Question 63547: I'm not sure how to go about this one
cos( tan-1(1/2) + tan-1(1/3))
Thanks
Found 2 solutions by venugopalramana, joyofmath:
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! I'm not sure how to go about this one
cos( tan-1(1/2) + tan-1(1/3))= I SAY
Thanks
PUT
TAN^(-1)[1/2]=X...HENCE
TAN(X)=1/2..................1
LET
TAN^(-1)[3]=Y....HENCE
TAN(Y)=1/3.........................2
WE HAVE
I = COS(X+Y)=COS(X)COS(Y)-SIN(X)SIN(Y)
COS(X) = 1/SEC(X) = 1/SQRT[(1+TAN^2(X)]=1/SQRT[1+1/4]=2SQRT(5)/5
SIN(X) = 1/COSEC(X)=1/SQRT[1+COT^2(X)]=1/SQRT[1+{1/TAN^2(X)}]
=1/SQRT(5)=SQRT(5)/5
SIMILARLY
COS(Y) = 3SQRT(10)/10
SIN(Y) = SQRT(10)/10
HENCE
I = [2SQRT(5)*3SQRT(10)/(5*10)] - [SQRT(5)*SQRT(10)/(5*10)]
=[6SQRT(50)/50]-[SQRT(50)/50]=5SQRT(50)/50 =25SQRT(2)/50=SQRT(2)/2
Answer by joyofmath(189)  (Show Source):
You can put this solution on YOUR website! cos( tan-1(1/2) + tan-1(1/3))
Use the angle sum identity: cos(x+y) = cos(x)cos(y)-sin(x)sin(y)
Then, cos( tan-1(1/2) + tan-1(1/3)) = cos(tan-1(1/2))cos(tan-1(1/3))-sin(tan-1(1/2))sin(tan-1(1/3))
Now, if A is the angle whose tangent = 1/2 then A is the angle in a right triangle where the opposite side = 1, the adjacent side = 2, and the hypotenuse, by the Pythagorean Theorem, = sqrt(5). Then cos(A) = adjacent/hypotenuse = 2/sqrt(5). And, sin(A) = opposite/hypotenuse = 1/sqrt(5).
Similarly, if B is the angle whose tangent = 1/2 then B is the angle in a right triangle where the opposite side = 1, the adjacent side = 3, and the hypotenuse = sqrt(10). Then cos(B) = adjacent/hypotenuse = 3/sqrt(10). And, sin(B) = opposite/hypotenuse = 1/sqrt(10).
So, cos(tan-1(1/2)) = 2/sqrt(5).
cos(tan-1(1/3)) = 3/sqrt(10).
sin(tan-1(1/2)) = 1/sqrt(5).
sin(tan-1(1/3)) = 1/sqrt(10).
So, cos( tan-1(1/2) + tan-1(1/3)) = cos(tan-1(1/2))cos(tan-1(1/3))-sin(tan-1(1/2))sin(tan-1(1/3)) =
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