How do you find the exact value of arccos ⁡ ( sin ⁡ ( π 3 ) )?

knowing that sin ⁡ ( π 3 ) = 3 2 arccos ⁡ ( sin ⁡ ( π 3 ) ) = arccos ⁡ ( 3 2 )we know that cos ⁡ ( π 6 ) = 3 2 so, π 6 = arccos ⁡ ( 3 2 ) arccos ⁡ ( sin ⁡ ( π 3 ) ) = arccos ⁡ ( 3 2 ) = π 6

By definition, cos ⁡ ( 1 2 π − θ ) = sin ⁡ θ f o r a l l θ ∴ arccos ⁡ ( sin ⁡ ( 1 3 π ) ) = arccos ⁡ ( cos ⁡ ( 1 2 π − 1 3 π ) ) = arccos ⁡ ( cos ⁡ ( 1 6 π ) ) = 1 6 π

knowing that sin ⁡ ( π 3 ) = 3 2 arccos ⁡ ( sin ⁡ ( π 3 ) ) = arccos ⁡ ( 3 2 )we know that cos ⁡ ( π 6 ) = 3 2 so, π 6 = arccos ⁡ ( 3 2 ) arccos ⁡ ( sin ⁡ ( π 3 ) ) = arccos ⁡ ( 3 2 ) = π 6

By definition, cos ⁡ ( 1 2 π − θ ) = sin ⁡ θ f o r a l l θ ∴ arccos ⁡ ( sin ⁡ ( 1 3 π ) ) = arccos ⁡ ( cos ⁡ ( 1 2 π − 1 3 π ) ) = arccos ⁡ ( cos ⁡ ( 1 6 π ) ) = 1 6 π

Expert Answer to "How do you find the exact value of..."

iarc6io

Answered question

2022-07-19

How do you find the exact value of

\arccos (\sin (\frac{π}{3}))

Answer & Explanation

eishale2n

Beginner2022-07-20Added 15 answers

knowing that

\sin (\frac{π}{3}) = \frac{\sqrt{3}}{2}

\arccos (\sin (\frac{π}{3})) = \arccos (\frac{\sqrt{3}}{2})

we know that

\cos (\frac{π}{6}) = \frac{\sqrt{3}}{2}

so,

\frac{π}{6} = \arccos (\frac{\sqrt{3}}{2})

\arccos (\sin (\frac{π}{3})) = \arccos (\frac{\sqrt{3}}{2}) = \frac{π}{6}

valtricotinevh

Beginner2022-07-21Added 3 answers

By definition,

\cos (\frac{1}{2} π - θ) = \sin θ f o r a l l θ

∴ \arccos (\sin (\frac{1}{3} π)) = \arccos (\cos (\frac{1}{2} π - \frac{1}{3} π)) = \arccos (\cos (\frac{1}{6} π))

= \frac{1}{6} π

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