What is the value of sin^2(30)-cot^2(60) ?

The value of sin^2(30)-cot^2(60) is -1/12

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sin^2(30)-cot^2(60)

Solution

sin^{2} (3 0^{\circ}) - cot^{2} (6 0^{\circ})

Solution

- \frac{1}{12}

Decimal Notation

- 0.08333 \dots

Solution steps

sin^{2} (3 0^{\circ}) - cot^{2} (6 0^{\circ})

Use the following trivial identity:

sin (3 0^{\circ}) = \frac{1}{2}

sin (3 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{1}{2}

Use the following trivial identity:

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

cot (6 0^{\circ})

cot (x)

periodicity table with

18 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} cot (x) \mp \infty 31 \frac{3}{3} 0 - \frac{3}{3} - 1 - 3

= \frac{3}{3}

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

Simplify

(\frac{1}{2})^{2} - (\frac{3}{3})^{2} : - \frac{1}{12}

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

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

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

Apply exponent rule:

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

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

Cancel the common factor:

3

= \frac{1}{3}

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

2^{2} = 4

= \frac{1}{4} - \frac{1}{3}

Least Common Multiplier of

4, 3 : 12

4, 3

Least Common Multiplier (LCM)

Prime factorization of

4 : 2 \cdot 2

4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2

Prime factorization of

3 : 3

3

3

is a prime number, therefore no factorization is possible

= 3

Multiply each factor the greatest number of times it occurs in either

4

3

= 2 \cdot 2 \cdot 3

Multiply the numbers:

2 \cdot 2 \cdot 3 = 12

= 12

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

12

For

\frac{1}{4} :

multiply the denominator and numerator by

3

\frac{1}{4} = \frac{1 \cdot 3}{4 \cdot 3} = \frac{3}{12}

For

\frac{1}{3} :

multiply the denominator and numerator by

4

\frac{1}{3} = \frac{1 \cdot 4}{3 \cdot 4} = \frac{4}{12}

= \frac{3}{12} - \frac{4}{12}

Since the denominators are equal, combine the fractions:

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

= \frac{3 - 4}{12}

Subtract the numbers:

3 - 4 = - 1

= \frac{- 1}{12}

Apply the fraction rule:

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

= - \frac{1}{12}

= - \frac{1}{12}

Popular Examples

1+2cos(60)

1 + 2 cos (6 0^{\circ})

sin((6pi)/(12))

sin (\frac{6 π}{12})

5*sin(20)

5 \cdot sin (2 0^{\circ})

arctan(2.90696E20)

arctan (2.90696 E 20)

cos(pi/2)-sin((5pi)/3)+tan((9pi)/4)-cos((5pi)/6)+tan((7pi)/6)

cos (\frac{π}{2}) - sin (\frac{5 π}{3}) + tan (\frac{9 π}{4}) - cos (\frac{5 π}{6}) + tan (\frac{7 π}{6})

Frequently Asked Questions (FAQ)

What is the value of sin^2(30)-cot^2(60) ?
The value of sin^2(30)-cot^2(60) is -1/12