Q. 19 B5.0( 2 Votes )

# Prove that

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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