In $Delta ABC$, find $cotdfrac{B}{2}.cotdfrac{C}{2}$ if $b+c=3a$
$begingroup$
If in a triangle ABC, $b+c=3a$, then $cotdfrac{B}{2}.cotdfrac{C}{2}$ is equal to ?
My reference gives the solution $2$, but I have no clue of where to start ?
My Attempt
$$
cotdfrac{B}{2}.cotdfrac{C}{2}=frac{cosfrac{B}{2}.cosfrac{C}{2}}{sinfrac{B}{2}.sinfrac{C}{2}}=frac{cos(frac{A-B}{2})+cos(frac{A+B}{2})}{cos(frac{A-B}{2})-cos(frac{A+B}{2})}
$$
geometry trigonometry triangle
edited Jan 4 at 15:59
Michael Rozenberg
108k1895200
asked Jan 4 at 12:23
ss1729ss1729
2,01511124
$endgroup$
add a comment |
$begingroup$
If in a triangle ABC, $b+c=3a$, then $cotdfrac{B}{2}.cotdfrac{C}{2}$ is equal to ?
My reference gives the solution $2$, but I have no clue of where to start ?
My Attempt
$$
cotdfrac{B}{2}.cotdfrac{C}{2}=frac{cosfrac{B}{2}.cosfrac{C}{2}}{sinfrac{B}{2}.sinfrac{C}{2}}=frac{cos(frac{A-B}{2})+cos(frac{A+B}{2})}{cos(frac{A-B}{2})-cos(frac{A+B}{2})}
$$
geometry trigonometry triangle
edited Jan 4 at 15:59
Michael Rozenberg
108k1895200
asked Jan 4 at 12:23
ss1729ss1729
2,01511124
$endgroup$
add a comment |
$begingroup$
If in a triangle ABC, $b+c=3a$, then $cotdfrac{B}{2}.cotdfrac{C}{2}$ is equal to ?
My reference gives the solution $2$, but I have no clue of where to start ?
My Attempt
$$
cotdfrac{B}{2}.cotdfrac{C}{2}=frac{cosfrac{B}{2}.cosfrac{C}{2}}{sinfrac{B}{2}.sinfrac{C}{2}}=frac{cos(frac{A-B}{2})+cos(frac{A+B}{2})}{cos(frac{A-B}{2})-cos(frac{A+B}{2})}
$$
geometry trigonometry triangle
edited Jan 4 at 15:59
Michael Rozenberg
108k1895200
asked Jan 4 at 12:23
ss1729ss1729
2,01511124
$endgroup$
If in a triangle ABC, $b+c=3a$, then $cotdfrac{B}{2}.cotdfrac{C}{2}$ is equal to ?
My reference gives the solution $2$, but I have no clue of where to start ?
My Attempt
$$
cotdfrac{B}{2}.cotdfrac{C}{2}=frac{cosfrac{B}{2}.cosfrac{C}{2}}{sinfrac{B}{2}.sinfrac{C}{2}}=frac{cos(frac{A-B}{2})+cos(frac{A+B}{2})}{cos(frac{A-B}{2})-cos(frac{A+B}{2})}
$$
geometry trigonometry triangle
geometry trigonometry triangle
edited Jan 4 at 15:59
Michael Rozenberg
108k1895200
asked Jan 4 at 12:23
ss1729ss1729
2,01511124
edited Jan 4 at 15:59
Michael Rozenberg
108k1895200
asked Jan 4 at 12:23
ss1729ss1729
2,01511124
edited Jan 4 at 15:59
Michael Rozenberg
108k1895200
edited Jan 4 at 15:59
Michael Rozenberg
108k1895200
edited Jan 4 at 15:59
Michael Rozenberg
108k1895200
108k1895200
asked Jan 4 at 12:23
ss1729ss1729
2,01511124
asked Jan 4 at 12:23
ss1729ss1729
2,01511124
asked Jan 4 at 12:23
ss1729ss1729
2,01511124
2,01511124
add a comment |
add a comment |
2 Answers
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$begingroup$
Hint:
$$b+c=3aimpliessin B+sin C=3sin A$$
$$iff2sindfrac{B+C}2cosdfrac{B-C}2=6sindfrac A2cosdfrac A2$$
