AN IDENTITY IS AN EQUALITY 

that is true for any value of the variable. (An equation is an equality that is true only for
certain values of the variable.)
In algebra, for example, we have this identity:
(x + 5)(x − 5) = x² − 25.
The significance of an identity is that, in calculation, we may replace either member of the identity with the
other.  We use an identity to give an expression a more convenient form.  In calculus and all its applications, the
trigonometric identities are of central importance.
On this page, we will present the main identities.  The student will have no better way of practicing algebra
than by proving them.  Links to the proofs appear below.
Reciprocal identities
   1  
csc     1   
sin θ  =  csc           =  
θ sin θ
θ
 
   1      1   
cos  sec 
 =  sec           =  cos 
θ θ
θ θ
 
   1      1   
tan  cot 
 =  cot           =  tan 
θ θ
θ θ
The proof of each of those follows from the definitions of the trigonometric functions, Topic 17.
Proof of the reciprocal relations
By definition:
y
sin  csc  r
  =   ,    =   . 
θ θ y
r
Therefore, sin θ is the reciprocal of csc θ:
   1    
sin 
  = csc 
θ
θ
where 1-over any quantity is the symbol for its reciprocal; Lesson 5 of Algebra.  Similarly for the remaining functions.

Again, the point about an identity is that, in calculation, we may replace either member of the identity with
the other.  Thus if we see
   1  
  "sin θ", then we may, if we wish, replace it
csc  ";  and,
with "
θ
   1   
  symmetrically, if we see
csc  ", then we may replace it with "sin θ".
"
θ

Tangent and cotangent identities

sin θ cos 
tan θ  = cos          cot θ  =  θ
θ sin θ
To prove:
tan θ  = sin θ and  cot θ  = cos  .
 cos  θ
θ sin θ
Proof.   By definition,
y
tan θ  = .
x
Therefore, on dividing both numerator and denominator by r,
sin θ
tan  y/r
 =    =  cos  .
θ x/r
θ
 
   1    cos 
cot 
  =  tan    =  θ .
θ
θ sin θ
These are the two identities.

Pythagorean identities
sin²θ +
a)  = 1
cos²θ
 
b) 1 + tan²θ  = sec²θ
 
csc
c) 1 + cot²θ  =
²θ

a')     sin²θ  =  1 − cos²θ.       cos²θ  =  1 − sin²θ.

These are called Pythagorean identities, because, as we will see in theirproof, they are the trigonometric
version of the Pythagorean theorem.
The two identities labeled a') -- "a-prime" -- are simply different versions of a).  The first shows how we can
express sin θ in terms of cosθ; the second shows how we can express cos θ in terms of sin θ.
Note:  sin²θ -- "sine squared theta" -- means (sin θ)².
Proof of the Pythagorean identities
To prove:
sin²θ +
a)  = 1
cos²θ
 
b) 1 + tan²θ  = sec²θ
 
csc
c) 1 + cot²θ  =
²θ

Proof.   According to the Pythagorean theorem,

x² + y² = r².  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .(1)

Therefore, on dividing both sides by r²,
x² y² r²
  +     =    =   1.
r² r² r²
 That is, according to the definitions,
cos²θ  +  sin²θ  =  1.
Apart from the order of the terms, this is the first Pythagorean identity, a).  To derive b), divide line (1) by x²;
and to derive c), divide by y².

Sum and difference formulas

sin (  + β)  =   sin   cos β + cos   sin β
sin (  − β)  =   sin   cos β − cos   sin β
cos (  + β)  =   cos   cos β − sin   sin β
cos (  − β)  =   cos   cos β + sin   sin β
Note:  In the sine formulas, + or − on the left is also + or − on the right.  But in the cosine formulas, + on the
left becomes − on the right; and vice-versa.
Since these identities are proved directly from geometry, the student is not normally required to master the
proof.  However, all the identities that follow are based on these sum and difference formulas.  The student should
definitely know them.

Theorem. sin (  + β)  =  sin   cos β + cos   sin β

and    

  cos (  + β)  =  cos   cos β − sin   sin β.

