Question

Please can you help me
sin^2A + sin^2B + sin^2C = 2 + 2cosAcosBcosC

Answer 1

What are the directions to this problem? @sladi4koto_rumi

Answer 2

If A+B+C=180 prove that sin^2A + sin^2B + sin^2C = 2 + 2cosAcosBcosC

Answer 3

ummm
write sin^2A = (1-cos(2A))/2 and similar to other and try to solve?

Answer 4

http://en.wikipedia.org/wiki/List_of_trigonometric_identities
go down down down :P

Answer 5

to power reduction formula

Answer 6

(1-cos(2A))/2+(1-cos(2B))/2+(1-cos(2C))/2
solveee it??
atleast tell me where are u stuck o_o

Answer 7

i think she is away?
sigh

Answer 8

(3- (cos2A + cos2B +cos2C) )/2 = (3 - ( -4cosAcosBcosC - 1))/2 = (4+4cosAcosBcosC)/2 = 2 + 2cosAcosBcosC
right?

Answer 9

tell me
whats the formula for cos2A + cos2B

Answer 10

let me remind you some formulas which are going to be very useful
cos(x+y) + cos(x-y) = 2 cos(x)cos(y)
so let's start from the right side

2 + 2cos(A)cos(B)cos(C) = 2 + (cos(A+B) + cos(A-B))cos(C)
and now from: A + B + C = 180, we can write that: A + B = 180 - C
and we plug it
2 + (cos(180-C) + cos(A-B))cos(C)
and cos(180-C) = -cos(C)
and we use it
2 + (-cos(C) + cos(A-B))cos(C)
now we distribute it
2 - cos^2(C) + cos(A-B)cos(C)
we know use for the last C, C = 180 - A - B
2 - cos^2(C) + cos(A-B)cos(180 - A - B)

and we apply again for the last term
and as cos(180 - A - B) = - cos(A + B)
we get
2 - cos^2(C) - cos(A-B)cos(A+B)

now we apply again the first property
2cos(x)cos(y) = cos(x + y) + cos(x - y)
in this case x = A - B and y = A + B
cos(A - B)cos(A + B) = ( cos(A - B + A + B) + cos(A - B - A - B) ) /2
so let's plug it
first let's reduce it

cos(A - B)cos(A + B) = ( cos(2A) + cos(-2B) ) / 2
and as cos(- X ) = cos( x)
we have

cos(A - B)cos(A + B) = ( cos(2A) + cos(2B) ) / 2
2 - cos^2(C) - ( cos(2A) + cos(2B) ) /2
and here you just need to apply the definition of the double angle
cos(2X) = cos^2(X) - sin^2(X) = 1 - 2sin^2(X)

let's apply it

2 - cos^2(C) - ( 1 - 2sin^2(A) + 1 - 2sin^2(B) ) /2
and we apply the Pythagorean theore for the cos^2(C)
cos^2(C) = 1 - sin^2(C)

then we get:
2 - ( 1 - sin^2(C) ) - ( 2 - 2 sin^2(A) - 2sin^2(B) ) /2
distributing the signs
2 - 1 + sin^2(C) - 1 + sin^2(A) + sin^2(B)

and as you can see 2 - 1 - 1 = 0
and we get the left side of the original expression:
sin^2(A) + sin^2(B) + sin^2(C)

Answer 11

O_O yes the hot chick solved the answer

Answer 12

i mean the question

Answer 13

thank you so much

Answer 14

i mean proved it

Answer 15

No worries, good luck! @sladi4koto_rumi

Answer 16

thank you

Answer 17

=)

Answer 18

Method 2 : starting from left side

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

...

2 - 2cosAcosB[cosAcosB - sinAsinB]
2 - 2cosAcosB[cos(A+B)]
2 + 2cosAcosBcosC

Answer 19

Method 3 : take both sides
try it :)