What is the value of cos(160)cos(40)+sin(160)sin(40) ?

The value of cos(160)cos(40)+sin(160)sin(40) is -1/2

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cos(160)cos(40)+sin(160)sin(40)

cos (16 0^{\circ}) cos (4 0^{\circ}) + sin (16 0^{\circ}) sin (4 0^{\circ})

- \frac{1}{2}

Decimal Notation

- 0.5

Solution steps

cos (16 0^{\circ}) cos (4 0^{\circ}) + sin (16 0^{\circ}) sin (4 0^{\circ})

Rewrite using trig identities:

cos (12 0^{\circ})

cos (16 0^{\circ}) cos (4 0^{\circ}) + sin (16 0^{\circ}) sin (4 0^{\circ})

Use the Angle Difference identity:

cos (s) cos (t) + sin (s) sin (t) = cos (s - t)

= cos (16 0^{\circ} - 4 0^{\circ})

Simplify

= cos (12 0^{\circ})

= cos (12 0^{\circ})

Use the following trivial identity:

cos (12 0^{\circ}) = - \frac{1}{2}

cos (12 0^{\circ})

cos (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= - \frac{1}{2}

= - \frac{1}{2}

0.5 sin (3 0^{\circ})

28 sin (4 5^{\circ})

arccos (\frac{8}{12})

(sin (4 5^{\circ}))^{2} + (cos (4 5^{\circ}))^{2}

8 cos (30 0^{\circ})

What is the value of cos(160)cos(40)+sin(160)sin(40) ?
The value of cos(160)cos(40)+sin(160)sin(40) is -1/2