Conditional Identities

What are conditional identities?

In mathematics, trigonometric identities can be considered as equalities that can involve any function of trigonometry. These identities are true for every value of the variable given in the equation. We may find them useful in situations where we get trigonometric functions present in any equation or group of expressions. Identities that can involve a function for one or more than one angle can be considered as trigonometric identities.

When we talk about the trigonometric conditional identities, trigonometric identities that can satisfy the given additional condition can be considered conditional trigonometric identities. In geometry, these are the identities in which certain relationships and conditions exist between the angles involved.

Formulas of conditional identities

Let’s say A, B, and C are three angles of a triangle then they need to follow a basic condition that states A + B + C = 1800  or π. This condition or relation helps us in making many of the important identities. Below are some of the basic examples of conditional trigonometric identity based on the above condition.

From the above-given condition, we can say there will always be a complementary angle for the sum of the other two angles. For example A + B =  1800  – C.

Now using this condition we can make following identities:

• sin(A + B) = sin(1800   – C) = sinC
• cos(A + B) = cos(1800  – C) = -cosC
• tan(A + B) = tan(1800 – C) = -tanC

Also this is very simple to understand that three angles that are following A + B + C =1800 will also follow A/2 + B/2 + C/2 = 900.

So.

• sin(A/2 + B/2) = sin( 900. – C/2) = cosC/2
• cos(A/2 + B/2) = cos(900 – C/2) = sinC/2
• tan(A/2 + B/2) = tan(900 – C/2) = cotC/2

These expressions will be similar to the angles of any kind of triangle. That is why we were considering conditional trigonometric identities based on the additional condition. Whatever the value of the variable these identities need to follow a certain given condition.

Let’s understand conditional trigonometric identities using some of the solutions to problems.

Examples

Let’s start with considering  that there is a triangle where the angles are A, B, and C. we can say that  A + B + C =  π  and there is an expression

sin2A + sin2B + sin2C.

We can write this expression as following:

sin2A + sin2B + sin2C = 2sin[(2A + 2B)/2]cos[(2A – 2B)/2] + sin2C                             ___(1)

As we know,

sinA + sinB =2[{sin(a+b)/2}{cos(A-B)/2}], and sin2C = 2sinCcosC

We also know that A + B + C = π, so we can interchange (A + B) with (π – C)

From equation (1),

2sin[(2A + 2B)/2]cos[(2A – 2B)/2] + sin2C = 2sin(1800 – C)cos(A-B) + 2sinCcosC

= 2sinCcos(A – B) + 2sinCcosC

= 2sinC[ cos(A – B) + cosC ]

= 2sinC[ cos(A – B) + cos(A + B)]    ___(2)

As we know,

cosA – cosB  =  -2sin(A + B)/2sin(A – B)/2

From equation (2),

2sin[(2A + 2B)/2]cos[(2A – 2B)/2] + sin2C = 2sinC[ sin{( A – B + A + B )/2}sin{(A – B – A – B)/2}]

= 2sinC[-2sin(2A/2)sin(-2B/2) ]

Since sin(-A) = -sinA

2sinC[ -2sin(2A/2)sin(-2B/2) ] = 4sinAsinBsinC

Here we can see that,

sin2A + sin2B + sin2C = 4sinAsinBsinC

This can be considered as an example of conditional trigonometry identity, because these expressions can be satisfied to all the values of variables A, B, and C with a condition which states that A + B + C = π.

There can be many more examples of this type. Let’s take a look at some of them:

•  Lets say A + B + C = π then expression sin2A + sin2B − sin2C can be converted in following way:

sin2A + sin2B − sin2C = 2sin[(2A + 2B)/2]cos[(2A – 2B)/2] – sin2C

= 2sin(A + B)cos(A – B) – 2SinCcosC

= 2sinC[ cos(A – B) – cos(180-(A + B)]

= 2sinC[ cos(A – B) + cos(A + B)]

= 2sinC[2cosAcosB]

= 4sinCcosAcosB

• Lets say this time we have the expression cos2A + cos2B + cos2C with the similar condition as in above. We can convert it them in the following way:

cos2A + cos2B + cos2C = 2cos{(2A + 2B)/2}cos{(2A – 2B)/2} + cos2C

= 2cos(A + B)cos(A – B) + cos2C

As we know cos2A = 1 – 2cos²A,

2cos(A + B)cos(A – B) + cos2C =  2cos(A + B)cos(A – B) + 1 – 2cos2C

= 2cosC(cos(A + B) – cos(A – B)) – 1

= -1 – 4cosAcosBcosC

• Let’s say this time we have an expression, sinA + sinB + sinC with the same condition as in above. Than we can transform them using the following way:

sinA + sinB + sinC = 2sin{(A + B)/2}cos{(A-B)/2} + sinC

Let’s say sinC = sin(2C/2), than,

2sin{(A + B)/2}cos{(A-B)/2} + sinC = 2cos(C/2)cos{(A-B)/2} + 2sin(C/2)cos(C/2)

=  2cos(C/2)[cos{(A-B)/2} + sin(C/2)]

= 2cos(C/2)[cos{(A-B)/2} + sin{ π/2 – (A+B)/2 }]

= 2cos(C/2)[cos{(A-B)/2} + cos{(A+B)/2 }]

Because, 2cosAcosB = cos(A + B) + cos(A – B),

2cos(C/2)[cos{(A-B)/2} + cos{(A+B)/2 }] = 4cosA/2cosB/2cosC/2

So we can say that.

sinA + sinB + sinC = 4cosA/2cosB/2cosC/2

• Let’s say for example we have the expression sin²A + sin²B – sin²C, with the condition A + B + C =  π. We can transform the expression in the following way:

• sin²A+sin²B-sin²C=sin²A+sin(B + C)sin(B – C)

We know that sin²A-sin²B

= sin(B + C)sin(B – C). So,

= sin²A + sin(π – A)sin(B -C)

= sin²A + sinA sin(B -C)

= sinA(sin(B+C) + sin(B-C))

We know that sin(A+B) + sin(A-B) = 2sinAsinB,

sinA(sin(B+C) + sin(B-C)) = 2sinAsinBsinC

The above expression can also be written as,

sin²A + sin²B-sin²C= 2sinAsinBsinC.

Here, we can see how these identities work using such conditions.

Conclusion

In this article, we have discussed what are the conditional trigonometric identities. We can say that we use the trigonometric conditional identities  to satisfy the given additional condition. Along with this we also discussed conditions that can be there if  the angles are given from a triangle. We have seen some of the examples to make our concepts about conditional identities clear. Conditional identities are used for some specific conditions only while the global identities can be use for all the values of angle. There is a limited range of values of angle for which conditional identities are applicable.