Given to find the value of :-

cos20° + cos40° + cos60° +... cos180°

To know :-

COS(180-θ) = -cosθ

cos180° = cos(90°+90°)

= -sin 90 [cos(90+θ) = sinθ]

cos 180°= -1

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SOLUTION:-

cos20° + cos40° + cos60° + cos80° + cos100° + cos120° + cos 140° + cos160° + cos180°

cos20° can be written as cos(180°-160°)

cos40° can be written as cos(180°-140°)

cos60° can be written as cos(180°-120°)

cos80° can be written as cos(180°-100°)

If you observe these I have written all interms of after terms that MEANS 160°, 140°, 120°, 100°

So,

cos20° + cos40° + cos60° + cos80° + cos100° + cos120° + cos 140° + cos160° + cos180°

cos(180°-160°) + cos(180°-140°) + cos(180°-120°) + cos(180°-100°) + cos100° + cos120° + cos 140° + cos160° + cos180°

As we know ,

cos(180-θ) = -cosθ

= -cos160° - cos140° -cos120° - cos100° + cos100° + cos120° + cos140° + cos160° + cos180°

= -cos160° + cos160° -cos140° +cos140° -cos120° +cos120° -cos100°+cos100° + cos180°

All cancelled except cos180°

= cos180°

= -1

Note :-

The PURPOSE of writing 20°,40° 60° in terms of 120°,140° by using Quadrant angles is to be cancelled opposite signs And we can get the value