Prove that sin^272° - cos^230° = (√5 - 1)/8

Question

Prove that

sin²72° - cos²30° = \(\frac{\sqrt{5}\,-\,1}{8}\)

sin²72° - cos²30° = (√5 - 1)/8

Answer 1

To Prove: sin²72° - cos²30° = \(\frac{\sqrt{5}\,-\,1}{8}\)

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

now applying it in the equation

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

∴ sin 18° = \(\frac{\sqrt{5}\,-\,1}{4}\)

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°

= \(\frac{\sqrt{5}\,-\,1}{8}\)

= RHS

∴ LHS = RHS

Hence Proved