Question

If A, B and C are interior angles of a ΔABC then cos $\sf{(\frac{B+C}{2})}$
is equal to ?

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Answer 1

Answer:

cos (∠B + ∠C)/2 = sin ∠A/2

Step-by-step explanation:

Given that A, B and C are interior angles of ∆ABC. We nedd to find out the value of cos (B + C)/2.

We know that sum of all angles of triangle is 180°.

→ ∠A + ∠B + ∠C = 180°

→ ∠B + ∠C = 180° - ∠A --------------- (eq 1)

As per given question, we need to find the value of cos (B + C)/2. We have angle B + angle, just convert it in the form of given question. To do this, multiply the both sides by cos and divide by 2.

On multiplying with cos and dividing by 2 we get,

→ cos (∠B + ∠C)/2 = cos (180° - ∠A)

→ cos (∠B + ∠C)/2 = cos (90° - ∠A/2)

As, cos (90° - Ø) = sinØ so,

→ cos (∠B + ∠C)/2 = sin ∠A/2

Therefore, if A, B and C are interior angles of a ΔABC then cos (∠B + ∠C)/2 = sin ∠A/2.

TRIGONOMETRY FORMULAS:

secØ = 1/cosØ, cosØ = 1/secØ

cosecØ = 1/sinØ, sinØ = 1/cosecØ

tanØ = sinØ/cosØ, cotØ = cosØ/sinØ

cotØ = 1/tanØ, tanØ = 1/cotØ

sin²Ø + cos²Ø = 1

tan²Ø + 1 = sec²Ø

1 + cot²Ø = cosec²Ø

sinØ = cos(90° - Ø)

cosØ = sin(90° - Ø)

tanØ = cot(90° - Ø)

Answer 2

Answer:

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