Q. 4 B5.0( 8 Votes )

Consider ∆ACB, right angled at C, in which AB = 29 units, BC = 21 units and ABC=θ. Determine the values of

a. cos2 θ+ sin2 θ

b. cos2 θ – sin2 θ

Answer :


(a) Cos2θ +sin2 θ


Firstly we have to find the value of AC.


So, we can find the value of AC with the help of Pythagoras theorem.


According 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)2


Using the identity a2 –b2 = (a+b) (a – b)


(AC)2 = (29–21)(29+21)


(AC)2 = (8)(50)


(AC)2 = 400


AC =400


AC =±20


But side AC can’t be negative. So, AC = 20units


Now, we will find the sin θ and cos θ



In ∆ACB, Side opposite to angle θ = AC = 20


and Hypotenuse = AB = 29


So,


Now, We know that



In ∆ACB, Side adjacent to angle θ = BC = 21


and Hypotenuse = AB = 29


So,


So




=1


Cos2θ +sin2 θ = 1


(b) Cos2θ – sin2 θ


Putting values, we get





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