Math, asked by swarisgshuja, 1 year ago

Cos(A+B)COSB+sin(A+B)sinB=cosA

Answers

Answered by tony47
3
Verify the following identity:
cos(A+B) cos(B)+sin(A+B) sin(B) = cos(A)
cos(B) cos(A+B) = 1/2 (cos(B-(A+B))+cos(B+(A+B))) = 1/2 (cos(-A)+cos(A+2 B)):
(cos(-A)+cos(A+2 B))/(2)+sin(B) sin(A+B) = ^?cos(A)
Use the identity cos(-A) = cos(A):
(cos(A)+cos(A+2 B))/2+sin(B) sin(A+B) = ^?cos(A)
(cos(A)+cos(A+2 B))/2 = (cos(A))/2+1/2 cos(A+2 B):
(cos(A))/(2)+(cos(A+2 B))/(2)+sin(B) sin(A+B) = ^?cos(A)
sin(B) sin(A+B) = 1/2 (cos(B-(A+B))-cos(B+(A+B))) = 1/2 (cos(-A)-cos(A+2 B)):
(cos(A))/2+(cos(A+2 B))/2+(cos(-A)-cos(A+2 B))/(2) = ^?cos(A)
Use the identity cos(-A) = cos(A):
(cos(A))/2+(cos(A+2 B))/2+(cos(A)-cos(A+2 B))/2 = ^?cos(A)
(cos(A)-cos(A+2 B))/2 = (cos(A))/2-1/2 cos(A+2 B):
(cos(A))/2+(cos(A+2 B))/2+(cos(A))/(2)-(cos(A+2 B))/(2) = ^?cos(A)
(cos(A))/2+(cos(A+2 B))/2+(cos(A))/2-(cos(A+2 B))/(2) = cos(A):
cos(A) = ^?cos(A)
The left hand side and right hand side are identical:
Answer: |
| (identity has been verified)


or


Verify the following identity:
cos(A+B) cos(B)+sin(A+B) sin(B) = cos(A)
cos(A+B) = cos(A) cos(B)-sin(A) sin(B):
cos(B) cos(A) cos(B)-sin(A) sin(B)+sin(B) sin(A+B) = ^?cos(A)
sin(A+B) = cos(B) sin(A)+cos(A) sin(B):
cos(B) (cos(A) cos(B)-sin(A) sin(B))+sin(B) cos(B) sin(A)+cos(A) sin(B) = ^?cos(A)
cos(B) (cos(A) cos(B)-sin(A) sin(B)) = cos(A) cos(B)^2-cos(B) sin(A) sin(B):
cos(A) cos(B)^2-cos(B) sin(A) sin(B)+sin(B) (cos(B) sin(A)+cos(A) sin(B)) = ^?cos(A)
sin(B) (cos(B) sin(A)+cos(A) sin(B)) = cos(B) sin(A) sin(B)+cos(A) sin(B)^2:
cos(A) cos(B)^2-cos(B) sin(A) sin(B)+cos(B) sin(A) sin(B)+cos(A) sin(B)^2 = ^?cos(A)
sin(B)^2 = 1-cos(B)^2:
cos(A) cos(B)^2-cos(B) sin(A) sin(B)+cos(B) sin(A) sin(B)+cos(A) 1-cos(B)^2 = ^?cos(A)
cos(A) (1-cos(B)^2) = cos(A)-cos(A) cos(B)^2:
cos(A) cos(B)^2-cos(B) sin(A) sin(B)+cos(B) sin(A) sin(B)+cos(A)-cos(A) cos(B)^2 = ^?cos(A)
cos(A) cos(B)^2-cos(B) sin(A) sin(B)+cos(B) sin(A) sin(B)+cos(A)-cos(A) cos(B)^2 = cos(A):
cos(A) = ^?cos(A)
The left hand side and right hand side are identical:
Answer: |
| (identity has been verified)
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