4)Let A be an acute angle such that sin4A = SinA. What is the measure of A in degrees?
sin (4A) = sin A
sin ( 2 * 2A) = sin A
2sin 2A cos 2A = sin A
2 [ 2 sin A cos A ] * [ 2cos^2 - 1 ] = sin A
8sin A cos^3 A - 4sin A cos A - sin A = 0
sin A [ 8 cos^3 A - 4 cos A - 1 ] = 0
sin A = 0 ⇒ A = 0° [reject....A is > 0° ]
8cos^3 - 4cos A - 1 = 0 .... let cos A = x
8x^3 - 4x - 1 = 0
( 2x + 1) ( 4x^2 - 2x - 1 ) = 0
Setting the first factor to 0 and solving for x, we have that
x = -1/2 ⇒ cos A = -1/2 ⇒ A = 120° [reject....A is acute ]
Setting the second factor to 0 and solving for x, we have
4x^2 - 2x - 1 = 0
The solutions to this are :
x = [ 1 - √5 ] / 4 ⇒ cos A = [ 1 - √5] /4 ⇒ A = 108° [ reject....A is acute ]
x = [ 1 + √5 ] / 4 ⇒ cos A = [ 1 + √5 ] / 4 ⇒ A = 36°