Now use $dfrac{B+C}2=dfracpi2-dfrac A2,cosdfrac{B+C}2=?$
As $0<A<pi,sinsindfrac A2ne0$
$impliescosdfrac{B-C}2=3sindfrac A2=3cosdfrac{B+C}2$
answered Jan 4 at 12:28
lab bhattacharjeelab bhattacharjee
226k15158275
$endgroup$
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$begingroup$
In the standard notation we obtain: $$sinbeta+singamma=3sin(beta+gamma)$$ or
$$2sinfrac{beta+gamma}{2}cosfrac{beta-gamma}{2}=6sinfrac{beta+gamma}{2}cosfrac{beta-gamma}{2}$$ or
$$cosfrac{beta}{2}cosfrac{gamma}{2}+sinfrac{beta}{2}sinfrac{gamma}{2}=3left(cosfrac{beta}{2}cosfrac{gamma}{2}-sinfrac{beta}{2}sinfrac{gamma}{2}right)$$ or
$$cosfrac{beta}{2}cosfrac{gamma}{2}=2sinfrac{beta}{2}sinfrac{gamma}{2}$$ or
$$cotfrac{beta}{2}cotfrac{gamma}{2}=2.$$
Also, we can use the following way:
$$cotfrac{beta}{2}cotfrac{gamma}{2}=frac{a+b+c}{b+c-a}=frac{4a}{2a}=2.$$
answered Jan 4 at 15:43
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Hint:
$$b+c=3aimpliessin B+sin C=3sin A$$
$$iff2sindfrac{B+C}2cosdfrac{B-C}2=6sindfrac A2cosdfrac A2$$
Now use $dfrac{B+C}2=dfracpi2-dfrac A2,cosdfrac{B+C}2=?$
As $0<A<pi,sinsindfrac A2ne0$
$impliescosdfrac{B-C}2=3sindfrac A2=3cosdfrac{B+C}2$
answered Jan 4 at 12:28
lab bhattacharjeelab bhattacharjee
226k15158275
$endgroup$
add a comment |
$begingroup$
Hint:
$$b+c=3aimpliessin B+sin C=3sin A$$
$$iff2sindfrac{B+C}2cosdfrac{B-C}2=6sindfrac A2cosdfrac A2$$
Now use $dfrac{B+C}2=dfracpi2-dfrac A2,cosdfrac{B+C}2=?$
As $0<A<pi,sinsindfrac A2ne0$
$impliescosdfrac{B-C}2=3sindfrac A2=3cosdfrac{B+C}2$
answered Jan 4 at 12:28
lab bhattacharjeelab bhattacharjee
226k15158275
$endgroup$
add a comment |
$begingroup$
Hint:
$$b+c=3aimpliessin B+sin C=3sin A$$
$$iff2sindfrac{B+C}2cosdfrac{B-C}2=6sindfrac A2cosdfrac A2$$
Now use $dfrac{B+C}2=dfracpi2-dfrac A2,cosdfrac{B+C}2=?$
As $0<A<pi,sinsindfrac A2ne0$
$impliescosdfrac{B-C}2=3sindfrac A2=3cosdfrac{B+C}2$
answered Jan 4 at 12:28
lab bhattacharjeelab bhattacharjee
226k15158275
$endgroup$
Hint:
$$b+c=3aimpliessin B+sin C=3sin A$$
$$iff2sindfrac{B+C}2cosdfrac{B-C}2=6sindfrac A2cosdfrac A2$$
Now use $dfrac{B+C}2=dfracpi2-dfrac A2,cosdfrac{B+C}2=?$
As $0<A<pi,sinsindfrac A2ne0$
$impliescosdfrac{B-C}2=3sindfrac A2=3cosdfrac{B+C}2$
answered Jan 4 at 12:28
lab bhattacharjeelab bhattacharjee
226k15158275
answered Jan 4 at 12:28
lab bhattacharjeelab bhattacharjee
226k15158275
answered Jan 4 at 12:28
lab bhattacharjeelab bhattacharjee
226k15158275
answered Jan 4 at 12:28
lab bhattacharjeelab bhattacharjee
226k15158275
226k15158275
add a comment |
add a comment |
$begingroup$
In the standard notation we obtain: $$sinbeta+singamma=3sin(beta+gamma)$$ or
$$2sinfrac{beta+gamma}{2}cosfrac{beta-gamma}{2}=6sinfrac{beta+gamma}{2}cosfrac{beta-gamma}{2}$$ or
