Question

in ∆abc c = 2π/3, then prove that cos^2A + cos^2B - cosAcosB = 3/4​

Answer 1

Answer:

If A + B = 60° then how do I find the value of cos^2A + cos^2B - cosA cosB? 3 Answers. Bitan Sarkar, Btech Electronics and Communication 

Step-by-step explanation:

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Answer 2

Answer:

Step-by-step explanation:

c=2π/3=2×180/3=120

In triangle a+b+c=180

a+b=180-c=180-120=60

a+b=60--> 1

LHS

cos²A + cos²B - cosAcosB

cos²b+sin²b=1

cos²b=1-sin²b

cos²A + 1-sin²b - cosAcosB

cos²A -sin²b =cos(a+b)×cos(a-b)

cos(a+b)×cos(a-b)+1 - cosAcosB -->2

1 in 2

cos(60)×cos(a-b)+1 - cosAcosB

$\frac{1}{2\\}$(cos a-b)+1 - cosAcosB -->3

(cos a-b)=cos a cos b + sin a sin b-->4

4 in 3

$\frac{cos a cos b + sin a sin b}{2}$+1 - cosAcosB

$\frac{cos a cos b + sin a sin b-2cos a cos b}{2}$+1

$\frac{sinasinb-cosacosb}{2\\}$+1

-(cos a cos b - sin a sin b)/2+1---> 5

cos(a+b)= cos a cos b- sin a sin b--> 6

6 in 5

-(cos(a+b))/2+1--> 7

1 in 7

-(1/2)/2+1

1-1/4

4-1/4=3/4