Trigonometric Functions
There are six trigonometric functions, of which the most commonly used are the sine and cosine functions. The other four functions can be expressed in terms of these two. Hence, once we know how to differentiate the sine and cosine, we can derive a formula for differentiating the remaining trigonometric functions. Whilst all of these functions are nonlinear, only the sine and cosine functions are continuous. This means that they can be differentiated for any value of x. In fact, as the illustration below shows, the sine and cosine functions are both cyclic. The graphs of the sine and cosine functions are identical in shape, but are separated by a ninetydegree difference in phase (note that for the remainder of this discussion, the value of x is expressed in radians and not degrees).
The sine and cosine functions are nonlinear, continuous and cyclic
The sine function
The derivative of sin(x) is simply cos(x). But how do we arrive at this result? In order to understand how this works, we need to go back to first principles. You should be familiar with the way in which the derivative of a function y = ƒ(x) is defined. We express this as:
dy  =  lim  ƒ(x + δx)  ƒ(x)  
dx  δx→0  δx 
The function we are interested in is the sine function. We will therefore define the derivative of the sine function as follows:
d  (sin(x)) =  lim  sin(x + δx)  sin(x)  
dx  δx→0  δx 
Assuming you have a reasonable grasp of trigonometry, you should be familiar with the following trigonometric identity:
sin(α + β) = sin(α) · cos(β) + cos(α) · sin(β)
We will now use this trigonometric identity to rewrite our definition of the derivative of the sine function:
d  (sin(x)) =  lim  sin(x) · cos(δx) + cos(x) · sin(δx)  sin(x)  
dx  δx→0  δx 
Factorising for the term sin(x), we get:
d  (sin(x)) =  lim  sin(x) · (cos(δx)  1) + cos(x) · sin(δx)  
dx  δx→0  δx 
We can split this result into two separate fractions and, since the limit applies only to the term δx, we can remove the terms sin(x) and cos(x) from the numerator of each fraction and place them in front of the limit:
d  (sin(x)) = sin(x)  lim  cos(δx)  1  + cos(x)  lim  sin(δx)  
dx  δx→0  dx  δx→0  dx 
We will now make use of the limits of two functions. The first function is sin(θ)/θ. As the value of θ (measured in radians) approaches zero, the value of sin(θ)/θ approaches one. This is expressed as:
lim  sin(θ)  = 1  
θ→0  θ 
The second function is (cos(θ)  1)/θ. As the value of θ (measured in radians) approaches zero, the value of (cos(θ)  1)/θ approaches zero. This is expressed as:
lim  cos(θ)  1  = 0  
θ→0  θ 
You should be able to see that, if we allow the value of δx to become zero, our definition of the derivative of the sine function will simplify to:
d  (sin(x)) = sin(x) · 0 + cos(x) · 1 = cos(x) 
dx 
The cosine function
The derivative of cos(x) is sin(x). We obtain this result in more or less the same way we obtained the derivative of the sine function, using first principles. We can define the derivative of the cosine function as follows:
d  (cos(x)) =  lim  cos(x + δx)  cos(x)  
dx  δx→0  δx 
This time we will use the following trigonometric identity:
cos(α + β) = cos(α) · cos(β)  sin(α) · sin(β)
Using this trigonometric identity to rewrite our definition of the derivative of the cosine function, we get:
d  (cos(x)) =  lim  cos(x) · cos(δx)  sin(x) · sin(δx)  cos(x)  
dx  δx→0  δx 
Factorising for the term cos(x), we get:
d  (cos(x)) =  lim  cos(x) · (cos(δx)  1)  sin(x) · sin(δx)  
dx  δx→0  δx 
Once again, we can split this result into two separate fractions and, since the limit applies only to the term δx, we can remove the terms cos(x) and sin(x) from the numerator of each fraction and place them in front of the limit:
d  (cos(x)) = cos(x)  lim  cos(δx)  1   sin(x)  lim  sin(δx)  
dx  δx→0  dx  δx→0  dx 
We can again make use of the fact that, as the value of θ (measured in radians) approaches zero, the value of sin(θ)/θ approaches one, and the value of (cos(θ)  1)/θ approaches zero. Thus, if we allow the value of δx to become zero, our definition of the derivative of the cosine function simplifies to:
