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Sahdev Sharma 8 years, 3 months ago
Let A = 18°
Therefore, 5A = 90°
2A + 3A = 90˚
=> 2A = 90˚ - 3A
Taking sine on both sides, we get
sin 2A = sin (90˚ - 3A) = cos 3A
=> {tex}2 sin A cos A = 4cos^3 A - 3 cos A {/tex}
=> {tex} 2 sin A cos A - 4cos^3A + 3 cos A = 0 {/tex}
=>{tex}cos A (2 sin A - 4 cos^2 A + 3) = 0 {/tex}
Dividing both sides by cos A = cos 18˚ ≠ 0, we get
=> {tex}2 sin A - 4 (1 - sin^2 A) + 3 = 0 {/tex}
=> {tex}4 sin^2 A + 2 sin A - 1 = 0{/tex}
which is a quadratic in sin A
Therefore, {tex}Sin A = {-2\pm \sqrt{(2)^2-4(4)(-1)}\over 2(4)}{/tex}
{tex}=> sin A = {−2\pm \sqrt {20}\over 8 } {/tex}
{tex}=> Sin A={-1\pm \sqrt 5\over 4}{/tex}
Now sin 18° is positive, as 18° lies in first quadrant.
Therefore, {tex}=> Sin 18^°={-1+ \sqrt 5\over 4}{/tex}
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