Question

If, A+B+C=π
(i) Prove that, SinA+SinB-SinC = 4SinA/2. SinB/2. CosC/2
(ii) Prove that, Cos^2A + Cos^2B-Cos^2C = 1-2SinA.SinB.Cos.C
(iii) Prove that, CosA+CosB-CosC = -1+4CosA/2. CosB/2.SinC/2

"^" indicates the Power.. I Don't know how to represent it.

Answer 1

what math class is this from?

Answer 2

Peter, Im in 11th... I asked that level question.. can anybody please help?

Answer 3

from the start, if A+B+C=pi, it looks like you have a triangle with angles A, B, and C

Answer 4

yeah.. you're right..
but i'm confused , how to solve them..

Answer 5

for i)
do you know half angle identities?

Answer 6

no.. i dont

Answer 7

then you probably wouldn't be expected to do anything with them.

Answer 8

what was the last thing you learned in class?

Answer 9

Formulas.. and their derivations.. up to Exercise 3.3 of NCERT book.

Answer 10

I don't think I can help you.

Answer 11

It's okay: )

Answer 12

@jim_thompson5910

Answer 13

@UsukiDoll

Answer 14

hello! I have been summoned from the chat

Answer 15

trig identities ohhhhhhhh boy.

Answer 16

that was long ago.

Answer 17

Can anybody please help me..
i need it for my homework..
please?

Answer 18

any one if anybody can solve?

Answer 19

what class is this from? trigonometry?

Answer 20

because I'm seeing trig identites

Answer 21

yes, yes!! Exactly..

Answer 22

crud you know that was a topic that I tackled in 2009. And my brain or coconut head forgot.

Answer 23

:/

Answer 24

if you could build a time machine to 2009, I would gladly help you.

Answer 25

I'm seeing double angle identities too. AWWW man! frustrating for me. I used to know this!@

Answer 26

Well.. It's okay. May be, I could get someone who can help me out..
14hours are left for my school..

Answer 27

I do know that pi is 180 in degree mode

Answer 28

A + B + C = 180
Angle A + Angle B + Angle C = 180

Answer 29

Yeah,, that is understood.

Answer 30

180/ 3 angles = 60 degrees per letter.

Answer 31

Assuming A = 60 degrees B = 60 degrees and C = 60 degrees.

Answer 32

you want the result to equal 180 so trying plugging in 60 degrees for the letters and calculate.

Answer 33

hmm..

Answer 34

oh I wish I still retained that info...that's what happens when I don't practice, but I do see identities.

Answer 35

It's okay. :)

Answer 36

sina+sinb-sinc= 2sin(a+b)/2 * cos(a-b)/2 - sinc
= 2sin(90-c/2)*cos(a-b)/2-2sinc/2*cosc/2
=2cos c/2*cos(a-b)/2 - 2sinc/2*cos c/2
=2cos c/2 {cos(a-b)/2 - sin c/2}
=2cos c/2 {cos(a-b)/2 - cos (a+b)/2)}
=2cos c/2 {2 sina/2 * sin b/2}
=4 sin a/2 * sin b/2 * cos c/2..

Is this so?? for the 1st one?

Answer 37

oh what!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Answer 38

plug in 60 degrees for a b c
calculate all of this yeah

Answer 39

What? I'm not confirm.. im asking

Answer 40

: (

Answer 41

what happeneD?

Answer 42

This is definitely a hard problem

Answer 43

But luckily I managed to find help here

http://www.enotes.com/homework-help/b-c-pi-prove-that-identity-true-sin-sin-b-sin-c-154655

Answer 44

and I also used these trig identities

http://en.wikipedia.org/wiki/Trigonometric_identity
http://www.sosmath.com/trig/Trig5/trig5/trig5.html

Answer 45

I got the 1st one.
i need.. 2nd and 3rd equation solbed.
well.
thank you a lot, Sir.

