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Question

Consider △ACB, right-angled at C, in which AB=29 units, BC=21 units and ∠ABC=θ. Determine the values of cos2θ−sin2θ.

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Solution

Firstly we have to find the value of AC with the help of Pythagoras theoremAccording to Pythagoras theorem(Hypotenuse)2=(Base)2+(Perpendicular)2⇒(AC)2+(BC)2=(AB)2⇒(AC)2+(21)2=(29)2⇒(AC)2=(29)2−(21)2Using the identity a2−b2=(a+b)(a−b)⇒(AC)2=(29−21)(29+21)⇒(8)(50)=400⇒AC=√400=±20But AC can't be negative, so, AC=20 unitsNow, we will find the sinθ and cosθsinθ=sideoppositetoangleθhypotenuseIn △ACB, side opposite to angle θ=AC=20 and Hypotenuse =AB=29So, sinθ=ACAB=2029Now we know thatcosθ=sideadjacenttoangleθhypotenuseIn △ACB, side adjacent to angle θ=BC=21 and Hypotenuse =AB=29So, cosθ=BCAB=2129Putting values we getcos2θ−sin2θ=(2129)2−(2029)2⇒441−40029×29=41841

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