$$cosfrac{beta}{2}cosfrac{gamma}{2}+sinfrac{beta}{2}sinfrac{gamma}{2}=3left(cosfrac{beta}{2}cosfrac{gamma}{2}-sinfrac{beta}{2}sinfrac{gamma}{2}right)$$ or
$$cosfrac{beta}{2}cosfrac{gamma}{2}=2sinfrac{beta}{2}sinfrac{gamma}{2}$$ or
$$cotfrac{beta}{2}cotfrac{gamma}{2}=2.$$
Also, we can use the following way:
$$cotfrac{beta}{2}cotfrac{gamma}{2}=frac{a+b+c}{b+c-a}=frac{4a}{2a}=2.$$
answered Jan 4 at 15:43
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
add a comment |
$begingroup$
In the standard notation we obtain: $$sinbeta+singamma=3sin(beta+gamma)$$ or
$$2sinfrac{beta+gamma}{2}cosfrac{beta-gamma}{2}=6sinfrac{beta+gamma}{2}cosfrac{beta-gamma}{2}$$ or
$$cosfrac{beta}{2}cosfrac{gamma}{2}+sinfrac{beta}{2}sinfrac{gamma}{2}=3left(cosfrac{beta}{2}cosfrac{gamma}{2}-sinfrac{beta}{2}sinfrac{gamma}{2}right)$$ or
$$cosfrac{beta}{2}cosfrac{gamma}{2}=2sinfrac{beta}{2}sinfrac{gamma}{2}$$ or
$$cotfrac{beta}{2}cotfrac{gamma}{2}=2.$$
Also, we can use the following way:
$$cotfrac{beta}{2}cotfrac{gamma}{2}=frac{a+b+c}{b+c-a}=frac{4a}{2a}=2.$$
answered Jan 4 at 15:43
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
add a comment |
$begingroup$
In the standard notation we obtain: $$sinbeta+singamma=3sin(beta+gamma)$$ or
$$2sinfrac{beta+gamma}{2}cosfrac{beta-gamma}{2}=6sinfrac{beta+gamma}{2}cosfrac{beta-gamma}{2}$$ or
$$cosfrac{beta}{2}cosfrac{gamma}{2}+sinfrac{beta}{2}sinfrac{gamma}{2}=3left(cosfrac{beta}{2}cosfrac{gamma}{2}-sinfrac{beta}{2}sinfrac{gamma}{2}right)$$ or
$$cosfrac{beta}{2}cosfrac{gamma}{2}=2sinfrac{beta}{2}sinfrac{gamma}{2}$$ or
$$cotfrac{beta}{2}cotfrac{gamma}{2}=2.$$
Also, we can use the following way:
$$cotfrac{beta}{2}cotfrac{gamma}{2}=frac{a+b+c}{b+c-a}=frac{4a}{2a}=2.$$
answered Jan 4 at 15:43
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
In the standard notation we obtain: $$sinbeta+singamma=3sin(beta+gamma)$$ or
$$2sinfrac{beta+gamma}{2}cosfrac{beta-gamma}{2}=6sinfrac{beta+gamma}{2}cosfrac{beta-gamma}{2}$$ or
$$cosfrac{beta}{2}cosfrac{gamma}{2}+sinfrac{beta}{2}sinfrac{gamma}{2}=3left(cosfrac{beta}{2}cosfrac{gamma}{2}-sinfrac{beta}{2}sinfrac{gamma}{2}right)$$ or
$$cosfrac{beta}{2}cosfrac{gamma}{2}=2sinfrac{beta}{2}sinfrac{gamma}{2}$$ or
$$cotfrac{beta}{2}cotfrac{gamma}{2}=2.$$
Also, we can use the following way:
$$cotfrac{beta}{2}cotfrac{gamma}{2}=frac{a+b+c}{b+c-a}=frac{4a}{2a}=2.$$
answered Jan 4 at 15:43
Michael RozenbergMichael Rozenberg
108k1895200
edited Jan 4 at 15:49
answered Jan 4 at 15:43
Michael RozenbergMichael Rozenberg
108k1895200
answered Jan 4 at 15:43
Michael RozenbergMichael Rozenberg
108k1895200
answered Jan 4 at 15:43
Michael RozenbergMichael Rozenberg
108k1895200
108k1895200
add a comment |
add a comment |
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