Question

In a triangle ABC write cos(B+c/2)in terms of A

Answer 1

Answer:

SinA/2

Step-by-step explanation:

Question is incomplete it is as follows

In a ΔABC find Value of Cos(B+C/2)

To find--->Value of cos(B+C/2)

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Solution---->We know that sum of angles

------------- of triangle is 180⁰

So , A+B+C=180⁰

=> B+C=180⁰-A

Now Cos(B+C/2)=Cos(180⁰-A/2)

=Cos(180⁰/2 - A/2)

= Cos(90⁰- A/2)

We know that Cos(90⁰-θ)=sinθ

=sin(A/2)

Additional information--->

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1)Sin(90⁰-θ)=Cosθ

2)tan(90⁰-θ)=Cotθ

3)Cot(90⁰-θ)=tanθ

4)Sec(90⁰-θ)=Cosecθ

5)Cosec(90⁰-θ)=Secθ

6)Sin²θ+Cos²θ=1

7)1+tan²θ=sec²θ

8)1+Cot²θ=cosec²θ

Answer 2

Step-by-step explanation:

Angle sum property of a triangle

angle A+ Angle B + angle B = 180 degree

• = Angle B + angle C = 180 - angle A
• B+C/2 = 180- A/2
• B+C/2 = 90- A/2
• COS (b+c/2) - cos (90°- A/2)
• Cos (B+C/2) = Sin A/2