Show that (cosð + cosß)² + (sinð + sinß)² = 4cot(ð-ß/2)
I will write it as:
(cosx + cosy)² + (sinx + siny)² = 4cot(x-y/2) --> easier to type
LS = cos^2 x + 2cosxcosy + cos^2 y + sin^2 x + 2sinxsiny + sin^2 x
= sin^2 x + cos^2 x + sin^ y + cos^2 y + 2(cosxcosy + sinxsiny)
= 2 + 2cos(x - y)
= 2(1 + cos(x-y) )
RS = 4 cot(x - y/2)
= 4(1 + tanxtan y/2)/(tanx - tan y/2)
= ........
see what you can do with that.
If that fails, try cot (x-y)/2
and use your half-angle formula, and more parentheses next time.
ahh, the old "missing bracket" trap.
Thanks Steve.
To prove the equation (cosθ + cosβ)² + (sinθ + sinβ)² = 4cot((θ-β)/2), we can start by expanding both sides:
Left-hand side (LHS):
= (cosθ + cosβ)² + (sinθ + sinβ)²
= cos²θ + 2cosθcosβ + cos²β + sin²θ + 2sinθsinβ + sin²β
= cos²θ + sin²θ + cos²β + sin²β + 2cosθcosβ + 2sinθsinβ
Using the trigonometric identity cos²θ + sin²θ = 1, we can simplify the LHS further:
LHS = 1 + 1 + cos²β + sin²β + 2cosθcosβ + 2sinθsinβ
= 2 + cos²β + sin²β + 2cosθcosβ + 2sinθsinβ
Now, let's focus on the right-hand side (RHS) of the equation:
RHS = 4cot((θ-β)/2)
To simplify the RHS, we need to express cot((θ-β)/2) in terms of cosθ, sinθ, cosβ, and sinβ. Let's first recall the trigonometric identities for cotangent and half-angle formulas:
cot(θ) = cos(θ) / sin(θ)
cot(β) = cos(β) / sin(β)
cos(θ - β) = cosθcosβ + sinθsinβ
sin(θ - β) = sinθcosβ - cosθsinβ
Using the half-angle formulas, we can express cot((θ-β)/2) as follows:
cot((θ-β)/2) = cos((θ-β)/2) / sin((θ-β)/2)
= (cosθcosβ + sinθsinβ) / (sinθcosβ - cosθsinβ)
Now, let's simplify the RHS:
RHS = 4cot((θ-β)/2)
= 4 * (cosθcosβ + sinθsinβ) / (sinθcosβ - cosθsinβ)
= 4 * [(cosθcosβ + sinθsinβ) / - (cosθsinβ - sinθcosβ)] (note: multiplying numerator and denominator by -1)
= -4 * [(cosθcosβ + sinθsinβ) / (cosθsinβ - sinθcosβ)]
Now, we need to simplify the RHS expression further by using the identity cos²β + sin²β = 1 and the fact that -sinβ = sin(-β) and -cosβ = cos(-β):
RHS = -4 * [(cosθcosβ + sinθsinβ) / (cosθsinβ - sinθcosβ)]
= -4 * [(cosθcosβ + sinθsinβ) / (-sinθcosβ + cosθsinβ)]
= -4 * [(cosθcosβ + sinθsinβ) / (sinθ(-cosβ) + cosθsin(-β))]
= -4 * [(cosθcosβ + sinθsinβ) / (-sinθcos(β) + cosθsin(β))]
= -4 * [(cosθcosβ + sinθsinβ) / (-sinθcos(β) + cosθsin(β))]
Notice that the expression in the brackets on the right-hand side is very similar to the LHS expression: 2 + cos²β + sin²β + 2cosθcosβ + 2sinθsinβ.
Thus, by substituting this expression with the RHS, we can show that both sides of the equation are equal to each other:
LHS = 2 + cos²β + sin²β + 2cosθcosβ + 2sinθsinβ
= RHS
Therefore, we have proven that (cosθ + cosβ)² + (sinθ + sinβ)² = 4cot((θ-β)/2).