Q. 43 A4.3( 4 Votes )

# If , prove that : 3 cos B – 4cos3 B = 0

Given:

We know that,

Or

Let,

Perpendicular =AB =k

and Hypotenuse =AC =2k

where, k is any positive integer

So, by Pythagoras theorem, we can find the third side of a triangle

In right angled ABC, we have

(AB)2 + (BC)2 = (AC)2

(k)2 + (BC)2 = (2k)2

k2 + (BC)2 = 4k2

(BC)2 = 4k2 –k2

(BC)2 = 3k2

BC =3k2

BC =k3

So, BC = k√3

Now, we have to find the value of cos B

We know that,

Side adjacent to angle B or base = BC = k√3

Hypotenuse = AC =2k

So,

Now, LHS = 3 cos B – 4cos3 B

=RHS

Hence Proved

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