d  (cos(x)) = cos(x) · 0  sin(x) · 1 = sin(x) 
dx 
The tangent function
We could of course apply first principles to obtain the derivatives of the tangent, cosecant, secant and cotangent functions. However, these functions can all be defined in terms of the sine and cosine functions. We will therefore use the definitions we already have, together with the appropriate rules of differentiation, to obtain their derivatives. We will start by looking at the tangent function, which is defined as follows:
tan(x) =  sin(x) 
cos(x) 
We can find the derivative of the tangent function by using the derivatives of the sine and cosine functions together with the quotient rule. The quotient rule states that the derivative of the quotient of two functions is equal to the product of the denominator and the derivative of the numerator, minus the product of the numerator and the derivative of the denominator, all over the denominator squared. Applying this rule, we get:
d  (tan(x)) =  d  (  sin(x)  )  =  cos^{2}(x) + sin^{2}(x) 
dx  dx  cos(x)  cos^{2}(x) 
The trigonometric identity known as the Pythagorean identity states:
cos^{2}(x) + sin^{2}(x) = 1
We therefore have:
d  (tan(x)) =  1 
dx  cos^{2}(x) 
Finally, we will use one of the reciprocal trigonometric identities,
sec(x) =  1 
cos(x) 
to obtain the result:
d  (tan(x)) =  1  = sec^{2}(x) 
dx  cos^{2}(x) 
The cosecant function
The cosecant function is one of the socalled reciprocal functions. It is in fact the reciprocal of the sine function, and is defined as:
csc(x) =  1 
sin(x) 
You may remember that we can obtain the derivative of the reciprocal of any function u using the following formula:
d  (  1  )  = 


dx  u  u^{2} 
Applying this to the definition of the cosecant function, we get:
d  (csc(x)) =  cos(x) 
dx  sin^{2}(x) 
We can rearrange this result as follows:
d  (csc(x))  =   1  ·  cos(x) 
dx  sin(x)  sin(x) 
This result now contains the definition for the cosecant function in terms of the sine function, which we saw above. It also contains the definition for the cotangent function in terms of the sine and cosine functions:
cot(x)  =  cos(x) 
sin(x) 
So finally, we have:
d  (csc(x))  = csc(x) · cot(x) 
dx 
The secant function
The secant function is another reciprocal function. It is the reciprocal of the cosine function, and is defined as:
sec(x) =  1 
cos(x) 
Just as we did with the cosecant function, we can use the formula for obtaining the derivative of the reciprocal of a function to obtain the derivative of the secant function:
d  (sec(x)) =  sin(x) 
dx  cos^{2}(x) 
We can rearrange this result as follows:
d  (sec(x))  =  1  ·  sin(x) 
dx  cos(x)  cos(x) 
The result now contains the definition for the secant function in terms of the cosine function, which we saw above. It also contains the definition for the tangent function in terms of the sine and cosine functions, which we also saw above. We therefore have:
d  (sec(x))  = sec(x) · tan(x) 
dx 
The cotangent function
The cotangent function is the reciprocal of the tangent function, and is defined as:
cot(x) =  1 
tan(x) 
The tangent function itself can be defined as the quotient of the sine function and the cosine function. We have already seen (see above) that we can define the cotangent function as:
cot(x)  =  cos(x) 
sin(x) 
We can therefore find the derivative of the cotangent function using the same method we used to find the derivative of the tangent function, namely by using the derivatives of the sine and cosine functions together with the quotient rule. Applying this rule here, we get:
d  (cot(x)) =  d  (  cos(x)  )  =  sin^{2}(x)  cos^{2}(x) 
dx  dx  sin(x)  sin^{2}(x) 
We saw previously that the trigonometric identity known as the Pythagorean identity states:
cos^{2}(x) + sin^{2}(x) = 1
We therefore have:
d  (cot(x)) =   1 
dx  sin^{2}(x) 
Finally, we will again use one of the reciprocal trigonometric identities,
csc(x) =  1 
sin(x) 
to obtain the result:
d  (cot(x)) =   1  = csc^{2}(x) 
dx  sin^{2}(x) 