a b c cos 1 2 sin 2 is applicable in any triangle, verify.

a+b)/c

by the sine theorem.

sina + sinb )/sinc

Thereinto. by and difference product formulas.

sina + sinb = 2sin[(a+b)/2]cos[(a-b)/2]

by the formula of the double angle.

sinc = 2sin(c 2)cos(c 2) again because a + b + c = 180

a/2 + b/2 + c/2 = 90

So sin[(a+b) 2] = cos(c 2) so primitive. cos[(a-b)2] sin(c2)ab c are the three corresponding angles.

Let the radius of the circumscribed circle of abc be r, and the three sides are a,,b,c, which are known by the sinusoidal theorem a sina=b sinb=c sinc=2rsin a sin b sin c, so (a 2r) 2+(b 2r) 2=(c 2r) 2 is sorted out: a 2+b 2=c 2, satisfies the Pythagorean theorem, so abc is a right triangle.

These two interchanges are based on the sine theorem.

a sina=b sinb=c sinc=ka=ksina,b=ksinb,c=ksinc and then substitute k to verify your problem.

Let the radius of the circumscribed circle of abc be r, and the corresponding edges are a, b, and c respectively, and know a sina=b sinb=c sinc=2r from the sinusoidal theorem (generally found in high school books), and know sin a sin b sin c from the question, that is, (a 2r) 2+(b 2r) 2=(c 2r) 2, and the sorted a 2+b 2=c 2, satisfies the Pythagorean theorem, so abc is a right triangle.

Using the corollary of the sine theorem, a

sina=b sinb=c sinc=2r, which is processed by multiplying 4r on both sides of the known sin a sin b sin c to obtain a +b =c, which is obviously a right triangle.

According to the sinusoidal theorem: a sina = b sinb = c sinc = 2rSo the original formula is equal to a 2 + b 2 = c 2, so the triangle is a right triangle.

There is still a problem with the title, it should be "cos +cos +2cos cos = 1 2" I lost the root number 2???

cos²α+cos²β+cosαcosβ

1+cos2 ) 2+(1+cos2) 2+cos cos (powered).

1 + cos2α+cos2β)/2 + cosαcosβ

1 + 2 + cos cos (obtained from the above equation "sum difference product").

1 + cos(α+cos(α-cosαcosβ

1 - 2 2)cos( -2cos cos (cos( += - 2 2 substitution)

1 - 2 2) (cos cos + sin sin) 2cos cos (cosine formula with the sum of two angles).

1 + 2/2)(cosα*cosβ-sinα*sinβ)

1 + 2 2) cos ( + invert the cosine formula of the difference between the two angles).

1-1 2 ( cos( +=- 2 2 substitution)

Wrong question, right?

For example, if you want = =3 8, then: cos +cos +cos cos =3cos = is not true.

This question is wrong, there are a lot of things missing, it is known that it should be + =3 4 , and the proof should be cos +cos +2 cos cos cos =1 2

It can be proved that the product of the first two items after the reduction and the difference between the product and the product of the latter item can be proved after offsetting.

If there is a problem with this question, please send me the correct one.

Left = (sin a-sin b) sin c = sin(a+b) sin(a-b) sin c=sin(a-b) sinc=right.

Simplification is achieved using the formula sin -sin =sin( +sin( - and sin(a+b)=sinc in the triangle.

Use the cosine theorem and trigonometric formulas to find the right.

Isosceles right triangle.

Hello classmates, I hope the answer is helpful to you, please give a good review and the upper right corner.

I know I wish you a happy life and study, thank you!!