In a Δ ABC, ∠ C = 90°, ∠ ABC = θ° , BC = 21units and AB = 29units. Show that cos2θ –sin2θ = 41/841

Δ ABC is a right angled triangle



By Pythagoras theorem, (hypotenuse)2 = (perpendicular)2 + (base)2


∴AB2 = AC2 + BC2


= (29)2 = AC2 + (21)2


= 841 = AC2 + 441


= AC2 = 400


→ AC = 20


∴ sinθ = AC/AB = 20/29


cosθ = BC/AB = 21/29


cos2θ –sin2θ = (21/29)2 - (20/29)2


=


= 41/841


= RHS


HENCE PROVED


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