What is the value of cos((3pi)/4+(5pi)/6) ?

The value of cos((3pi)/4+(5pi)/6) is (sqrt(2)(sqrt(3)-1))/4

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cos((3pi)/4+(5pi)/6)

cos (\frac{3 π}{4} + \frac{5 π}{6})

\frac{2 ( 3 - 1 )}{4}

Solution steps

cos (\frac{3 π}{4} + \frac{5 π}{6})

Rewrite using trig identities:

cos (\frac{5 π}{6}) cos (\frac{3 π}{4}) - sin (\frac{5 π}{6}) sin (\frac{3 π}{4})

= cos (\frac{5 π}{6}) cos (\frac{3 π}{4}) - sin (\frac{5 π}{6}) sin (\frac{3 π}{4})

Use the following trivial identity:

cos (\frac{5 π}{6}) = - \frac{3}{2}

Use the following trivial identity:

cos (\frac{3 π}{4}) = - \frac{2}{2}

Use the following trivial identity:

sin (\frac{5 π}{6}) = \frac{1}{2}

Use the following trivial identity:

sin (\frac{3 π}{4}) = \frac{2}{2}

= (- \frac{3}{2}) (- \frac{2}{2}) - \frac{1}{2} \cdot \frac{2}{2}

Simplify

(- \frac{3}{2}) (- \frac{2}{2}) - \frac{1}{2} \cdot \frac{2}{2} : \frac{2 ( 3 - 1 )}{4}

= \frac{2 ( 3 - 1 )}{4}

- (\frac{2}{π}) (4) cos (1 0^{\circ}) - cos (8)

tan^{2} (3 0^{\circ}) + cos^{2} (6 0^{\circ})

\frac{sin ( 6 0 ^{\circ} ) cos ( 3 0 ^{\circ} )}{tan ( 4 5 ^{\circ} )}

\frac{1}{e ^{0.01} ( sin ( 0.01 ) + cos ( 0.01 ) )}

\frac{37. 2 ^{2} sin ( 2 \cdot 14.1 )}{9.8}

What is the value of cos((3pi)/4+(5pi)/6) ?
The value of cos((3pi)/4+(5pi)/6) is (sqrt(2)(sqrt(3)-1))/4