Answer: see proof below
Step-by-step explanation:
Use the Cofunction Identity: sin (90 - A) = cos (A)
Use the Double Angle Identity: sin(2A) = 2 sinA cosA
Use the Triple Angle Identity: cos (3A) = 4cos³A - 3cosA
Use the Zero Product Property: If a · b = 0, then a = 0 or b = 0
Use Pythagorean Identity: cos²A + sin²A = 1 → cos²A = 1 - sin²A
Use Quadratic Formula: [tex]x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
Proof LHS → RHS:
Assumption: Let A = 18°
Multiplication: then 5A = 5(18°)
→ 5A = 90°
Expand: 2A + 3A = 90°
→ 2A = 90° - 3A
Apply sin: sin (2A) = sin (90° - 3A)
Cofunction: sin (2A) = cos (3A)
Double Angle: 2 sinA cos A = cos (3A)
Triple Angle: 2 sin A cos A = 4cos³A - 3cos A
→ -4cos³A + 3cos A + 2sinA cos A = 0
Factor: cos A(-4cos² A + 3 + 2sin A) = 0
Zero Product Property: cos A = 0 or -4cos² A + 3 + 2sin A = 0
↓
Disregard (since cos A ≠ 0)
Pythagorean: -4cos² A + 3 + 2sin A = 0
→ -4(1 - sin² A) + 3 + 2sin A = 0
→ -4 + 4sin² A + 3 + 2sin A = 0
Simplify: 4sin² A + 2sin A - 1 = 0
a=4 b=2 c= -1
[tex]\text{Quadratic Formula:}\qquad \sin A=\dfrac{-(2)\pm \sqrt{2^2-4(4)(-1)}}{2(4)}[/tex]
[tex]=\dfrac{-2\pm \sqrt{4+16}}{8}[/tex]
[tex]=\dfrac{-2\pm \sqrt{20}}{8}[/tex]
[tex]=\dfrac{-2\pm 2\sqrt{5}}{2(4)}[/tex]
[tex]=\dfrac{-1\pm \sqrt{5}}{4}[/tex]
[tex]=\dfrac{ \sqrt{5}-1}{4}[/tex]
[tex]\text{Substitute A = 18:}\qquad \sin 18=\dfrac{ \sqrt{5}-1}{4}[/tex]
