Can you help prove this Trig Identity?

Question

sin(x+y+z)=sin x cos y cos z+cos x sin y cos z
+cos x cos y sin z-sin x sin y sin z

Answer 1

The trick is grouping two of the variables.

sin (x+y+z)
= sin [(x+y)+z]
= sin (x+y) cos z + cos (x+y) sin z
= (sin x cos y + cos x sin y) cos z + (cos x cos y - sin x sin y) sin z
= sin x cos y cos z + cos x sin y cos z + cos x cos y sin z - sin x sin y sin z

Answer 2

cos=756b hfn-0 jf98 dj== kf6 fhhjd= fl cos - tan- sin=h764 sin=35.67 (56=780=93) tan cat 35df 6=o uycos98.4

Answer 3

I forget the specific identities but if you let y+z= u, you have sin(x+u)= terms in cos and sin x and u.
Now substitute u=y+z in terms with u.

Answer 4

Use the formula sin(A + B) = sinAcosB + cosAsinB
sin(x+y+z)=sinxcos(y+z)+cosxsin(y+z)
Again, also using the formula
cos(A + B) = cosAcosB - sinAsinB

sin(x+y+z) = sinx(cosycosz-sinysinz)+cosx(sinycosz+cosysinz)
=sinxcosycosz-sinxsinysinz+cosxsinycosz+cosxcosysinz
Rearrange the terms
=sinxcosycosz+cosxsinycosz+cosxcosysinz-sinxsinysinz

Answer 5

Expand the brackets of the sine
sin(x+y+z) = sin(x+y)cosz + cos(x+y)sinz
= cosz * (sinxcosy + cosx siny) + sinz*(cosx cosy - sinx siny)
You could finish it from there.

Answer 6

sin(x+y+z) use sin (a+b) theorem twice
a= x+y
sin(a+z) = sin(a)cos(z) +cos(a)sin(z)
replace a
sin(x+y)cos(z) +cos(x+y)sin(z)
cos(z) (sin(x)cos(y) +cos(x)sin(y))+ sin(z) (cos(x)cos(y) -sin(x)sin(y))
distribute
sin(x)cos(y)cos(z) +cos(x)sin(y)cos(z) +sin(z) cos(x)cos(y) -sin(x)sin(y)sin(z)