What is the value of (sin(60))/(cos(45)) ?

The value of (sin(60))/(cos(45)) is sqrt(3/2)

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(sin(60))/(cos(45))

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

\frac{3}{2}

Decimal Notation

1.22474 \dots

Solution steps

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

Use the following trivial identity:

sin (6 0^{\circ}) = \frac{3}{2}

sin (6 0^{\circ})

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{3}{2}

Use the following trivial identity:

cos (4 5^{\circ}) = \frac{2}{2}

cos (4 5^{\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{2}{2}

= \frac{\frac{3}{2}}{\frac{2}{2}}

Simplify

\frac{\frac{3}{2}}{\frac{2}{2}} : \frac{3}{2}

\frac{\frac{3}{2}}{\frac{2}{2}}

Divide fractions:

\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{a \cdot d}{b \cdot c}

= \frac{3 \cdot 2}{2 2}

Cancel the common factor:

2

= \frac{3}{2}

Combine same powers :

\frac{x}{y} = \frac{x}{y}

= \frac{3}{2}

= \frac{3}{2}

- 2 sec (\frac{π}{4})

cot (29 0^{\circ})

tan (2 + \frac{π}{4})

sec^{2} (\frac{3 π}{2})

\frac{10}{cos ( 6 8 ^{\circ} )}

What is the value of (sin(60))/(cos(45)) ?
The value of (sin(60))/(cos(45)) is sqrt(3/2)