Question

If A+B+C=180 prove that cos^2a+cos^2B+cos^2C=1-COSACOSBCOSC

Answer 1

SRY 1-2COSACOSBCOSC

Answer 2

helppppppp

Answer 3

If A+B+C=180, then A+B=180-C, and hence
\[\cos(A+B) = \cos(180-C) = -\cos C\]
Also, we're going to use the following trigonometric identities:
\begin{eqnarray*}
\cos x\cos y &=& \cos(x+y) + \cos(x-y) \\
\cos 2x &=& 2\cos^2x-1
\end{eqnarray*}
Let's start from 1 - 2cosAcosBcosC.
\begin{eqnarray*}
1-2\cos A\cos B\cos C&=&1-\cos C(\cos A\cos B) \\
&=&1-\cos C(\cos(A+B) + \cos(A-B)) \\
&=&1-\cos C(-\cos C + \cos(A-B)) \\
&=&1+\cos^2C - \cos C\cos(A-B) \\
&=&1+\cos^2C + \cos(A+B)\cos(A-B) \\
&=&1+\cos^2C + \frac{1}{2}(\cos(A+B+A-B)+\cos(A+B-A+B)) \\
&=&1+\cos^2C + \frac{1}{2}(\cos 2A + \cos 2B) \\
&=&1+\cos^2C + \frac{1}{2}(2\cos^2A-1+2\cos^2B-1) \\
&=&1+\cos^2C + \frac{1}{2}(2\cos^2A+2\cos^2B-2) \\
&=&1+\cos^2C + \cos^2A + \cos^2B - 1 \\
&=&\cos^2A+\cos^2B+\cos^2C
\end{eqnarray*}
as required.

Answer 4

wow another one

Answer 5

If A+B+C=180 prove that tan (A/2)*tan(B/2)+tan(B/2)*tan(C/2)+tan(C/2)*tan(A/2)=1

Answer 6

xactxx one doubt 2cosxcosy=cos(X+y)+cos(x-y)
and not cosx cosy=cos(X+y)+cos(x-y)

Answer 7

You're right, there should be a 2 on the LHS or 1/2 on the RHS

Answer 8

xactxx fantastic job!

Answer 9

plz rectify it

Answer 10

How do I edit my own post?

Answer 11

myininaya y u say fantastic ??????myininaya he made a mistake

Answer 12

but still it is an awesome job

Answer 13

gn8 guys ..plz post the answer

Answer 14

i wouldn't have thought to do A+B=180-C
then say cos(A+B)=cos(180-C)=-1cos(-C)=-cos(-C)=-cos(C)

Answer 15

is there a way to edit my post please?

Answer 16

no

Answer 17

but you can copy the latex above and post it again and change something if you want

Answer 18

right click and then say show source

Answer 19

Well I'll just repost the answer with the fixes.

If A+B+C=180, then A+B=180-C, and hence
\[\cos(A+B)=\cos(180−C)=−\cos C\]
Also, we're going to use the following trigonometric identities:
\begin{eqnarray*}
2\cos x\cos y &=& \cos(x+y)+\cos(x-y) \\
\cos 2x&=&2cos^2x−1
\end{eqnarray*}
Let's start from 1 - 2cosAcosBcosC.
\begin{eqnarray*}
1−2\cos A\cos B\cos C&=&1−\cos C(2\cos A\cos B) \\
&=&1−\cos C(\cos(A+B)+\cos(A−B)) \\
&=&1−\cos C(−\cos C+\cos(A−B)) \\
&=&1+\cos^2C−\cos C\cos(A−B) \\
&=&1+\cos^2C+\cos(A+B)\cos(A−B) \\
&=&1+\cos^2C+\frac{1}{2}(\cos(A+B+A−B)+\cos(A+B-A+B)) \\
&=&1+\cos^2C+\frac{1}{2}(\cos 2A+\cos 2B) \\
&=&1+\cos^2C+\frac{1}{2}(2\cos^2A−1+2\cos^2B−1) \\
&=&1+\cos^2C+\frac{1}{2}(2\cos^2A+2\cos^2B−2) \\
&=&1+\cos^2C+\cos^2A+\cos^2B−1 \\
&=&\cos^2A+\cos^2B+\cos^2C
\end{eqnarray*}
as required.