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

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

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sin^2((5pi)/3)+cos((5pi)/3)

Solution

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

Solution

\frac{5}{4}

Decimal Notation

1.25

Solution steps

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

Rewrite using trig identities:

sin^{2} (\frac{5 π}{3}) = 1 - cos^{2} (\frac{5 π}{3})

sin^{2} (\frac{5 π}{3})

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

sin^{2} (x) = 1 - cos^{2} (x)

= 1 - cos^{2} (\frac{5 π}{3})

= 1 - cos^{2} (\frac{5 π}{3}) + cos (\frac{5 π}{3})

Rewrite using trig identities:

cos (\frac{5 π}{3}) = \frac{1}{2}

cos (\frac{5 π}{3})

Rewrite using trig identities:

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

cos (\frac{5 π}{3})

Write

cos (\frac{5 π}{3})

cos (π + \frac{2 π}{3})

= cos (π + \frac{2 π}{3})

Use the Angle Sum identity:

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

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

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

Use the following trivial identity:

cos (π) = (- 1)

cos (π)

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}

= (- 1)

Use the following trivial identity:

cos (\frac{2 π}{3}) = - \frac{1}{2}

cos (\frac{2 π}{3})

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{1}{2}

Use the following trivial identity:

sin (π) = 0

sin (π)

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}

= 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}

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

Simplify

= \frac{1}{2}

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

Simplify

1 - (\frac{1}{2})^{2} + \frac{1}{2} : \frac{5}{4}

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

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

Convert element to fraction:

1 = \frac{1}{1}

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

2^{2} = 4

2^{2}

2^{2} = 4

= 4

= \frac{1}{1} - \frac{1}{4} + \frac{1}{2}

Least Common Multiplier of

1, 4, 2 : 4

1, 4, 2

Least Common Multiplier (LCM)

Prime factorization of

1

Prime factorization of

4 : 2 \cdot 2

4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2

Prime factorization of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

Compute a number comprised of factors that appear in at least one of the following:

1, 4, 2

= 2 \cdot 2

Multiply the numbers:

2 \cdot 2 = 4

= 4

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

4

For

\frac{1}{1} :

multiply the denominator and numerator by

4

\frac{1}{1} = \frac{1 \cdot 4}{1 \cdot 4} = \frac{4}{4}

For

\frac{1}{2} :

multiply the denominator and numerator by

2

\frac{1}{2} = \frac{1 \cdot 2}{2 \cdot 2} = \frac{2}{4}

= \frac{4}{4} - \frac{1}{4} + \frac{2}{4}

Since the denominators are equal, combine the fractions:

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

= \frac{4 - 1 + 2}{4}

Add/Subtract the numbers:

4 - 1 + 2 = 5

= \frac{5}{4}

= \frac{5}{4}

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Frequently Asked Questions (FAQ)

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