Question

Please someone help me...​

Answer 1

2cos pi/13 cos 9pi/13+ cos 3pi/13 +cos 5pi/13

=cos 10 pi/13 +cos 8 pi/13 +cos 3pi/13 +cos 5pi/13

=cos 10 pi/13 +cos 3pi/13 +cos 8pi/13 +cos 5pi/13

=2 cos pi/2 .cos 7 pi/26 +2 cos pi/2 .cos 3 pi /26

=2 (0)cos 7 pi /26 + 2(0) cos 3pi/26

=0 =R.H.S.

Answer 2

Answer: see proof below

Explanation:

Use the following identities:

2cos x · cos y = cos(x + y) + cos(x - y)

cos x + cos y = 2 cos (x + y)/2 · cos(x - y)/2

Use the Unit Circle to evaluate: cos(π/2) = 0

Proof LHS → RHS

`\text{LHS:}\qquad \qquad \qquad 2\cos (9\pi)/(13)\cos (\pi)/(13)\quad +\quad \cos (3\pi)/(13)+\cos (5\pi)/(13)\\\\\text{Identity:}\qquad \quad \cos\bigg((9\pi)/(13)+(\pi)/(13)\bigg)+ \cos\bigg((9\pi)/(13)-(\pi)/(13)\bigg)+\quad \cos\bigg((3\pi)/(13)\bigg)+\cos \bigg((5\pi)/(13)\bigg)`

`\text{Simplify:}\qquad \qquad \cos\bigg( (10\pi)/(13)\bigg)+\cos \bigg((8\pi)/(13)\bigg)+\qquad \cos\bigg((3\pi)/(13)\bigg)+\cos \bigg((5\pi)/(13)\bigg)`

`\text{Regroup:}\qquad \qquad \cos\bigg( (10\pi)/(13)\bigg)+\cos \bigg((3\pi)/(13)\bigg)\quad +\quad \cos\bigg((8\pi)/(13)\bigg)+\cos \bigg((5\pi)/(13)\bigg)`

`\text{Identity:}\qquad 2\cos \bigg((10\pi+3\pi)/(13\cdot 2)\bigg)\cdot \cos \bigg((10\pi-3\pi)/(13\cdot 2)\bigg)+\quad 2\cos\bigg((8\pi+5\pi)/(13\cdot 2)\bigg)\cdot \cos\bigg((8\pi-5\pi)/(13\cdot 2)\bigg)`

`\text{Simplify:}\qquad 2\cos \bigg((13\pi)/(26)\bigg)\cdot \cos \bigg((7\pi)/(26)\bigg)+\quad 2\cos\bigg((13\pi)/(26)\bigg)\cdot \cos\bigg((3\pi)/(26)\bigg)\\\\\\.\qquad \qquad =2\cos \bigg((\pi)/(2)\bigg)\cdot \cos \bigg((7\pi)/(26)\bigg)+\quad 2\cos\bigg((\pi)/(2)\bigg)\cdot \cos\bigg((3\pi)/(26)\bigg)\\\\\\\text{Factor:}\qquad =2\cos\bigg((\pi)/(2)\bigg)\bigg[ \cos \bigg((7\pi)/(26)\bigg)+ \cos \bigg((3\pi)/(26)\bigg)\bigg]`

`\text{Unit Circle:}\quad 2(0)\bigg[ \cos \bigg((7\pi)/(26)\bigg)+ \cos \bigg((3\pi)/(26)\bigg)\bigg]`

Product of Zero 0

LHS = RHS: 0 = 0
`\checkmark`