What is the value of sin((3pi)/4)-cos((3pi)/4) ?

The value of sin((3pi)/4)-cos((3pi)/4) is square root of 2

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sin((3pi)/4)-cos((3pi)/4)

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

2

Decimal Notation

1.41421 \dots

Solution steps

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

Use the following trivial identity:

sin (\frac{3 π}{4}) = \frac{2}{2}

sin (\frac{3 π}{4})

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{2}{2}

Use the following trivial identity:

cos (\frac{3 π}{4}) = - \frac{2}{2}

cos (\frac{3 π}{4})

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} 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{2}{2} - (- \frac{2}{2})

Simplify

\frac{2}{2} - (- \frac{2}{2}) : 2

\frac{2}{2} - (- \frac{2}{2})

Apply rule

- (- a) = a

= \frac{2}{2} + \frac{2}{2}

Add similar elements:

\frac{2}{2} + \frac{2}{2} = 2 \cdot \frac{2}{2}

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

Multiply fractions:

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

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

Cancel the common factor:

2

= 2

= 2

\frac{6.4}{sin ( 4 2 ^{\circ} )}

sin (- \frac{17 π}{12})

\frac{1}{cos ( 2 )}

arctan (- 1.6)

sin (- 39 0^{\circ})

What is the value of sin((3pi)/4)-cos((3pi)/4) ?
The value of sin((3pi)/4)-cos((3pi)/4) is square root of 2