Question

FAN AND MEDAL
Easier trig conversion here
Ill insert the formula now

Answer 1

\[1-\cos2A = (\cos2A(\sin2C-\sin2B)) \div 4\]

Answer 2

@ikram002p

Answer 3

@ganeshie8

Answer 4

@ganeshie8 any ideas?

Answer 5

not sure, see if this helps :
SinC - SinD = 2Cos[(C+D)/2]Sin[(C-D)/2]

Answer 6

@mathmate

Answer 7

The fact that B and C appear only once tells me that you are not proving an identity. Can you confirm that this is an equation to be solved, possibly with another condition such as A+B+C=90 or A+B+C=180?

Answer 8

sorry i keep forgetting to add it is
A+B+C=180

Answer 9

So it is an identity we need to prove, right?
I'll work on it and get back to you, unless someone else gets at it before!

Answer 10

thanks!

Answer 11

By the way, the left-hand side is equal to
\(2sin^2(A)=2sin^2(B+C)\)

Answer 12

Please check if the identity to be proved is actually:
\(\large 1−cos(2A)=\frac{cos(2A)(sin(2C)−sin(2B))}{4}\)

If I designate:
\(\large f(A,B,C)=1−cos(2A)-\frac{cos(2A)(sin(2C)−sin(2B))}{4}\)

I get
f(30,60,90)=\(\large \frac{\sqrt 3}{16}+\frac 1 2\) for A+B+C=180
and
f(30,30,30)=\(\large \frac 1 2 \) for A+B+C=90

This means that the above idetity is not.
Could you please check for typos?

Answer 13

@crashonce

Answer 14

thanks for your help @mathmate sorry about it there might be a typo
but i got to solve it (not this but the actual question but this was a part)
thanks anyway!

Answer 15

You're welcome! :)

For the future, it would be nice if you would specify that this is a part of the problem you derived so helpers would understand the context.
:)

Answer 16

yep sorrry about that mathmate

Answer 17

No problem! :)