If A+B+C = 180 degrees then what is the value of: cos squareA + cos square
B + cos square C
Answers
Answer:
Step-by-step explanation:
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If cos square A + cos square b + cos square C + 2 cos a × cos b × Cosc= 1 if a + b + C = 180 degree
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Answers : (1)
Solving RHS
2cosAcosBcosC =(2cosAcosB)cosC
=(cos(A+B) + cos(A-B))cosC
=cos(pi-C)cosC + cos(A-B)cos(pi-(A+B))
=-2cos^2C - 2cos(A+B)cos(A-B)
=-cos^2C - cos^2A + sin^2B
so,
1 - 2cosAcosBcosC = 1 + cos^2C + cos^2A - sin^2B
= 1 - sin^2B + cos^2C + cos^2A
= cos^2B + cos^2C + cos^2A
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Step-by-step explanation:
We know that :cosA+cosB=2cos(
2
A+B
)cos(
2
A−B
)
thereforecos2A+cos2B=2cos(
2
2A+2B
)cos(
2
2A−2B
)
=2cos(A+B)cos(A−B)
cos2A+cos2B−cos2c=2cos(A+B)cos(A−B)−cos2C
cos2A+cos2B−cos2C=2cos(π−c)cos(A−B)−cos2C
(∵A+B+C=π so A+B=π−C)
$$\cos 2 A+\cos 2 B-\cos 2 C &=2 \cos (\pi-c) \cos (A-B)-\left(2 \cos ^{2} C-1\right) $$
=−2cosCcos(A−B)−2cos
2
C+1
=−2cosC{cos(A−B)+cosC}+1
=−2cosC{2cos(
2
A−B+C
)cos(
2
A−B−C
)}+1
=−2cosC{2cos(
2
π−2B
)cos(
2
2A−π
)}+1
=1−2cosC{2sinBsinA}
=1−4sinAsinBcosC
=1−4sinA⋅sinB⋅cosC