Verify Identities using Basic Identities

Verify Identities using Basic Identities

In this lecture, we verify individual trigonometric equations as identities.

In previous lectures, we encountered some basic trigonometric identities.

FUNDAMENTAL/BASIC IDENTITIES Quotient Identities:

tan sin

cos

 

  , cos   0 cot ^cos sin

 

  , sin   0 Reciprocal Identities:

csc 1

 sin

  , sin   0 sin ¹

 csc

  , csc   0 sec 1

 cos

  , cos   0 1 cos  sec

  , sec   0 cot 1

 tan

  , tan   0 tan ¹

 cot

  , cot   0 Odd Identities:

   

sin x sin x

   ^{ csc}   ^{x  csc}   ^{ x}

   

tan x tan x

   ^{ cot}   ^{x  cot}   ^{ x}

Even Identities:

   

cos x  cos  x ^sec   ^{x  sec}   ^{ x}

Co-function Identities:

 2 

sin ^  x  cos x cos  ^ 2  x   sin x

 2 

csc ^  x  sec x sec  ^ 2  x   csc x

 2 

tan ^  x  cot x cot  ^ 2  x   tan x Primary Objective: Students verify trigonometric identities using fundamental identities.

An identity is a statement involving variables that is true regardless of the variable values.

The fundamental identities include the three Pythagorean Identities.

Deriving the second two Pythagorean identities is quite simple based on the first. For instance, consider the first Pythagorean identity. If we divide through by cos ²  we arrive at the second Pythagorean identity as shown below.

2 2

2 2 2

2 2

sin cos 1

cos cos cos

sin cos 1

^ _ ^ _{ }

   

 

Of course, dividing by a quantity introduces the difficulty of worrying about whether or not the divisor-quantity equals zero, but in this case tangent and secant are not defined if cos   0 , so we simply divide for all values of  except where cos   0 . To conclude, we employ the fundamental and reciprocal identities as below.

2 2

2 2 2

sin cos 1

cos cos cos

2 2

tan 1 sec

 

  

 

 

 

We leave the derivation for 1 cot  ²   csc ²  to the reader.

The work above is a derivation of one identity from another. The remainder of this lecture demonstrates how to use the fundamental identities to verify other proposed identities. For example, we will show that csc   cos cot    sin  is an identity.

First, we arbitrarily choose one side of the equation to manipulate. We select the left-hand side (LHS).

LHS  csc   cos cot  

Our goal is to manipulate this side of the identity until we arrive at the right-hand side (RHS). In this case, we begin by applying the Reciprocal Identity for cosecant.

1 sin

LHS csc cos cot cos cot



  

 

 

 

Next, we apply the Quotient Identity for cotangent.

cos 1

sin _ cos  sin _ ^

  

Pythagorean Identities:

2 2

sin   cos   1

2 2

1 tan    sec 

2 2

1 cot    csc 

Next, we add the resulting fractions.

2

2

cos 1

sin sin

1 cos

sin sin

1 cos sin

cos ^

 

  

 





  

 



Noting that sin ²   cos ²   1 implies sin ²    1 cos ²  , we conclude this verification.

2

2

1 cos sin sin

sin sin sin

sin sin sin

 

 

 



 













 _sin

sin RHS







The perceptive reader detects that verifying identities requires some foresight and creative thinking. Here are a few tips.

Suggestions for Verifying Identities 1. Work with the more complicated side.

2. Make substitutions using known identities.

3. Try rewriting expressions in terms of sine or cosine.

4. Perform indicated algebraic operations such as adding fractions or multiplying polynomials.

5. Reverse operations like factoring polynomials or decomposing fractions.

6. Check each result against the other side of the identity.

Example Exercise 1 Verifying Identities

Recall the algebraic identity called the Difference of Squares Rule.

 ^{a b a b }  ^  ^{ a 2  b 2}

Looking at

  

1

1 tan 1 sin 1 sin

x x x

    , we start with the right-hand side of the identity, and we apply the Difference of Squares Rule.

  

   ²  ²    ^{2 2}

2

2 2 2 2 2

2

RHS 1 sin 1 sin

1 sin

1 sin Square the terms

cos sin sin Pythagorean Identity: cos sin 1 cos Subtract

x x

x a b a b a b

x

x x x x x

x

  

     

 

    



2

2

1 sec

2 2

1 1 tan

ion

Reciprocal Identity for Cosine

Pythagorean Identity: 1 tan sec LHS

x

x x x





  



Verify: ¹ ₂  1 sin  1 sin 

1 tan x x

x   



Example Exercise 2 Verifying Identities

Multiply the left side by a propitious choice of “1” and apply the Difference of Squares Rule.

 

2 2

LHS 1

tan sec

1 tan sec

tan sec tan sec

1 tan sec

tan sec

 

 

   

 

 

 



 

 

 

 

 

The Pythagorean Identity, 1 tan  ²   sec ²  implies tan ²   sec ²   1 .