Answer 46

here's what I got

sin(A) + sin(B) = 2*sin(A+B)/2 * cos(A-B)/2


sin(C) = sin(C/2 + C/2) = 2*sin(C/2)*cos(C/2)


A+B+C = pi


A+B = pi-C


-------------------------------------


sin(A)+sin(B)-sin(C)


2*sin((A+B)/2) * cos((A-B)/2) - 2*sin(C/2)*cos(C/2)


2*sin((pi-C)/2) * cos((A-B)/2) - 2*sin(C/2)*cos(C/2)


2*sin(pi/2-C/2) * cos((A-B)/2) - 2*sin(C/2)*cos(C/2)


2*cos(C/2) * cos((A-B)/2) - 2*sin(C/2)*cos(C/2)


cos(C/2)[2cos((A-B)/2) - 2*sin(C/2)]


2cos(C/2)[cos((A-B)/2) - cos(pi/2-C/2)]


2cos(C/2)[-2sin(((A-B)/2+pi/2-C/2)/2)sin(((A-B)/2-pi/2+C/2)/2)]


2cos(C/2)[-2sin((A-B)/4+pi/4-C/4)sin((A-B)/4-pi/4+C/4)]


2cos(C/2)[-2sin((A-B+pi-C)/4)sin((A-B-pi+C)/4)]


2cos(C/2)[-2sin((A-B+A+B)/4)sin((A-B-(A+B))/4)]


2cos(C/2)[-2sin((2A)/4)sin((-2B)/4)]


2cos(C/2)[-2sin(A/2)sin(-B/2)]


-4cos(C/2)sin(A/2)sin(-B/2)


4cos(C/2)sin(A/2)sin(B/2)


4sin(A/2)sin(B/2)cos(C/2)


--------------------------------------------------------


So this shows us that


sin(A)+sin(B)-sin(C)


turns into


4sin(A/2)sin(B/2)cos(C/2)


So this verifies that


sin(A)+sin(B)-sin(C) = 4sin(A/2)sin(B/2)cos(C/2)


is an identity

Answer 47

thank you, sir. thanks a lot..
Please. try 2nd and 3rd equations too..

Answer 48

I knew they were identities but since that topic was 4 years old in my head, I just couldn't pin point anymore.

Answer 49

yeah yeah.. now you'll say that..

Answer 50

at least you're not writing proofs. Those are a nightmare

Answer 51

whatever..
and thanks..
atleast you tried. and helped me a lil.

Answer 52

yeah

Answer 53

Hard problem! For the (ii), I found a way. From left hand side,
\[\cos^2A+\cos^2B-\cos^2C=1-\sin^2A+\frac{1+\cos(2B)}{2}-\frac{1+\cos(2C)}{2}=\\
=1-\sin^2B+\frac{1}{2}(\cos(2B)-\cos(2C))=\\
=1-\sin^2A-\sin(B+C)\sin(B-C)=\\
=1-\sin^2A-\sin(\pi-A)\sin(B-C)=\\
=1-\sin^2A-\sin(A)\sin(B-C)=\\
=1-\sin A(\sin A+\sin(B-C))=\\
=1-\sin A(\sin (\pi-(B+C))+\sin(B-C))=\\
=1-\sin A(\sin (B+C)+\sin(B-C))=\\
=1-2\sin A\sin (B)\cos(C)\]
Note that I have used,
\[\cos(2B)=\cos(B+B+C-C)=\cos((B+C)+(B-C))=\\
=\cos(A+C)\cos(B-C)-\sin(B+C)\sin(B-C)\]
And the same for cos(2C).
\[\cos(2C)=\cos(C+C+B-B)=\cos((B+C)-(B-C))=\\
=\cos(A+C)\cos(B-C)+\sin(B+C)\sin(B-C)\]

Answer 54

Thank you sir.. Thanks a lot for your help..

Answer 55

Took me forever, but I figured out (ii) with the help of these trig identities



http://en.wikipedia.org/wiki/Trigonometric_identity
http://www.sosmath.com/trig/Trig5/trig5/trig5.html



and this is the page that gave me a good hint


http://www.goiit.com/posts/list/trignometry-if-a-b-c-pi-prove-that-cosa-2-cosb-2-cosc-2-1-2cosacosbcosc-1065034.htm;jsessionid=1F13CD3666FAAFA880CAD2508F845599.node2#1416589


--------------------------------------------------------

A+B+C = pi


B+C = pi-A


sin(B+C) = sin(pi-A)


sin(B)cos(C)+cos(B)sin(C) = sin(pi)cos(A)-cos(pi)sin(A)


sin(B)cos(C)+cos(B)sin(C) = sin(A)


sin(B)cos(C) - sin(A) = -cos(B)sin(C)


