What is the value of 6cos(7pi) ?

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6cos(7pi)

6 cos (7 π)

- 6

Solution steps

6 cos (7 π)

Rewrite using trig identities:

cos (7 π) = - 1

cos (7 π)

cos (7 π) = cos (π)

cos (7 π)

Rewrite

7 π

2 π \cdot 3 + π

= cos (2 π 3 + π)

Apply the periodicity of

cos

cos (x + 2 π \cdot k) = cos (x)

cos (2 π \cdot 3 + π) = cos (π)

= cos (π)

= cos (π)

Use the following trivial identity:

cos (π) = (- 1)

cos (π)

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}

= (- 1)

= - 1

= 6 (- 1)

Simplify

= - 6

38 cos (6 0^{\circ})

- 50 \cdot 9.81 \cdot sin (3 7^{\circ}) \cdot 3

2 (- 2.19) cos (2 (- 2.19)) - ((- 2.19) + 1)^{2}

sin (\frac{5}{3}) π

1 + 2 cos (\frac{4 π}{3})