In mathematics, we often come across with functions. These are utilized in each and every branch of mathematics. A mathematical function is said to be a special kind of relation between inputs and outputs, where every input value is connected with exactly one output value by the means of a particular property. We deal with several different kind of functions in mathematics. The trigonometric functions are one of the most important ones.

We know that there are six basic trigonometric functions. They are functions of angles which relate angles of right triangles with its sides. In trigonometry, we also learn about the concept of cofunctions which is applied to the complementary angles (a pair of angles whose sum is right angle). The trigonometry is typically based on the right-angled triangles. In a right triangle, one angle is 90$^{\circ}$ and other two are eventually acute angles. Here, the sum of both acute angles is 90$^{\circ}$. **In the diagram below, a right-angled triangle is shown:**

The two acute angles are $\phi$ and $\psi$, then we have:

$\phi + \psi = 90^{\circ}$ which gives

$\phi = 90^{\circ} - \psi$ and $\psi = 90^{\circ} - \phi$

Such angles play a very important role in trigonometry.

The trigonometric functions of these angles have cofunctions. Let us go ahead in this article and learn about trigonometric cofunctions in detail, their representations and problems based on those.

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In mathematics, especially in trigonometry, a function f is defined as a cofunction if and only if

f(X) = g(Y) provided that X and Y are complementary angles. Recall that complementary angles are a pair of angles whose sum is equal to a right angle. The cofunctions are generally applied to the trigonometric functions.

We can say that two trigonometric function are known as cofunctions of each other if value of an angle for one function is equal to the value of the complement of that angle for other function.

**For Instance:**** **The sine and cosine are said to cofunctions of each other. Hence, the sine of a given acute angle is equivalent to the cosine of the complement of this angle. Also, the cosine of given acute angle is same as sine of the complement.

The trigonometric identities which express the relationships between sine and cosine, secant and cosecant, tangent and cotangent are known as cofunction identities. They represent that the value of one trig function of some acute angle is equal to the value of the cofunction of complement of that angle. f(X) = g(Y) provided that X and Y are complementary angles. Recall that complementary angles are a pair of angles whose sum is equal to a right angle. The cofunctions are generally applied to the trigonometric functions.

We can say that two trigonometric function are known as cofunctions of each other if value of an angle for one function is equal to the value of the complement of that angle for other function.

Let $\alpha$ be one of the acute angles in a right-angled triangle, then the cofunciton identities are given below:

$sin (90^{\circ} - \alpha) = cos \alpha$

and $cos (90^{\circ} - \alpha) = sin \alpha$

$sin (\frac{\pi}{2} - \alpha) = cos \alpha$

and $cos (\frac{\pi}{2} - \alpha) = sin \alpha$

$sec (90^{\circ} - \alpha) = cosec \alpha$

and $cosec (90^{\circ} - \alpha) = sec \alpha$

$sec (\frac{\pi}{2} - \alpha) = cosec \alpha$

and $cosec (\frac{\pi}{2} - \alpha) = sec \alpha$

$tan (90^{\circ} - \alpha) = cot \alpha$

and $cot (90^{\circ} - \alpha) = tan \alpha$

In Radians

$tan (\frac{\pi}{2} - \alpha) = cot \alpha$

and $cot (\frac{\pi}{2} - \alpha) = tan \alpha$

If we draw graph of y = sin $\alpha$ and shift it by 90$^{\circ}$ to the left, then it is exactly similar to the graph of y = cos $\alpha$. This is true for secant and cosecant as well as tangent and cotangent. According to the cofunction identities, the one function reflects same value as when it is shifted to the left by a right angle.

cos (2A + 16$^{\circ}$) = sin (A + 11$^{\circ}$)

cos (2A + 16$^{\circ}$) = sin (A + 11$^{\circ}$)

cos (2A + 16$^{\circ}$) = cos (90$^{\circ}$ - (A + 11$^{\circ}$)

2A + 16$^{\circ}$ = 90$^{\circ}$ - (A + 11$^{\circ}$)

2A + 16$^{\circ}$ = 90$^{\circ}$ - A - 11$^{\circ}$

3A = 90$^{\circ}$ - 11$^{\circ}$ - 16$^{\circ}$

3A = 63$^{\circ}$

A = 21$^{\circ}$

$tan$ ($\frac{\pi}{2}$ - $\theta$) = $tan$($\theta$ + $\frac{\pi}{6}$)

$\frac{\pi}{2}$ - $\theta$ = $\theta$ + $\frac{\pi}{6}$

$\frac{\pi}{2}$ - $\frac{\pi}{6}$ = 2$\theta$

$2\theta$ = $\frac{\pi}{2}$ - $\frac{\pi}{6}$

$2\theta$ = $\frac{\pi}{3}$

$\theta$ = $\frac{\pi}{6}$

**Example 3:** Prove that $sin^{2} 18^{\circ} + sin^{2} 40^{\circ} + Sin^{2} 50^{\circ} + sin^{2} 72 ^{\circ}$ = 2

= $sin^{2} 18^{\circ} + sin^{2} 40^{\circ} + sin^{2} 50^{\circ} + sin^{2} 72 ^{\circ}$

= $sin^{2} 18^{\circ} + sin^{2} 72^{\circ} + sin^{2} 40^{\circ} + sin^{2} 50^{\circ}$

= $sin^{2} 18^{\circ} + sin^{2} (90^{\circ} - 18 ^{\circ}) + sin^{2} 40^{\circ} + sin^{2} (90^{\circ} - 40 ^{\circ})$

Using the identity $sin (90^{\circ} - \alpha) = cos \alpha$

= $sin^{2} 18^{\circ} + cos^{2}18 ^{\circ} + sin^{2} 40^{\circ} + cos^{2} 40 ^{\circ}$

We have Pythagorean identity $sin^{2} \alpha + cos^{2} \alpha$ = 1

= 1 + 1 = 2 = R.H.S

Hence proved.

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