Question

Help with proof in trig

Answer 1

If ABC is any triangle with a, b, and c and opposite angles A, B, and C respectively show the following is true:

\[\frac{ CosA }{ a }+\frac{ CosB }{ b }+\frac{ CosC }{ c }=\frac{ a^2+b^2+c^2 }{2abc }\]

Answer 2

so lefts start with the left side

Answer 3

@ganeshie8

Answer 4

Should we give the triangle numbers to work with?

Answer 5

i think cosine law works nicely here...

Answer 6

\[a^2 = b^2+c^2-2bc \cos A\]

isolating \(\cos A\) gives
\[\cos A = \dfrac{b^2+c^2-a^2}{2bc}\]

Answer 7

So could we take the inverse of cosine to get A alone ?

Answer 8

or is that what CosA is in numerator?

Answer 9

why do you want A alone ?
simply replace \(\cos A\) by above expression and see if it simplifies :)

Answer 10

Wait *facepalm* we basically have every thing we need on the right side in the form it is in the original

Answer 11

you want to start with left hand side, yes ?

Answer 12

Yes

Answer 13

Left hand side :
\[\frac{ CosA }{ a }+\frac{ CosB }{ b }+\frac{ CosC }{ c }\]

Answer 14

replace \(Cos A\) by previous expression

Answer 15

\[\frac{ b^2+c^2-a^2}{ \frac{ 2bc }{ a } }\]