# Prove that Cot4A (Sin5A + Sin3A) = CotA (Sin5A – Sin3A)

## Simplifying Left Hand Side Cot4A (Sin5A + Sin3A)

Cot4A (Sin5A + Sin3A)

Using formula
SinC + SinD = 2 Sin(C + D)/2 Cos(C – D)/2

Replacing
C = 5A
D = 3A

(C + D)/2 = (5A + 3A)/2 = 8A/2 = 4A

(C – D)/2 = (5A – 3A)/2 = 2A/2 = A

Therefore SinC + SinD = 2 Sin(C + D)/2 Cos(C – D)/2 becomes

Sin5A + Sin3A = 2 Sin4A CosA

Cot4A (Sin5A + Sin3A) = Cot4A (2 Sin4A CosA)

$$Cot4A (Sin5A + Sin3A) = Cot4A (2 Sin4A \text{ } CosA) \\ \text{ } \\ Cot4A (Sin5A + Sin3A) = \frac{Cos4A}{Sin4A} (2 Sin4A \text{ } CosA) \\ \text{ } \\ Cot4A (Sin5A + Sin3A) = 2 Cos4A \text{ } CosA$$

Therefore Cot4A (Sin5A + Sin3A) = 2 Cos4A CosA

## Simplifying CotA (Sin5A – Sin3A)

CotA (Sin5A – Sin3A)

Using formula
SinC – SinD = 2 Cos(C + D)/2 Sin(C – D)/2

Replacing
C = 5A
D = 3A

(C + D)/2 = (5A + 3A)/2 = 8A/2 = 4A

(C – D)/2 = (5A – 3A)/2 = 2A/2 = A

Therefore SinC – SinD = 2 Cos(C + D)/2 Sin(C – D)/2 can be written as

Sin5A – Sin3A = 2 Cos4A SinA

Therefore CotA (Sin5A – Sin3A) = CotA (2 Cos4A SinA)

$$CotA (Sin5A – Sin3A) = CotA \text{ } (2 \text{ } Cos4A \text{ } SinA) \\ \text{ } \\ CotA (Sin5A – Sin3A) = \frac{CosA}{SinA} \text{ } (2 \text{ } Cos4A \text{ } SinA) \\ \text{ } \\ CotA (Sin5A – Sin3A) = 2Cos4A \text{ } CosA$$

Therefore CotA (Sin5A – Sin3A) = 2 Cos4A CosA

## Putting together

• Cot4A (Sin5A + Sin3A) = 2 Cos4A CosA
• CotA (Sin5A – Sin3A) = 2 Cos4A CosA

Therefore Cot4A (Sin5A + Sin3A) = CotA (Sin5A – Sin3A) = 2 Cos4A CosA

Hence it’s proved that Cot4A (Sin5A + Sin3A) = CotA (Sin5A – Sin3A)