Please ask a math problem, very urgent, a math problem, very urgent

sin sin =-1 2[cos( +cos( -cos cos =1 2[cos( +cos( -sin cos =1 2[sin( +sin( -cos sin =1 2[sin( +sin( -cos sin =1 2[sin( +sin( -The formula for high school math selections is "Sum of Differences".

It's helpful to remember better.

Then the original formula = 1 2 [sin(12°+18°)+sin(12°-18°)]1 2[sin(78°+72°)+sin(78°-72°)].

1/2[sin30°+sin(-6°)]1/2[sin150°+sin6°]

1 2 (1 2-sin6 ° + 1 2 + sin6 °) where sin30 ° = sin150 ° = 1 2

sin(-6°)=-sin6°

You should know these two common senses.

The formula of accumulation and difference, the teacher should have talked about it.

sin sin =-1 2[cos( +cos( -cos cos =1 2[cos( +cos( -sin cos =1 2[sin( +sin( -cos sin =1 2[sin( +sin( - thus. Original.

1/2[sin(12°+18°)+sin(12°-18°)]1/2[sin(78°+72°)+sin(78°-72°)]

1/2[sin30°+sin(-6°)]1/2[sin150°+sin6°]

1 2 (1 2-sin6°+1 2+sin6°) The above formula actually has a good memory method, sc ss plus, which means that sin cos = 1 2[sin( +sin( -

It's very simple, brother! Original formula = cos78sin72 + sin78 cos72sin (78 + 72).

sin150sin30

I omitted the power of the note, cos78=sin12 cos18=sin72

Accumulation and difference formulas.

sin sin =-1 2[cos( +cos( -cos cos =1 2[cos( +cos( -sin cos =1 2[sin( +sin( -cos sin =1 2[sin( +sin( - thus. Original.

1/2[sin(12°+18°)+sin(12°-18°)]1/2[sin(78°+72°)+sin(78°-72°)]

1/2[sin30°+sin(-6°)]1/2[sin150°+sin6°]

1/2（1/2-sin6°+1/2+sin6°）

Draw first: two right-angled triangles are similar according to having an equal acute angle.

cd:ef yields: bc=3

By: CD1AB=6

Write slowly, think slowly.

Send it to see if I will, if you don't send it I will teach you.

Oooo

This is a number series problem, 9 to 5 difference 4, 17 to 9 difference 8 points, you can know that it is a multiple of 4, then bn=bn-1+4n.

bn is equal to n of 2 plus 1 to the power plus 1

Namely. 5=2*2+1