Question 35940: Suppose that an isosceles triangle has sides of length a,a, and b and is such that cos(2A)=sin(B). Show that sin^2(B)= -2sin(A)cos(A)cos(B).
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! Suppose that an
isosceles triangle has sides of length a, a, and b and
is such that cos(2A)=sin(B). Show that
sin^2(B)=-2sin(A)cos(A)cos(B).
1 solutions
Answer 22050 by venugopalramana(1719) About Me on
2006-05-03 01:18:38 (Show Source):
suppose that an isosceles triangle has sides of length
a, a, and b and is
> such that cos(2A)=sin(B). Show that
sin^2(B)=-2sin(A)cos(A)cos(B).
COS(2A)=SIN(B).............................................................I
IN A TRIANGLE
A+B+C=180BUT SINCE C=A
B+2A=180...................................................................II
2A=180-B
COS(2A)=COS(180-B)=-COS(B)....FROM EQN.I,WE GET
-COS(B)=SIN(B)....OR......COS(B)=-SIN(B)..................................III
WE HAVE T.S.T.
SIN^2B=-2SIN(A)COS(A)COS(B),SUBSTITUTING FROM III IN
RHS
TST....SIN^2B=-2SIN(A)COS(A)(-SIN(B))....OR.......
TST....SIN(B)=2SIN(A)COS(A)=SIN(2A)
USING EQN.II..............2A=180-B.....HENCE
SIN(2A)=SIN(180-B)=SIN(B)......THUS WE PROVED THE
RESULT.
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