Math, asked by revathi726, 1 year ago

if A+B=60 then cos^2A+cos^B-cosAcosB= ?

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Answered by Divyasamota
14
If A + B = 60° then how do I find the value of cos^2A + cos^2B - cosA cosB?

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3 ANSWERS



Bitan Sarkar, Btech Electronics and Communication Engineering, Indian Institute of Technology, Bhubaneswar (2021)

Answered Jun 1, 2017

If I need to find the value very quickly, then I would prefer putting [math]A = 60° and B = 0°[/math]
[math]cos^2A + cos^2B - cosAcosB = 1 + 1/4 -1/2 = 3/4[/math]
Now take [math]A = 30° and B = 30°[/math]
[math]cos^2A + cos^2B - cosAcosB = 3/4 + 3/4 - 3/4[/math]
Thus I can roughly say, yes [math]cos^2A + cos^2B - cosAcosB = 3/4[/math], for [math]A + B = 60°[/math]

Now, if you are looking for a proper method,
then also its easy,
[math]A + B = 60°[/math] 
[math]=> cos(A + B) = 1/2[/math] 
[math]=> cosA.cosB - sinA.sinB = 1/2[/math]
[math]=> 2cosA.cosB - 1 = 2sinA.sinB[/math]
squaring both sides,
[math]=> 4cos^2A.cos^2B -4cosA.cosB + 1 = 4sin^2A.sin^2B[/math]
[math]=> 4cos^2A.cos^2B -4cosA.cosB + 1 = 4(1 - cos^2A)(1 - cos^2B)[/math]
[math]=> 4cos^2A.cos^2B -4cosA.cosB + 1 = 4 - 4cos^2A - 4cos^2B + 4cos^2A.cos^2B[/math]
[math]=> -4cosA.cosB = 3 - 4cos^2A - 4cos^2B[/math]
Thus
[math]4cos^2A + 4cos^2B -4cosA.cosB = 3[/math]or [math]cos^2A + cos^2B -cosA.cosB = 3/4[/math]

Answered by anurag432
2

Answer:

cos²A + cos²B - cosA cosB = 3/4

Step-by-step explanation:

Given A+B=60

A = 60 - B

Apply cos on both sides

cos(A) = cos(60 - B)

cos(A - B) = cosAcosB + sinAsinB

cosA = cos60cosB + sin60sinB

cos60 = 1/2, sin60 =√3/2

cosA = 1/2 cosB  + √3/2 sinB

2cosA = cosB + √3 sinB

2cosA - cosB = √3 sinB

Squaring on both the sides.

(2cosA - cosB)² = (√3 sinB)²

4cos²A + cos²B - 4cosA cosB = 3sin²B

here we can write sin²B = 1 - cos²B

4cos²A + cos²B - 4cosA cosB = 3(1 - cos²B)

4cos²A + cos²B - 4cosA cosB = 3 - 3cos²B

4cos²A + cos²B + 3cos²B - 4cosA cosB = 3

4cos²A + 4cos²B - 4cosA cosB = 3

cos²A + cos²B - cosA cosB = 3/4

therefore the value of cos²A + cos²B - cosA cosB = 3/4.

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