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Find an acute angle A such that \sin 4A = \sin A + \sin 2A. Express your answer in degrees.

 
 Jun 20, 2025
 #1
avatar+15102 
+1

\(\sin 4A = \sin A + \sin 2A\)     Find acute angle A.

 

 

\( \sin 4A=8\sin A\cos^3 A-4\sin A\cos A\\ \cos^3 A=\\ \sin 2A=2\sin A\cos A\\ 8\sin A\cos^3 A-4\sin A\cos A=\sin A+2\sin A\cos A\)

 

will be continued

laugh !

 Jun 20, 2025
 #3
avatar+2 
0

A such that \sin 4A = \sin A + \sin 2A @smashy road free.

elongrimer  Jun 24, 2025
 #2
avatar+130503 
+1

\( \sin 4A = \sin A + \sin 2A \)

sin ( 2(2A))  = sin A + sin 2A

2sin2Acos2A  = sinA + sin2A

2sin2A ( 1 - 2sin^2A)  = sinA + sin2A

2sin2A  - 4sin2A(sin^2A) = sinA + sin2A

sin(2A) - 4sin(2A)* (sin^2( A))  = sin(A)

2sinAcosA - 8sinAcosA (sin^2A) = sinA

sinA [ 2cosA - 8cosA (1 -cos^2A) ] - 1 =  0

sinA  = 0     A =  0”

2cosA - 8cosA (1 -cos^2 A) = 1

2cosA - 8cosA + 8cos^3A - 1 = 0

8cos^3 A - 6cosA - 1 =  0

 

Let  A  = x

 

8x^3 - 6x - 1   =  0

 

Using WolframAlpha to solve this  cubic   gives  x ≈  .93969

 

cos A  =.93969

 

arccos (.93969)  = A  =  20°

 

cool cool cool

 Jun 20, 2025
edited by CPhill  Jun 20, 2025
 #4
avatar+15102 
+1

Let cos A = x

laugh  !

asinus  Jun 25, 2025

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