Evaluate sin(2arccos(3/5))

Problem

sin(2*arccos(3/5))

Solution

1. Identify the inner expression and let θ=arccos(3/5) This implies cos(θ)=3/5 where 0≤θ≤π

2. Determine the value of sin(θ) using the Pythagorean identity sin2(θ)+cos2(θ)=1 Since θ is in the first quadrant (where cosine is positive), sin(θ) must be positive.

3. Calculate sin(θ) by substituting the known value of cos(θ)

sin(θ)=√(,1−(3/5)2)

sin(θ)=√(,1−9/25)

sin(θ)=√(,16/25)

sin(θ)=4/5

4. Apply the double-angle formula for sine, which states sin(2*θ)=2*sin(θ)*cos(θ)

5. Substitute the values of sin(θ) and cos(θ) into the formula.

sin(2*θ)=2⋅4/5⋅3/5

sin(2*θ)=24/25

Final Answer

sin(2*arccos(3/5))=24/25

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