=> arcsin(2x) = arcsin(x) + pi/12 (in radians 15° = pi/12)
Take sin(...) of both sides and use sin(x+y)=sinx cosy + cosx siny
=> 2x = x cos(pi/12) + cos(arcsin(x)) sin(pi/12)
as arcsin(x) between -pi/2 and pi/2 => cos(arcsin(x)) > 0
therefore cos(arcsin(x)) = + sqrt(1-x²)
=> 2x = x cos(pi/12) + sqrt(1-x²) sin(pi/12)
=> 2x - x cos(pi/12) = sqrt(1-x²) sin(pi/12)
=> x(2 - cos(pi/12)) / sin(pi/12) = sqrt(1-x²)
Now take square of both sides :
=> x²(2-cos(pi/12))²/sin²(pi/12) = 1-x²
=> x² ( (2-cos(pi/12))² + sin²(pi/12) ) / sin²(pi/12) = 1
=> x² ( 4 + cos²(pi/12) - 4 cos(pi/12) + sin²(pi/12) ) / sin²(pi/12) = 1
=> x² ( 5 - 4 cos(pi/12) ) = sin²(pi/12)
=> x² = sin²(pi/12) / (5 - 4 cos(pi/12))
=> x = +- sin(pi/12) / sqrt(5 - 4 cos(pi/12))
= +- 0.2428
Only the + sign gives a real solution of the equation, so we have
x = +0.2428 !
Remark :
we can find exact values for sin(pi/12) and cos(pi/12), through
sin(pi/6) = 1/2 and cos(pi/6) = sqrt(3)/2
and the doubling formula
cos(2x) = 2cos²(x)-1
=> sqrt(3)/2 = 2 cos²(pi/12) - 1
=> 1/2 + sqrt(3)/4 = cos²(pi/12)
=> (4 + 2sqrt(3))/8 = cos²(pi/12)
=> (3 + 2sqrt(3) + 1)/8 = cos²(pi/12)
=> (sqrt(3) + 1)² / 8 = cos²(pi/12)
=> (sqrt(3) + 1)/sqrt(8) = cos(pi/12)
=> (sqrt(3) - 1)/sqrt(8) = sin(pi/12)
It may be a bit easier to rearrange the equation to the form
.. arcsin(2x) = arcsin(x) + 15°
Taking the sine of both sides, we get
.. 2x = x*cos(15°) + â(1-x^2)*sin(15°) ... use the formula for sum of angles, and for cos in terms of sin
.. x(2-cos(15°)) = sin(15°)*â(1-x^2) ... collect x terms on the left, leave the root on the right
.. x^2*(2 - cos(15°))^2 = sin(15°)^2*(1-x^2) ... square both sides
This quadratic can be solved after some tedious algebra to give
.. x = sin(15°)/â(5-4*cos(15°)) â .242800937
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Verified answer
=> arcsin(2x) = arcsin(x) + pi/12 (in radians 15° = pi/12)
Take sin(...) of both sides and use sin(x+y)=sinx cosy + cosx siny
=> 2x = x cos(pi/12) + cos(arcsin(x)) sin(pi/12)
as arcsin(x) between -pi/2 and pi/2 => cos(arcsin(x)) > 0
therefore cos(arcsin(x)) = + sqrt(1-x²)
=> 2x = x cos(pi/12) + sqrt(1-x²) sin(pi/12)
=> 2x - x cos(pi/12) = sqrt(1-x²) sin(pi/12)
=> x(2 - cos(pi/12)) / sin(pi/12) = sqrt(1-x²)
Now take square of both sides :
=> x²(2-cos(pi/12))²/sin²(pi/12) = 1-x²
=> x² ( (2-cos(pi/12))² + sin²(pi/12) ) / sin²(pi/12) = 1
=> x² ( 4 + cos²(pi/12) - 4 cos(pi/12) + sin²(pi/12) ) / sin²(pi/12) = 1
=> x² ( 5 - 4 cos(pi/12) ) = sin²(pi/12)
=> x² = sin²(pi/12) / (5 - 4 cos(pi/12))
=> x = +- sin(pi/12) / sqrt(5 - 4 cos(pi/12))
= +- 0.2428
Only the + sign gives a real solution of the equation, so we have
x = +0.2428 !
Remark :
we can find exact values for sin(pi/12) and cos(pi/12), through
sin(pi/6) = 1/2 and cos(pi/6) = sqrt(3)/2
and the doubling formula
cos(2x) = 2cos²(x)-1
=> sqrt(3)/2 = 2 cos²(pi/12) - 1
=> 1/2 + sqrt(3)/4 = cos²(pi/12)
=> (4 + 2sqrt(3))/8 = cos²(pi/12)
=> (3 + 2sqrt(3) + 1)/8 = cos²(pi/12)
=> (sqrt(3) + 1)² / 8 = cos²(pi/12)
=> (sqrt(3) + 1)/sqrt(8) = cos(pi/12)
=> (sqrt(3) - 1)/sqrt(8) = sin(pi/12)
It may be a bit easier to rearrange the equation to the form
.. arcsin(2x) = arcsin(x) + 15°
Taking the sine of both sides, we get
.. 2x = x*cos(15°) + â(1-x^2)*sin(15°) ... use the formula for sum of angles, and for cos in terms of sin
.. x(2-cos(15°)) = sin(15°)*â(1-x^2) ... collect x terms on the left, leave the root on the right
.. x^2*(2 - cos(15°))^2 = sin(15°)^2*(1-x^2) ... square both sides
This quadratic can be solved after some tedious algebra to give
.. x = sin(15°)/â(5-4*cos(15°)) â .242800937