What is the value of cos(40)cos(10)+sin(40)sin(10) ?

The value of cos(40)cos(10)+sin(40)sin(10) is (sqrt(3))/2

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cos(40)cos(10)+sin(40)sin(10)

cos (4 0^{\circ}) cos (1 0^{\circ}) + sin (4 0^{\circ}) sin (1 0^{\circ})

\frac{3}{2}

Decimal Notation

0.86602 \dots

Solution steps

cos (4 0^{\circ}) cos (1 0^{\circ}) + sin (4 0^{\circ}) sin (1 0^{\circ})

Rewrite using trig identities:

cos (3 0^{\circ})

cos (4 0^{\circ}) cos (1 0^{\circ}) + sin (4 0^{\circ}) sin (1 0^{\circ})

Use the Angle Difference identity:

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

= cos (4 0^{\circ} - 1 0^{\circ})

Simplify

= cos (3 0^{\circ})

= cos (3 0^{\circ})

Use the following trivial identity:

cos (3 0^{\circ}) = \frac{3}{2}

cos (3 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{3}{2}

= \frac{3}{2}

tan (- \frac{19 π}{6})

1 - 2 arctan (1)

tan (66 0^{\circ})

arcsin (0.23)

600 cos (3 0^{\circ})

What is the value of cos(40)cos(10)+sin(40)sin(10) ?
The value of cos(40)cos(10)+sin(40)sin(10) is (sqrt(3))/2