What is the value of sqrt(1-sin^2(5pi)) ?

The value of sqrt(1-sin^2(5pi)) is 1

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sqrt(1-sin^2(5pi))

1 - sin^{2} (5 π)

1

Solution steps

1 - sin^{2} (5 π)

Rewrite using trig identities:

sin (5 π) = 0

sin (5 π)

sin (5 π) = sin (π)

sin (5 π)

Rewrite

5 π

2 π \cdot 2 + π

= sin (2 π 2 + π)

Apply the periodicity of

sin

sin (x + 2 π \cdot k) = sin (x)

sin (2 π \cdot 2 + π) = sin (π)

= sin (π)

= sin (π)

Use the following trivial identity:

sin (π) = 0

sin (π)

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= 0

= 0

= 1 - 0^{2}

Simplify

= 1

7 \cdot tan (3 6^{\circ})

sin (\frac{2 5}{5})

- cos (2 π) - sin (2 π)

15 cos (6 8^{\circ})

19.8 cos (5 2^{\circ})