Question

Prove that cos (A+B\2) = sin C, if A + B +C = 180.​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Answer:

Correct Question:-

If A + B + C = 180°

$\bf To \: Prove :- \:cos(\dfrac{A + B}{2} ) = sin\dfrac{C}{2}$

$\bf\underbrace\orange{Answer:}$

Identity used :-

• cos(90° - x) = sinx

$\bf\underbrace\orange{Solution:}$

$\bf \:A + B + C = 180°$

$\bf\implies \:A + B = 180° - C$

$\bf \:Consider \: cos(\dfrac{A + B}{2} )$

Put the value of A + B, we get

$\bf\implies \:cos(\dfrac{180 - C}{2} )$

$\bf\implies \:cos(\dfrac{180}{2} - \dfrac{C}{2} )$

$\bf\implies \:cos(90 - \dfrac{C}{2} )$

$\bf\implies \:sin\dfrac{C}{2}$

$\bf \:Hence \: proved$

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Additional Information:-

Relation between sides & T - Ratios

• sin θ = Opposite Side/Hypotenuse
• cos θ = Adjacent Side/Hypotenuse
• tan θ = Opposite Side/Adjacent Side
• sec θ = Hypotenuse/Adjacent Side
• cosec θ = Hypotenuse/Opposite Side
• cot θ = Adjacent Side/Opposite Side

◇ Reciprocal Identities

• cosec θ = 1/sin θ
• sec θ = 1/cos θ
• cot θ = 1/tan θ
• sin θ = 1/cosec θ
• cos θ = 1/sec θ
• tan θ = 1/cot θ

◇ Co-function Identities

• sin (90°−x) = cos x
• cos (90°−x) = sin x
• tan (90°−x) = cot x
• cot (90°−x) = tan x
• sec (90°−x) = cosec x
• cosec (90°−x) = sec x

◇ Fundamental Trigonometric Identities

• sin²θ + cos²θ = 1
• sec²θ - tan²θ = 1
• cosec²θ - cot²θ = 1