Menu Close

a-3-cos-A-B-b-3-cos-B-C-c-3-cos-A-C-

Tinku Tara 0 Comments
Pin




Question Number 212232 by Davidtim last updated on 07/Oct/24
a^3 cos(A−B)+b^3 cos(B−C)+c^3 cos(A−C)=?
$${a}^{\mathrm{3}} {cos}\left({A}−{B}\right)+{b}^{\mathrm{3}} {cos}\left({B}−{C}\right)+{c}^{\mathrm{3}} {cos}\left({A}−{C}\right)=? \\ $$
Commented by Davidtim last updated on 07/Oct/24
I kindly request you for helping
$${I}\:{kindly}\:{request}\:{you}\:{for}\:{helping} \\ $$
Answered by Ghisom last updated on 08/Oct/24
this makes no sense. it should be a^3 cos (B−C) +b^3 cos (C−A) +c^3 cos (A−B) which gives 3abc use cos (x−y) =cos x cos y +sin x sin y and a^2 +b^2 −2abcos C =c^2 ⇒ cos C =((a^2 +b^2 −c^2 )/(2ab)) sin C =((√((a+b+c)(a+b−c)(a+c−b)(b+c−a)))/(2ab)) [for the other angles rotate a→b→c→a]
$$\mathrm{this}\:\mathrm{makes}\:\mathrm{no}\:\mathrm{sense}.\:\mathrm{it}\:\mathrm{should}\:\mathrm{be} \\ $$$${a}^{\mathrm{3}} \mathrm{cos}\:\left({B}−{C}\right)\:+{b}^{\mathrm{3}} \mathrm{cos}\:\left({C}−{A}\right)\:+{c}^{\mathrm{3}} \mathrm{cos}\:\left({A}−{B}\right) \\ $$$$\mathrm{which}\:\mathrm{gives} \\ $$$$\mathrm{3}{abc} \\ $$$$ \\ $$$$\mathrm{use} \\ $$$$\mathrm{cos}\:\left({x}−{y}\right)\:=\mathrm{cos}\:{x}\:\mathrm{cos}\:{y}\:+\mathrm{sin}\:{x}\:\mathrm{sin}\:{y} \\ $$$$\mathrm{and} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}\mathrm{cos}\:{C}\:={c}^{\mathrm{2}} \\ $$$$\Rightarrow \\ $$$$\mathrm{cos}\:{C}\:=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{ab}} \\ $$$$\mathrm{sin}\:{C}\:=\frac{\sqrt{\left({a}+{b}+{c}\right)\left({a}+{b}−{c}\right)\left({a}+{c}−{b}\right)\left({b}+{c}−{a}\right)}}{\mathrm{2}{ab}} \\ $$$$\left[\mathrm{for}\:\mathrm{the}\:\mathrm{other}\:\mathrm{angles}\:\mathrm{rotate}\:{a}\rightarrow{b}\rightarrow{c}\rightarrow{a}\right] \\ $$
Commented by Davidtim last updated on 08/Oct/24
by which law you have writen the relation for sinC? just a litle bit describe for me to know.
$$ \\ $$$${by}\:{which}\:{law}\:{you}\:{have}\:{writen}\:{the}\:{relation} \\ $$$${for}\:{sinC}?\:{just}\:{a}\:{litle}\:{bit}\:{describe}\:{for}\:{me}\:{to}\:{know}. \\ $$
Commented by mr W last updated on 08/Oct/24
Δ=((√((a+b+c)(−a+b+c)(a−b+c)(a+b−c)))/4) Δ=((ab sin C)/2) ((ab sin C)/2)=((√((a+b+c)(−a+b+c)(a−b+c)(a+b−c)))/4) ⇒sin C=((√((a+b+c)(−a+b+c)(a−b+c)(a+b−c)))/(2ab))
$$\Delta=\frac{\sqrt{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)\left({a}−{b}+{c}\right)\left({a}+{b}−{c}\right)}}{\mathrm{4}} \\ $$$$\Delta=\frac{{ab}\:\mathrm{sin}\:{C}}{\mathrm{2}} \\ $$$$\frac{{ab}\:\mathrm{sin}\:{C}}{\mathrm{2}}=\frac{\sqrt{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)\left({a}−{b}+{c}\right)\left({a}+{b}−{c}\right)}}{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{sin}\:{C}=\frac{\sqrt{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)\left({a}−{b}+{c}\right)\left({a}+{b}−{c}\right)}}{\mathrm{2}{ab}} \\ $$