 

 

 

2 2

2 2

2 2

1 tan sec

sec 1 sec

1 tan sec

Commutative Propeerty of Addition

sec sec 1

1 tan sec

sec sec 0

1 1

 

 

 

 

 

 

 

  

 

  

 

  



   ^{tan sec} 

1

  

 Reduce common factors  tan   sec 

Next, we substitute a Quotient Identity and a Reciprocal Identity.

sin 1

cos cos

tan sec ^ _{ }

 

 

 

Adding the fractions and using the commutative property of addition attains the right-hand side.

sin 1

cos cos

sin 1 cos 1 sin

cos

RHS

  

 

 





 







Verify: ^{1 1 sin}

tan sec cos



  

 

 

Example Exercise 3 Verifying Identities

Start with the left-hand side.

LHS  cot x  sin x  cos 2 x  sec x

By Quotient Identity, we have the following.

cos 2

 sin ^x _x  sin x  cos x  sec x Next, we reduce.

cos

 sin _x ^x  sin x ²

2

cos sec cos cos sec

x x

x x x

 

  

We conclude by applying the Reciprocal Identity, reducing common factor, and simplifying.

2

1 cos

cos cos sec cos cos cos cos cos cos

x

x x x

x x x

x x x

  

   

    _cos ¹ _x cos cos 1

cos cos 0

RHS

x x

x x

  

 





Verify: cot x  sin x  cos ² x  sec x  0

Example Exercise 4 Verifying Identities

Verifying an equation is not an identity simply requires finding a counter-example. Let   ^ ₂ .

    2 2   2

2 sin cos sin 2 sin cos sin 2 1 0 1 0 1

  

   



  



Obviously, the equation 2sin cos    sin  is not true for all values of theta, so it is not an identity.

Solve the equation for individual trigonometric functions to find conditions that make the equation true.

 

1 2

2 sin cos sin 2 sin cos sin 0 sin 2 cos 1 0 sin 0 or cos

  

  

 

 



 

 

 

The equation is conditionally true, meaning it is true on the condition that sin   0 or that

1

cos   2 .

Show that if 2sin cos    sin  is an identity then it is conditional.

#1) LHS  ^sin ₁ ^x  _cos ¹ _x  _cos ^{sin x} _x  tan x  RHS

#3) ^{RHS  tan 2 λ  sec 2 λ   1}  ^{sec λ  1 sec}  ^{λ   1}  ^LHS

#5)

^{  }

^    

tan sin sec 1 tan sin sec 1

tan sin sec 1 cos

sec ^x _x 1 ^x sec ^x _x 1 ^x sec ^x 1 ^{x x} tan ^{x x} sin sec 1 sin ^x _x sec 1 cos 1 cos

x ^ x ^ x x x x x

             

#7)

2 2 1 tan 1 tan 1

tan tan tan tan

LHS csc 1 cot 1 ^{x x} RHS

x x x x

x x ^

        

#9) LHS 1 tan   θ  ^cos _cos ^θ _θ  ^sin _cos ^θ _θ  ^cos _cos ^{θ  sin} _θ ^θ  RHS

#11) RHS  sin cos _x ¹ _x  ^{tan x} 1 ^{ cot x}  sin cos _x ¹ _x   ^sin cos ^x _x  ^cos sin ^x _x   sin cos ^sin _x ^x

_x  sin

^cos _x cos ^x _x  sec ² x  csc ² x  LHS

#13)

Application Exercise

Practice Problems Verify the identities.

#1) sin sec x x  tan x #2) ^{sin cos x x}  ^{tan x  cot x}  ^{ 1}

#3)  ^{sec   1 sec}  ^{   1}  ^{tan 2  #4) csc   sin   cos cot  }

#5) ^{tan sin} _sec ^{x } _{x  1} ^x   1 cos x #6) 2cot x  ^csc _sec ^x _x  ^cos _{sin x} ^x

#7)

2 1 tan

csc tan ^x

x  ^ x #8) ^cot _cos ^x _x  ^sec _cot ^x _x  sec ² x csc x

#9) 1 tan    ^cos _cos ^{ } _ ^{sin } #10) sec ⁴ x  sec ² x  tan ⁴ x  tan ² x

#11) sec ² x  csc ² x  ^tan _{sin cos} ^x _x ^ _ ^cot _x ^x #12) sec   2 tan   ^cot _csc ^ _ ^ _ ^2cos _{sin } ^ Graph the functions.

#13) y  sin   5 ^x #14) ^{y  2cos}   ^{ x}

A cannon fires a projectile with initial velocity v ₀ . Picture the projectile firing from the origin into the first quadrant, making an angle  with the positive x-axis.

If there is no significant air resistance, then at time t the coordinates of the projectile are

0 cos

x    v t  and y       16 t ² v t ₀ sin  . Show that

2 2

16 sec tan

y    v     x x  .