 738 + 1 = 73

8 = 73 18 = 72

= 9



Tan 4 = Cot 6Tan 4 = Cot Tan 4 = Tan (90 )

4 = 90 64 + 6 = 90

10 = 90

= 9



Sin 2 = Cos Sin 2 = Cos 2Sin Cos = Cos

Sin 2 = 2Sin Cos 2Sin = 1Sin = 1 /2

= 30 , 150 , 390 ,...



Cos 9 = 1 / Csc (3 + 6)

Cos 9 = 1 / Csc (3 + 6)Sin (90 9 ) = Sin (3 + 6)

90 9 = 3 + 690 6 = 3 + 9

84 = 1212 = 84

= 7



7.) Find the value of x in the followingconditions

Arctan x + Arctan (1/3) = /4Arctan (x /2) + Arctan (x /3) = 45Arctan (1 x) + Arctan (1 + x) = Arctan (1 /8)



Arctan x + Arctan (1/3) = /4

Let: A = Arctan x, B = Arctan (1/3)Tan A = x Tan B = 1 /3Then,Arctan x + Arctan (1/3) = /4

A + B = / 4Tan (A + B) = Tan ( / 4)

(Tan A + Tan B) / (1 Tan A Tan B) = Tan ( /4)(x + 1/3) / [1 (x)(1/3)] = 1

3x + 1 = 3 xx = 1 /2



Arctan (x /2) + Arctan (x /3) = 45Let: A = Arctan x/2, B = Arctan (x/3)

Tan A = x /2 Tan B = x /3Then,Arctan (x/2) + Arctan (x/3) = 45

A + B = 45Tan (A + B) = Tan 45

(Tan A + Tan B) / (1 Tan A Tan B) = Tan 45

(x/2 + x/3) / [1 (x/2)(x/3)] = 1x2 + 5x - 6 = 0

(x 1)(x + 6) = 0

x = 1, - 6



Arctan (1 x) + Arctan (1 + x) = Arctan (1 /8)Let: A = Arctan (1-x), B = Arctan (1+x)

C = Arctan (1/8)Tan A = (1-x), Tan B = (1+x), Tan C = 1 /8

Then,

Arctan (1-x) + Arctan (1+x) = Arctan(1/8)A + B = C

Tan (A + B) = Tan C

(Tan A + Tan B) / (1 Tan A Tan B) = 1 /8(1 x + 1 + x) / [1 (1-x)(1+x)] = 1 / 8

x2 = 16

x = ±4



8.) In the equations Arcsin(3x 4y) = /2 andArccos(x y) = /3, find

the value of xthe value of yArcsin(x y)



10.) Find the value of x from the givenrelation

x = Sin 0 + Sin 1 + Sin 2 +...+ Sin 90Cos 0 + Cos 1 + Cos 2 +...+ Cos 90

(Tan + Cot ) 2 Sin2 - Tan 2 = xSin x Cos x + Sin 2x = 1

2nd lesson in trigonometric equations

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