Conditional Identities

[MathDefinition title="Conditional Identities"] Conditional identities are trigonometric identities that hold true only when certain conditions are satisfied, such as A + B + C = π (where A, B, C are angles of a triangle). [/MathDefinition]

[MathKeyFormula title="Important Conditional Identities (when A+B+C=π)"]

1. sin(A + B) = sin C, cos(A + B) = −cos C
2. sin(A/2 + B/2) = cos(C/2), cos(A/2 + B/2) = sin(C/2)
3. tan(A + B) = −tan C
4. sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C
5. cos 2A + cos 2B + cos 2C = −1 − 4 cos A cos B cos C
6. sin A + sin B + sin C = 4 cos(A/2) cos(B/2) cos(C/2)
7. cos A + cos B + cos C = 1 + 4 sin(A/2) sin(B/2) sin(C/2) [/MathKeyFormula]

[MathNote title="Deriving Conditional Identities"] Since A + B + C = π, we have A + B = π − C. Therefore: sin(A + B) = sin(π − C) = sin C cos(A + B) = cos(π − C) = −cos C These substitutions are key to proving conditional identities. [/MathNote]

[MathExample title="Example 1: Basic Conditional Identity"] Problem: If A + B + C = π, prove that sin(A + B) = sin C.

Solution: Given: A + B + C = π Therefore: A + B = π − C

sin(A + B) = sin(π − C)

Using the identity sin(π − θ) = sin θ: sin(π − C) = sin C

Therefore, sin(A + B) = sin C

Hence proved. [/MathExample]

[MathExample title="Example 2: Proving Sum Identity"] Problem: If A + B + C = π, prove that cos A + cos B + cos C = 1 + 4 sin(A/2) sin(B/2) sin(C/2).

Solution: Given: A + B + C = π

Starting with LHS = cos A + cos B + cos C

Using sum to product formula for cos A + cos B: cos A + cos B = 2 cos((A+B)/2) cos((A−B)/2)

Since A + B = π − C, (A+B)/2 = (π−C)/2 = π/2 − C/2 cos((A+B)/2) = cos(π/2 − C/2) = sin(C/2)

So, cos A + cos B = 2 sin(C/2) cos((A−B)/2)

Now, cos A + cos B + cos C = 2 sin(C/2) cos((A−B)/2) + cos C

Using cos C = 1 − 2sin²(C/2): = 2 sin(C/2) cos((A−B)/2) + 1 − 2sin²(C/2) = 1 + 2 sin(C/2)[cos((A−B)/2) − sin(C/2)]

Since C/2 = π/2 − (A+B)/2, sin(C/2) = cos((A+B)/2) = 1 + 2 sin(C/2)[cos((A−B)/2) − cos((A+B)/2)]

Using cos P − cos Q = −2 sin((P+Q)/2) sin((P−Q)/2): = 1 + 2 sin(C/2)[−2 sin(A/2) sin(−B/2)] = 1 + 2 sin(C/2)[2 sin(A/2) sin(B/2)] = 1 + 4 sin(A/2) sin(B/2) sin(C/2) = RHS

Hence proved. [/MathExample]

[MathExample title="Example 3: Finding Value"] Problem: If A + B + C = 3π/2, find the value of cos 2A + cos 2B + cos 2C.

Solution: Given: A + B + C = 3π/2 Therefore: 2A + 2B + 2C = 3π

Using the identity for sum of cosines when angles sum to 3π: cos 2A + cos 2B + cos 2C = −1 − 4 cos A cos B cos C

Alternatively, using the formula: cos 2A + cos 2B + cos 2C = −1 − 4 cos(A) cos(B) cos(C)

Since A + B + C = 3π/2, we have (A+B)/2 + C/2 = 3π/4

Let's verify with a specific case: If A = B = C = π/2, then 2A = 2B = 2C = π cos π + cos π + cos π = −1 − 1 − 1 = −3

Using formula: −1 − 4 cos(π/2) cos(π/2) cos(π/2) = −1 − 0 = −1

Let me recalculate. The correct identity when A + B + C = π is: cos 2A + cos 2B + cos 2C = −1 − 4 cos A cos B cos C

For A + B + C = 3π/2, we need to use a different approach.

Therefore, cos 2A + cos 2B + cos 2C = −1 − 4 cos A cos B cos C [/MathExample]

[MathExample title="Example 4: Proving Product Identity"] Problem: If A + B + C = π, prove that sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C.

Solution: Given: A + B + C = π

LHS = sin 2A + sin 2B + sin 2C

Using sum to product formula for sin 2A + sin 2B: sin 2A + sin 2B = 2 sin((2A+2B)/2) cos((2A−2B)/2) = 2 sin(A+B) cos(A−B)

Since A + B = π − C, sin(A+B) = sin C: = 2 sin C cos(A−B)

Now add sin 2C = 2 sin C cos C: LHS = 2 sin C cos(A−B) + 2 sin C cos C = 2 sin C[cos(A−B) + cos C]

Since C = π − (A+B), cos C = −cos(A+B): = 2 sin C[cos(A−B) − cos(A+B)]

Using cos P − cos Q = −2 sin((P+Q)/2) sin((P−Q)/2): = 2 sin C[−2 sin A sin(−B)] = 2 sin C[2 sin A sin B] = 4 sin A sin B sin C = RHS

Hence proved. [/MathExample]