What is the value of cos^2((5pi)/4)-sin^2((2pi)/3) ?

The value of cos^2((5pi)/4)-sin^2((2pi)/3) is -1/4

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cos^2((5pi)/4)-sin^2((2pi)/3)

Solution

cos^{2} (\frac{5 π}{4}) - sin^{2} (\frac{2 π}{3})

Solution

- \frac{1}{4}

Decimal Notation

- 0.25

Solution steps

cos^{2} (\frac{5 π}{4}) - sin^{2} (\frac{2 π}{3})

Rewrite using trig identities:

cos^{2} (\frac{5 π}{4}) = \frac{1 + cos ( \frac{π}{2} )}{2}

cos^{2} (\frac{5 π}{4})

Use the following identity:

cos^{2} (θ) = \frac{1 + cos ( 2 θ )}{2}

Use the Double Angle identity

cos (2 θ) = 2 cos^{2} (θ) - 1

Switch sides

2 cos^{2} (θ) - 1 = cos (2 θ)

Add

1

to both sides

2 sin^{2} (θ) = 1 + cos (2 θ)

Divide both sides by

2

cos^{2} (θ) = \frac{1 + cos ( 2 θ )}{2}

= \frac{1 + cos ( 2 \cdot \frac{5 π}{4} )}{2}

Simplify:

2 \cdot \frac{5 π}{4} = \frac{5 π}{2}

2 \cdot \frac{5 π}{4}

Multiply fractions:

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

= \frac{5 π 2}{4}

Multiply the numbers:

5 \cdot 2 = 10

= \frac{10 π}{4}

Cancel the common factor:

2

= \frac{5 π}{2}

= \frac{1 + cos ( \frac{5 π}{2} )}{2}

cos (\frac{5 π}{2}) = cos (\frac{π}{2})

cos (\frac{5 π}{2})

Rewrite

\frac{5 π}{2}

2 π + \frac{π}{2}

= cos (2 π + \frac{π}{2})

Apply the periodicity of

cos

cos (x + 2 π) = cos (x)

cos (2 π + \frac{π}{2}) = cos (\frac{π}{2})

= cos (\frac{π}{2})

= \frac{1 + cos ( \frac{π}{2} )}{2}

= \frac{1 + cos ( \frac{π}{2} )}{2} - sin^{2} (\frac{2 π}{3})

\frac{1 + cos ( \frac{π}{2} )}{2} - sin^{2} (\frac{2 π}{3}) = \frac{1 + cos ( \frac{π}{2} ) - 2 sin ^{2} ( \frac{2 π}{3} )}{2}

\frac{1 + cos ( \frac{π}{2} )}{2} - sin^{2} (\frac{2 π}{3})

Convert element to fraction:

sin^{2} (\frac{2 π}{3}) = \frac{sin ^{2} ( \frac{2 π}{3} ) 2}{2}

= \frac{1 + cos ( \frac{π}{2} )}{2} - \frac{sin ^{2} ( \frac{2 π}{3} ) \cdot 2}{2}

Since the denominators are equal, combine the fractions:

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

= \frac{1 + cos ( \frac{π}{2} ) - sin ^{2} ( \frac{2 π}{3} ) \cdot 2}{2}

= \frac{1 + cos ( \frac{π}{2} ) - 2 sin ^{2} ( \frac{2 π}{3} )}{2}

Use the following trivial identity:

cos (\frac{π}{2}) = 0

cos (\frac{π}{2})

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}

= 0

Use the following trivial identity:

sin (\frac{2 π}{3}) = \frac{3}{2}

sin (\frac{2 π}{3})

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{3}{2}

= \frac{1 + 0 - 2 ( \frac{3}{2} ) ^{2}}{2}

Simplify

\frac{1 + 0 - 2 ( \frac{3}{2} ) ^{2}}{2} : - \frac{1}{4}

\frac{1 + 0 - 2 ( \frac{3}{2} ) ^{2}}{2}

Add the numbers:

1 + 0 = 1

= \frac{- 2 ( \frac{3}{2} ) ^{2} + 1}{2}

2 (\frac{3}{2})^{2} = \frac{3}{2}

2 (\frac{3}{2})^{2}

(\frac{3}{2})^{2} = \frac{3}{2 ^{2}}

(\frac{3}{2})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{( 3 ) ^{2}}{2 ^{2}}

(3)^{2} : 3

Apply radical rule:

a = a^{\frac{1}{2}}

= (3^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= 3^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

Multiply fractions:

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

= \frac{1 \cdot 2}{2}

Cancel the common factor:

2

= 1

= 3

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

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

Multiply fractions:

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

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

Multiply the numbers:

3 \cdot 2 = 6

= \frac{6}{2 ^{2}}

Factor

6 : 2 \cdot 3

Factor

6 = 2 \cdot 3

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

Cancel the common factor:

2

= \frac{3}{2}

= \frac{- \frac{3}{2} + 1}{2}

Join

- \frac{3}{2} + 1 : - \frac{1}{2}

- \frac{3}{2} + 1

Convert element to fraction:

1 = \frac{1 \cdot 2}{2}

= - \frac{3}{2} + \frac{1 \cdot 2}{2}

Since the denominators are equal, combine the fractions:

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

= \frac{- 3 + 1 \cdot 2}{2}

- 3 + 1 \cdot 2 = - 1

- 3 + 1 \cdot 2

Multiply the numbers:

1 \cdot 2 = 2

= - 3 + 2

Add/Subtract the numbers:

- 3 + 2 = - 1

= - 1

= \frac{- 1}{2}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{1}{2}

= \frac{- \frac{1}{2}}{2}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{\frac{1}{2}}{2}

Apply the fraction rule:

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

\frac{\frac{1}{2}}{2} = \frac{1}{2 \cdot 2}

= - \frac{1}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= - \frac{1}{4}

= - \frac{1}{4}

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Frequently Asked Questions (FAQ)

What is the value of cos^2((5pi)/4)-sin^2((2pi)/3) ?
The value of cos^2((5pi)/4)-sin^2((2pi)/3) is -1/4