What is the value of 2cos(pi/2)+sin(2(pi/2)) ?

The value of 2cos(pi/2)+sin(2(pi/2)) is 0

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2cos(pi/2)+sin(2(pi/2))

2 cos (\frac{π}{2}) + sin (2 (\frac{π}{2}))

0

Solution steps

2 cos (\frac{π}{2}) + sin (2 (\frac{π}{2}))

= 2 cos (\frac{π}{2}) + sin (2 \cdot \frac{π}{2})

Use the following trivial identity:

cos (\frac{π}{2}) = 0

cos (\frac{π}{2})

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= 0

Rewrite using trig identities:

sin (2 \cdot \frac{π}{2}) = 0

sin (2 \cdot \frac{π}{2})

Simplify:

2 \cdot \frac{π}{2} = π

2 \cdot \frac{π}{2}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{π 2}{2}

Cancel the common factor:

2

= 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

= 2 \cdot 0 + 0

Simplify

= 0

300 \cdot cos (5 5^{\circ}) \cdot 42

20 cos (5 5^{\circ})

sec (arcsin (\frac{2 5}{5}))

sin (3 5^{\circ}) cos (1 0^{\circ}) + cos (3 5^{\circ}) sin (1 0^{\circ})

- sin^{2} (0)

What is the value of 2cos(pi/2)+sin(2(pi/2)) ?
The value of 2cos(pi/2)+sin(2(pi/2)) is 0