(sin(B)cos(C) - sin(A))^2 = (-cos(B)sin(C))^2


sin^2(B)cos^2(C) + sin^2(A) - 2sin(A)sin(B)cos(C) = cos^2(B)sin^2(C)


-2sin(A)sin(B)cos(C) = cos^2(B)sin^2(C) - sin^2(B)cos^2(C) - sin^2(A)





-------------------------------------------------------


cos^2(A) + cos^2(B) - cos^2(C) = 1-2*sin(A)*sin(B)*cos(C)


cos^2(A) + cos^2(B) - cos^2(C) = 1+cos^2(B)sin^2(C) - sin^2(B)cos^2(C) - sin^2(A)


cos^2(A) + cos^2(B) - cos^2(C) = 1+(cos^2(B)sin^2(C) - sin^2(B)cos^2(C)) - sin^2(A)


cos^2(A) + cos^2(B) - cos^2(C) = 1+((cos(B)sin(C))^2 - (sin(B)cos(C))^2) - sin^2(A)


cos^2(A) + cos^2(B) - cos^2(C) = 1+(cos(B)sin(C)-sin(B)cos(C))(cos(B)sin(C)+sin(B)cos(C)) - sin^2(A)


cos^2(A) + cos^2(B) - cos^2(C) = 1+(sin(C)cos(B)-cos(C)sin(B))(sin(C)cos(B)+cos(C)sin(B)) - sin^2(A)


cos^2(A) + cos^2(B) - cos^2(C) = 1+sin(C-B)sin(C+B) - sin^2(A)


cos^2(A) + cos^2(B) - cos^2(C) = 1+(1/2)*(cos(2B)-cos(2C)) - sin^2(A)


cos^2(A) + cos^2(B) - cos^2(C) = 1+(1/2)*(2cos^2(B)-1-(2cos^2(C)-1)) - sin^2(A)


cos^2(A) + cos^2(B) - cos^2(C) = 1+(1/2)*(2cos^2(B)-1-2cos^2(C)+1) - sin^2(A)


cos^2(A) + cos^2(B) - cos^2(C) = 1+(1/2)*(2cos^2(B)-2cos^2(C)) - sin^2(A)


cos^2(A) + cos^2(B) - cos^2(C) = 1+cos^2(B)-cos^2(C) - sin^2(A)


cos^2(A) + cos^2(B) - cos^2(C) = 1 - sin^2(A) + cos^2(B) - cos^2(C)


cos^2(A) + cos^2(B) - cos^2(C) = cos^2(A) + cos^2(B) - cos^2(C)


So that verifies the identity

Answer 56

Thanks you, sir.. Thanks a lot..
it seems.. there are experts sitting over here..

Answer 57

There is also a way for (iii) using the ideas from (ii).
\[\cos A+\cos B-\cos C=2\cos^2\left(\frac{A}{2}\right)-1-2\sin\left(\frac{B+C}{2}\right) \sin\left(\frac{B-C}{2}\right)=\\
=2\cos^2\left(\frac{A}{2}\right)-1-2\sin\left(\frac{B+C}{2}\right) \sin\left(\frac{B-C}
{2}\right)=\\
=2\cos^2\left(\frac{A}{2}\right)-1-2\sin\left(\frac{\pi -A}{2}\right) \sin\left(\frac{B-C}{2}\right)=\\
=2\cos^2\left(\frac{A}{2}\right)-1-2\cos\left(\frac{A}{2}\right) \sin\left(\frac{B-C}{2}\right)=\\
=-1+2\cos\left(\frac{A}{2}\right)\left(\cos\left(\frac{A}{2}\right)- \sin\left(\frac{B-C}{2}\right)\right)=\\
=-1+2\cos\left(\frac{A}{2}\right)\left(\sin\left(\frac{B+C}{2}\right) -\sin\left(\frac{B-C}{2}\right)\right)=\\
=-1+4\cos\left(\frac{A}{2}\right)\cos\left(\frac{B}{2}\right) \sin\left(\frac{C}{2}\right)
\]

Answer 58

In this case I have used,
\[\cos B-\cos C=\cos\left(\frac{B+C+B-C}{2}\right)-\cos\left(\frac{B+C-(B-C)}{2}\right)=\\
=-2\sin\left(\frac{B+C}{2}\right)\sin\left(\frac{B-C}{2}\right)\]

Answer 59

thank you sir..
this is really gonna make me easy to solve these equations. :)