Answer :

To Prove:

Taking LHS,


= sin272° - cos230°


= sin2(90° - 18°) - cos230°


= cos2 18° - cos230° …(i)


Here, we don’t know the value of cos 18°. So, we have to find the value of cos 18°


Let x = 18°


so, 5x = 90°


Now, we can write


2x + 3x = 90°


so 2x = 90° - 3x


Now taking sin both the sides, we get


sin2x = sin(90° - 3x)


sin2x = cos3x [as we know, sin(90°- 3x) = Cos3x ]


We know that,


sin2x = 2sinxcosx


Cos3x = 4cos3x - 3cosx


2sinxcosx = 4cos3x - 3cosx


2sinxcosx - 4cos3x + 3cosx = 0


cosx (2sinx - 4cos2x + 3) = 0


Now dividing both side by cosx we get,


2sinx - 4cos2x + 3 = 0


We know that,


cos2x + sin2x = 1


or cos2x = 1 – sin2x


2sinx – 4(1 – sin2x) + 3 = 0


2sinx – 4 + 4sin2x + 3 = 0


2sinx + 4sin2x – 1 = 0


We can write it as,


4sin2x + 2sinx - 1 = 0


Now applying formula


Here, ax2 + bx + c = 0


So,


now applying it in the equation









Now sin 18° is positive, as 18° lies in first quadrant.



Now, we know that


cos2x + sin2x = 1


or cosx = √1 – sin2x


cos 18° = √1 –sin2 18°






Putting the value in eq. (i), we get


= cos2 18° - cos230°








= RHS


LHS = RHS


Hence Proved


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