Question

If A + B + C = 2π, then prove that cos 2A + cos 2B + cos 2C + cos 2D = 4 cos(A + B) cos(A + C) cos(A + D).

Answer 1

Answer:

$\bf\:cos2A+cos2B+cos2C+cos2D=4\:cos(A+B)\:cos(A+C)\:cos(A+D)$

Step-by-step explanation:

If A + B + C = 2π, then prove that cos 2A + cos 2B + cos 2C + cos 2D = 4 cos(A + B) cos(A + C) cos(A + D).

Given: $A+B+C+D=2\pi$

$cos2A+cos2B+cos2C+cos2D$

Using

$\boxed{cosC+cosD=2\:cos(\frac{C+D}{2})\:cos(\frac{C-D}{2})}$

$=2\:cos(A+B})\:cos(A-B)+2\:cos(C+D)\:cos(C-D)$

$=2\:cos(A+B)\:cos(A-B)+2\:cos(2\pi-(A+B))\:cos(C-D)$

$=2\:cos(A+B)\:cos(A-B)+2\:cos(A+B)\:cos(C-D)$

$=2\:cos(A+B)[cos(A-B)+cos(C-D)]$

Using

$\boxed{cosC+cosD=2\:cos(\frac{C+D}{2})\:cos(\frac{C-D}{2})}$

$=2\:cos(A+B)[2\:cos(\frac{A-B+C-D}{2})\:cos(\frac{A-B-C+D}{2})]$

$=4\:cos(A+B)\:cos(\frac{A+C-(B+D)}{2})\:cos(\frac{A+D-(B+C)}{2})$

$=4\:cos(A+B)\:cos(\frac{A+C-2\pi+(A+C)}{2})\:cos(\frac{A+D-2\pi+(A+D)}{2})$

$=4\:cos(A+B)\:cos(\frac{2(A+C)-2\pi}{2})\:cos(\frac{2(A+D)-2\pi}{2})$

$=4\:cos(A+B)\:cos((A+C)-\pi)\:cos((A+D)-\pi)$

$=4\:cos(A+B)\:cos(-(\pi-(A+C)))\:cos(-(\pi-(A+D)))$

Using

$\boxed{cos(-\theta)=cos\theta}$

$=4\:cos(A+B)\:cos(\pi-(A+C))\:cos(\pi-(A+D))$

Using

$\boxed{cos(\pi-\theta)=-cos\theta}$

$=4\:cos(A+B)\:(-cos(A+C))\:cos(-(A+D))$

$=4\:cos(A+B)\:cos(A+C)\:cos(A+D)$

$\implies\boxed{\bf\:cos2A+cos2B+cos2C+cos2D=4\:cos(A+B)\:cos(A+C)\:cos(A+D)}$

Answer 2

Answer:

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