What is the value of cos(190)cos(10)+sin(190)sin(10) ?

The value of cos(190)cos(10)+sin(190)sin(10) is -1

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cos(190)cos(10)+sin(190)sin(10)

cos (19 0^{\circ}) cos (1 0^{\circ}) + sin (19 0^{\circ}) sin (1 0^{\circ})

- 1

Solution steps

cos (19 0^{\circ}) cos (1 0^{\circ}) + sin (19 0^{\circ}) sin (1 0^{\circ})

Rewrite using trig identities:

cos (18 0^{\circ})

cos (19 0^{\circ}) cos (1 0^{\circ}) + sin (19 0^{\circ}) sin (1 0^{\circ})

Use the Angle Difference identity:

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

= cos (19 0^{\circ} - 1 0^{\circ})

Simplify

= cos (18 0^{\circ})

= cos (18 0^{\circ})

Use the following trivial identity:

cos (18 0^{\circ}) = (- 1)

cos (18 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}

= (- 1)

= - 1

arcsin (sin (\frac{11 π}{8}))

sin (17 0^{\circ})

arcsin (\frac{3}{4})

cot (0) \cdot sin (0)

2 cos^{2} (\frac{5 π}{8}) - 1

What is the value of cos(190)cos(10)+sin(190)sin(10) ?
The value of cos(190)cos(10)+sin(190)sin(10) is -1