Proof.   Let the straight line AB revolve to the point C and sweep out the

angle  , and let it continue to D and sweep out the angle β;

draw DE perpendicular to AB.
ED
  Then we are to determine sin (  + β), which is  ,
DA
AE
  and cos (  + β), which is  .
DA
Draw DF perpendicular to AC,
draw FG perpendicular to AB,
and draw FH perpendicular to ED.
 Then angle HDF is equal to angle  .
For, since the straight line AC crosses the parallel lines HF, AB, it makes the alternate angles equal (Theorem 8);
therefore angle HFA is equal to angle  .
And by the construction, angle DFH is the complement of angle HFA;
therefore angle HDF (the complement of DFH) is also equal to angle  .

Now,
  ED = GF + HD.
 
Therefore, on dividing by DA,
 
ED GF HD
sin (  + β)  =   = +
DA DA DA

GF AF HD FD
   =  + 
AF DA FD DA

    =  sin   cos β + cos   sin β.

Next,
  EA = GA − FH.
 
Therefore,
 
EA GA FH
cos (  + β)  =   = −
AD AD AD

GA AF FH DF
   =  − 
AF AD DF AD

    =  cos   cos β − sin   sin β.

 This is what we wanted to prove.
The difference formulas can be proved from the sum formulas, by replacing +β with +(−β), and using these
identities:
cos (−β) = cos β 
sin (−β) = −sin β.

 Prove:

   

Solution.    Tangent identity

      Formulas

  We will now construct  tan    by dividing the first term in the

numerator by cos   cos β.  But then we must divide every term by 
cos   cos β:
 

     

  

Double-angle formulas

Double-angle formulas
 Proof
The double-angle formulas are proved from the sum formulas by putting β =  .  We have
sin 2  = sin (  +  )= sin   cos   + cos   sin 
 
  = 2 sin   cos  .
 
cos 2  = cos (  +  )= cos   cos   − sin   sin 
 
cos 2 = cos²  − sin² .   .  .  .  .  .  . (1)
This is the first of the three versions of cos 2 .  To derive the second version, in line (1) use this Pythagorean
identity:
sin²  = 1 − cos² .
Line (1) then becomes

cos 2 = cos²  − (1 − cos² )

  = cos²  − 1 + cos² .

cos 2 = 2 cos²  − 1.  .  .  .  .  .  .  .  .  .  (2)

To derive the third version, in line (1) use this Pythagorean identity:
cos²  = 1 − sin² .
We have
cos 2 = 1 − sin²  − sin² ;.
 
cos 2 = 1 − 2 sin² .  .  .  .  .  .  .  .  .  .  (3)
These are the three forms of cos 2 .

We will now construct  tan    by dividing by cos  .  But to preserve the equality, we must also multiply by cos  .
 

     Lesson 5 of Algebra

 
     
 

     Reciprocal identities

 
     Pythagorean identities

Half-angle formulas
The following half-angle formulas are inversions of the double-angle formulas, because   is half of 2 .

The plus or minus sign will depend on the quadrant.  Under the radical, the cosine has the + sign; the sine, the
− sign.

Half−angle formulas

  .  .  .  .  .  .  .  (2')

  .  .  .  .  .  .  .  (3')
Whether we call the variable θ or   does not matter.  What matters is the form.  
Proof
Angle   is half of the angle 2 .  Therefore, in line (2), if we
θ
   put 2  = θ, then   will equal  :
2
θ
cos θ = 2 cos² , we will have the half-angle formula for the
2
cosine.
So, on exchanging sides and transposing 1, we have

2 cos² θ = 1 + cos θ
2

 cos² θ = ½(1 + cos θ)

2

cos  θ =
2

transposing, line (3) becomes

θ
  2 sin² = .
2
This is the half−angle formula for the sine.
 Products as sums
a)  sin   cos β   = ½[sin (  + β) + sin (  − β)]
 
b)  cos   sin β   = ½[sin (  + β) − sin (  − β)]
 
c)  cos   cos β   = ½[cos (  + β) + cos (  − β)]
 
d)  sin   sin β   = −½[cos (  + β) − cos (  − β)]

