Question

Prove that sinx+siny+sinz-sin(x+y+z)=4sin((x+y)/2)sin((y+z)/2)sin((z+x)/2)

Answer 1

Solution:

We have to prove that,

sinx+siny+sinz-sin(x+y+z)=4sin((x+y)/2)sin((y+z)/2)sin((z+x)/2)

Keep in mind

$sin A +sinB= 2 sin \frac{(A+B)}{2}\times cos\frac{(A-B)}{2}$

LHS:

As, in a triangle , sum of all interior angles is 180°.

x+y+z=180°

x+y=180°-z

$\frac{x+y}{2}=90 -\frac{z}{2}\\\\ sin (\frac{x+y}{2})=sin(90 -\frac{z}{2})\\\\ sin (\frac{x+y}{2})=cos(\frac{z}{2})$

Similarly, $cos{\frac{x}{2}=sin{\frac{z+y}{2}}{\text{and}} cos{\frac{y}{2}=sin{\frac{z+x}{2}}$

and, cos(x-y)+cos(x+y)=2 cos x cos y

=Sinx +Sin y +Sin z-sin (180°)

=Sin x +SIn y +Sin z-0

$=2sin {\frac{x+y}{2}}cos\frac {(x-y)}{2}+2Sin(\frac{z}{2})cos(\frac{z}{2})\\\\  2sin {\frac{x+y}{2}}cos\frac {(x-y)}{2}+2Sin(\frac{z}{2})sin {\frac{x+y}{2}}\\\\ 2sin {\frac{x+y}{2}}[cos\frac {(x-y)}{2}+cos\frac {(x+y)}{2}]\\\\ 2sin {\frac{x+y}{2}}\times 2cos(\frac{x}{2})\times cos(\frac{z}{2})\\\\4sin{\frac{x+y}{2}}sin{\frac{z+y}{2}}sin{\frac{x+z}{2}}$

Answer 2

Step-by-step explanation:

hope this helps

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