Commented by Davidtim last updated on 08/Oct/24
sorry, what doesΔ modify here?
$${sorry},\:{what}\:{does}\Delta\:{modify}\:{here}? \\ $$
Commented by Davidtim last updated on 08/Oct/24
sorry dear, what does mldify Δ here? because here is no any equation? or it′s area of triangle?
$${sorry}\:{dear},\:{what}\:{does}\:{mldify}\:\Delta\:{here}? \\ $$$${because}\:{here}\:{is}\:{no}\:{any}\:{equation}?\: \\ $$$${or}\:{it}'{s}\:{area}\:{of}\:{triangle}? \\ $$
Commented by mr W last updated on 08/Oct/24
your question is about trisngle, so i guess you know what Δ means here, especially in Δ=((√((a+b+c)(−a+b+c)(a−b+c)(a+b−c)))/4) this is the famous heron′s formula for area of triangle.
$${your}\:{question}\:{is}\:{about}\:{trisngle},\:{so}\:{i} \\ $$$${guess}\:{you}\:{know}\:{what}\:\Delta\:{means}\:{here}, \\ $$$${especially}\:{in} \\ $$$$\Delta=\frac{\sqrt{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)\left({a}−{b}+{c}\right)\left({a}+{b}−{c}\right)}}{\mathrm{4}} \\ $$$${this}\:{is}\:{the}\:{famous}\:{heron}'{s}\:{formula} \\ $$$${for}\:{area}\:{of}\:{triangle}. \\ $$
Commented by Frix last updated on 08/Oct/24
https://en.m.wikipedia.org/wiki/Heron%27s_formula
Commented by Davidtim last updated on 08/Oct/24
But the heron′s formula is bellow: Δ=(√(p(p−a)(p−b)(p−c)))
$${But}\:{the}\:{heron}'{s}\:{formula}\:{is}\:{bellow}: \\ $$$$\Delta=\sqrt{{p}\left({p}−{a}\right)\left({p}−{b}\right)\left({p}−{c}\right)} \\ $$
Commented by Davidtim last updated on 08/Oct/24
p=((a+b+c)/2)
$${p}=\frac{{a}+{b}+{c}}{\mathrm{2}} \\ $$
Commented by Frix last updated on 08/Oct/24
Use your brain and simple algebra: Δ^2 =p(p−a)(p−b)(p−c)= =((a+b+c)/2)(((a+b+c)/2)−a)(((a+b+c)/2)−b)(((a+b+c)/2)−c)= =(((a+b+c)(b+c−a)(a+c−b)(a+b−c))/(16)) Δ=((√((a+b+c)(a+b−c)(a+c−b)(b+c−a)))/4)
$$\mathrm{Use}\:\mathrm{your}\:\mathrm{brain}\:\mathrm{and}\:\mathrm{simple}\:\mathrm{algebra}: \\ $$$$\Delta^{\mathrm{2}} ={p}\left({p}−{a}\right)\left({p}−{b}\right)\left({p}−{c}\right)= \\ $$$$=\frac{{a}+{b}+{c}}{\mathrm{2}}\left(\frac{{a}+{b}+{c}}{\mathrm{2}}−{a}\right)\left(\frac{{a}+{b}+{c}}{\mathrm{2}}−{b}\right)\left(\frac{{a}+{b}+{c}}{\mathrm{2}}−{c}\right)= \\ $$$$=\frac{\left({a}+{b}+{c}\right)\left({b}+{c}−{a}\right)\left({a}+{c}−{b}\right)\left({a}+{b}−{c}\right)}{\mathrm{16}} \\ $$$$\Delta=\frac{\sqrt{\left({a}+{b}+{c}\right)\left({a}+{b}−{c}\right)\left({a}+{c}−{b}\right)\left({b}+{c}−{a}\right)}}{\mathrm{4}} \\ $$
Commented by Davidtim last updated on 08/Oct/24
or sir very nice
$${or}\:{sir}\:{very}\:{nice} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *