Question

in triangle ABC the value of cos((a+2b+3c)/2)+cos((a-c)=2)=​

Answer 1

$\frac{cos((a+2b+3c)}{2}+\frac{cos((a-c)}{2} =0$

Step-by-step explanation:

Given:

Here ABC is a triangle.

$\frac{cos((a+2b+3c)}{2}+\frac{cos((a-c)}{2} =?$

We know that in ΔABC, A + B + C = π

Also, $\frac{A + B + C}{2}=\frac{\pi}{2}$

Using formula

$cos C + cos D = 2 cos(\frac{C+D}{2} )cos \frac{cos (C-D)}{2}$

Now,

$\frac{cos((a+2b+3c)}{2}+\frac{cos((a-c)}{2}$

= $2cos[\frac{\frac{a+2b+3c}{2} }{2} +\frac{\frac{a-c}{2} }{2}] cos [\frac{\frac{a+2b+3c}{2} }{2}-\frac{\frac{(a-c)}{2} }{2}]$

= $2cos[\frac{\frac{a+2b+3c+a-c}{2} }{2} ]cos[\frac{\frac{a+2b+3c-a+c}{2} }{2} ]$

= $2cos[{\frac{2a+2b+2c}{4} ]cos[\frac{{2b+4c} }{4} ]$

= $2cos[{\frac{2(a+b+c)}{4} ]cos[\frac{{2(b+2c)} }{4} ]$

= $2cos[{\frac{(a+b+c)}{2} ]cos[\frac{{(b+2c)} }{2} ]$  

= $2cos[{\frac{{\pi}}{2} ]cos[\frac{{(b+2c)} }{2} ]$                  [given]

= $2\times 0 \times cos[\frac{{(b+2c)} }{2} ]$                 $[cos\frac{\pi}{2}=0 ]$

= 0

Answer 2

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