In $Delta ABC$, find $cotdfrac{B}{2}.cotdfrac{C}{2}$ if $b+c=3a$
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 13 hours ago
Michael Rozenberg
96.8k1589188
asked 16 hours ago
ss1729
1,8491723
add a comment |
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 13 hours ago
Michael Rozenberg
96.8k1589188
asked 16 hours ago
ss1729
1,8491723
add a comment |
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 13 hours ago
Michael Rozenberg
96.8k1589188
asked 16 hours ago
ss1729
1,8491723
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 13 hours ago
Michael Rozenberg
96.8k1589188
asked 16 hours ago
ss1729
1,8491723
edited 13 hours ago
Michael Rozenberg
96.8k1589188
asked 16 hours ago
ss1729
1,8491723
edited 13 hours ago
Michael Rozenberg
96.8k1589188
edited 13 hours ago
Michael Rozenberg
96.8k1589188
edited 13 hours ago
Michael Rozenberg
96.8k1589188
96.8k1589188
asked 16 hours ago
ss1729
1,8491723
asked 16 hours ago
ss1729
1,8491723
asked 16 hours ago
ss1729
1,8491723
1,8491723
add a comment |
add a comment |
2 Answers
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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 16 hours ago
lab bhattacharjee
223k15156274
add a comment |
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 13 hours ago
Michael Rozenberg
96.8k1589188
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
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active
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active
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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 16 hours ago
lab bhattacharjee
223k15156274
add a comment |
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 16 hours ago
lab bhattacharjee
223k15156274
add a comment |
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 16 hours ago
lab bhattacharjee
223k15156274
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 16 hours ago
lab bhattacharjee
223k15156274
answered 16 hours ago
lab bhattacharjee
223k15156274
answered 16 hours ago
lab bhattacharjee
223k15156274
answered 16 hours ago
lab bhattacharjee
223k15156274
223k15156274
add a comment |
add a comment |
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 13 hours ago
Michael Rozenberg
96.8k1589188
add a comment |
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 13 hours ago
Michael Rozenberg
96.8k1589188
add a comment |
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 13 hours ago
Michael Rozenberg
96.8k1589188
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 13 hours ago
Michael Rozenberg
96.8k1589188
edited 13 hours ago
answered 13 hours ago
Michael Rozenberg
96.8k1589188
answered 13 hours ago
Michael Rozenberg
96.8k1589188
answered 13 hours ago
Michael Rozenberg
96.8k1589188
96.8k1589188
add a comment |
add a comment |
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