Sums as products
e)  sin A + sin B   =  2 sin ½ (A + B) cos ½ (A − B)
 
f)  sin A − sin B   =  2 sin ½ (A − B) cos ½ (A + B)
 
cos A +
g)    =  2 cos ½ (A + B) cos ½ (A − B)
cos B
 
cos A − −2 sin ½ (A + B) sin ½
h)   = 
cos B (A − B)
In the proofs, the student will see that the identities e) through h) are inversions of a) through d) respectively,
which are proved first.  The identity f) is used to prove one of the main theorems of calculus, namely the derivative of
sin x.
The student should not attempt to memorize these identities.  Practicing their proofs -- and seeing that they
come from the sum and difference formulas -- is enough.

Proof of the product and sum formulas

Products as sums
a)  sin   cos β  =  ½[sin (  + β) + sin (  − β)]
 
b)  cos   sin β  =  ½[sin (  + β) − sin (  − β)]
 
c)  cos   cos β  =  ½[cos (  + β) + cos (  − β)]
 
d)  sin   sin β  =  −½[cos (  + β) − cos (  − β)]
Proof
These formulas are also derived from the sum and difference formulas.  To derive (a), write
sin (  + β)   =  sin   cos β + cos   sin β

 
sin (  − β)  =  sin   cos β − cos   sin β
and add vertically.  The last terms in each line will cancel:
sin (  + β) + sin (  − β) = 2 sin   cos β.
Therefore, on exchanging sides,
2 sin   cos β = sin (  + β) + sin (  − β),
so that
 sin   cos β = ½[sin (  + β) + sin (  − β)]. 
This is the identity (a)).
Formula (b) is derived in exactly the same manner, only instead of adding, subtract sin (  − β) from sin (  
+ β). 
Formulas (c) and (d) are derived similarly.  To derive (c), write
cos (  + β) = cos   cos β − sin   sin β,
cos (  − β) = cos   cos β + sin   sin β,
and add.  To derive (d), subtract.
Let us derive (d).  On subtracting, the first terms on the right will cancel.  We will have
cos (  + β) − cos (  − β) = −2 sin   sin β.
Therefore, on solving for sin   sin β,
sin   sin β = −½[cos (  + β) − cos (  − β)].

Sums as products
e)  sin A + sin B   =  2 sin ½ (A + B) cos ½ (A − B)
 
f)  sin A − sin B   =  2 sin ½ (A − B) cos ½ (A + B)
 
cos A +
g)    =  2 cos ½ (A + B) cos ½ (A − B)
cos B
 
cos A − −2 sin ½ (A + B) sin ½
h)   = 
cos B (A − B)
Proof
The formulas (e), (f), (g), (h) are derived from (a), (b), (c), (d)respectively; that is, (e) comes from (a), (f) comes
from (b), and so on. 
To derive (e), exchange sides in (a):
½[sin (  + β) + sin (  − β)] = sin   cos β,
so that
sin (  + β) + sin (  − β) = 2 sin   cos β.  .  .  .  .  . (1)
Now put
   + β= A
  and
   − β= B.  .   .  .  .  .  .  .  .  .  .  .  .  .  .  .(2)
The left-hand side of line (1) then becomes
sin A + sin B.
This is now the left-hand side of (e), which is what we are trying to prove.
To complete the right−hand side of line (1), solve those simultaneous equations (2) for   and β. 
On adding them,  2  = A + B,
so that
 = ½(A + B).
On subtracting those two equations,  2β = A − B,
so that
β = ½(A − B).
On the right−hand side of line (1), substitute those expressions for  and β.  Line (1) then becomes
sin A + sin B = 2 sin ½(A + B) cos ½(A − B).
This is the identity (e).
Read it as follows:
"sin A + sin B equals twice the sine of half their sum
times the cosine of half their difference."
Identities (f), (g), and (h) are derived in exactly the same manner from(b), (c), and (d) respectively.