+1661 Let theta be an acute angle such that sin 2 theta = sin 3 theta + sin 4 theta. What is the measure of theta in degrees?
+130575
sin2A = 2sinAcosA
sin3A = sin (2A + A) = sin2AcosA + sinAcos2A = 2sinAcos^2A + sinA (2cos^2A -1)
sin4A = sin (2A + 2A) = 2sin2Acos2A = 4sinAcosA *(2cos^2A - 1)
2sinAcosA = sinA [ 2cos^2A + 2cos^2A - 1] + 4sinAcosA *(2cos^2A -1)
(divide out sinA since sinA = 0 so angle A = 0 , but A is acute)
2cos A = 4cos^2A -1 + 8cos^3A - 4cosA
8cos^3A + 4cos^2A - 6cosA - 1 = 0
Let cos A =x
8x^3 + 4x^2 - 6x - 1 = 0
x ≈ .7406
cos A = .7406
arcos (.7406) = A ≈ 42.217°
+2 To find the angle AA from the equation:
8cos3A+4cos2A−6cosA−1=08cos3A+4cos2A−6cosA−1=0
Steps:
Let x=cosAx=cosA: 8x3+4x2−6x−1=08x3+4x2−6x−1=0
Approximate solution: Using numerical methods, you find: x≈0.7406x≈0.7406
Calculate angle: A=arccos(0.7406)≈42.22∘A=arccos(0.7406)≈42.22∘ (rounded to two decimal places).
Conclusion: drive mad
The angle AA is approximately 42.22∘42.22∘.
