What is the value of tan^2(36) ?

The value of tan^2(36) is 5-2sqrt(5)

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Popular >

tan^2(36)

Solution

tan^{2} (3 6^{\circ})

Solution

5 - 25

Decimal Notation

0.52786 \dots

Solution steps

tan^{2} (3 6^{\circ})

Rewrite using trig identities:

tan (3 6^{\circ}) = \frac{2 ( 5 - 1 ) 5 - 5}{4}

tan (3 6^{\circ})

Rewrite using trig identities:

\frac{sin ( 3 6 ^{\circ} )}{cos ( 3 6 ^{\circ} )}

tan (3 6^{\circ})

Use the basic trigonometric identity:

tan (x) = \frac{sin ( x )}{cos ( x )}

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

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

Rewrite using trig identities:

sin (3 6^{\circ}) = \frac{2 5 - 5}{4}

sin (3 6^{\circ})

Show that:

cos (3 6^{\circ}) - sin (1 8^{\circ}) = \frac{1}{2}

Use the following product to sum identity:

2 sin (x) cos (y) = sin (x + y) - sin (x - y)

2 cos (3 6^{\circ}) sin (1 8^{\circ}) = sin (5 4^{\circ}) - sin (1 8^{\circ})

Show that:

2 cos (3 6^{\circ}) sin (1 8^{\circ}) = \frac{1}{2}

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

sin (7 2^{\circ}) = 2 sin (3 6^{\circ}) cos (3 6^{\circ})

sin (7 2^{\circ}) sin (3 6^{\circ}) = 4 sin (3 6^{\circ}) sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

Divide both sides by

sin (3 6^{\circ})

sin (7 2^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

Use the following identity:

sin (x) = cos (9 0^{\circ} - x)

sin (7 2^{\circ}) = cos (9 0^{\circ} - 7 2^{\circ})

cos (9 0^{\circ} - 7 2^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

cos (1 8^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

Divide both sides by

cos (1 8^{\circ})

1 = 4 sin (1 8^{\circ}) cos (3 6^{\circ})

Divide both sides by

2

\frac{1}{2} = 2 sin (1 8^{\circ}) cos (3 6^{\circ})

Substitute

\frac{1}{2} = 2 sin (1 8^{\circ}) cos (3 6^{\circ})

\frac{1}{2} = sin (5 4^{\circ}) - sin (1 8^{\circ})

sin (5 4^{\circ}) = cos (9 0^{\circ} - 5 4^{\circ})

\frac{1}{2} = cos (9 0^{\circ} - 5 4^{\circ}) - sin (1 8^{\circ})

\frac{1}{2} = cos (3 6^{\circ}) - sin (1 8^{\circ})

Show that:

cos (3 6^{\circ}) + sin (1 8^{\circ}) = \frac{5}{4}

Use the factorization rule:

a^{2} - b^{2} = (a + b) (a - b)

a = cos (3 6^{\circ}) + sin (1 8^{\circ})

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - (cos (3 6^{\circ}) - sin (1 8^{\circ}))^{2} = ((cos (3 6^{\circ}) + sin (1 8^{\circ})) + (cos (3 6^{\circ}) - sin (1 8^{\circ}))) ((cos (3 6^{\circ}) + sin (1 8^{\circ})) - (cos (3 6^{\circ}) - sin (1 8^{\circ})))

Refine

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - (cos (3 6^{\circ}) - sin (1 8^{\circ}))^{2} = 2 (2 cos (3 6^{\circ}) sin (1 8^{\circ}))

Show that:

2 cos (3 6^{\circ}) sin (1 8^{\circ}) = \frac{1}{2}

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

sin (7 2^{\circ}) = 2 sin (3 6^{\circ}) cos (3 6^{\circ})

sin (7 2^{\circ}) sin (3 6^{\circ}) = 4 sin (3 6^{\circ}) sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

Divide both sides by

sin (3 6^{\circ})

sin (7 2^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

Use the following identity:

sin (x) = cos (9 0^{\circ} - x)

sin (7 2^{\circ}) = cos (9 0^{\circ} - 7 2^{\circ})

cos (9 0^{\circ} - 7 2^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

cos (1 8^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

Divide both sides by

cos (1 8^{\circ})

1 = 4 sin (1 8^{\circ}) cos (3 6^{\circ})

Divide both sides by

2

\frac{1}{2} = 2 sin (1 8^{\circ}) cos (3 6^{\circ})

Substitute

2 cos (3 6^{\circ}) sin (1 8^{\circ}) = \frac{1}{2}

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - (cos (3 6^{\circ}) - sin (1 8^{\circ}))^{2} = 1

Substitute

cos (3 6^{\circ}) - sin (1 8^{\circ}) = \frac{1}{2}

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - (\frac{1}{2})^{2} = 1

Refine

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - \frac{1}{4} = 1

Add

\frac{1}{4}

to both sides

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - \frac{1}{4} + \frac{1}{4} = 1 + \frac{1}{4}

Refine

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} = \frac{5}{4}

Take the square root of both sides

cos (3 6^{\circ}) + sin (1 8^{\circ}) = \pm \frac{5}{4}

cos (3 6^{\circ})

cannot be negative

sin (1 8^{\circ})

cannot be negative

cos (3 6^{\circ}) + sin (1 8^{\circ}) = \frac{5}{4}

Add the following equations

cos (3 6^{\circ}) + sin (1 8^{\circ}) = \frac{5}{2}

((cos (3 6^{\circ}) + sin (1 8^{\circ})) + (cos (3 6^{\circ}) - sin (1 8^{\circ}))) = (\frac{5}{2} + \frac{1}{2})

Refine

cos (3 6^{\circ}) = \frac{5 + 1}{4}

Square both sides

(cos (3 6^{\circ}))^{2} = (\frac{5 + 1}{4})^{2}

Use the following identity:

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

sin^{2} (3 6^{\circ}) = 1 - cos^{2} (3 6^{\circ})

Substitute

cos (3 6^{\circ}) = \frac{5 + 1}{4}

sin^{2} (3 6^{\circ}) = 1 - (\frac{5 + 1}{4})^{2}

Refine

sin^{2} (3 6^{\circ}) = \frac{5 - 5}{8}

Take the square root of both sides

sin (3 6^{\circ}) = \pm \frac{5 - 5}{8}

sin (3 6^{\circ})

cannot be negative

sin (3 6^{\circ}) = \frac{5 - 5}{8}

Refine

sin (3 6^{\circ}) = \frac{\frac{5 - 5}{2}}{2}

= \frac{\frac{5 - 5}{2}}{2}

Simplify

= \frac{2 5 - 5}{4}

Rewrite using trig identities:

cos (3 6^{\circ}) = \frac{5 + 1}{4}

cos (3 6^{\circ})

Show that:

cos (3 6^{\circ}) - sin (1 8^{\circ}) = \frac{1}{2}

Use the following product to sum identity:

2 sin (x) cos (y) = sin (x + y) - sin (x - y)

2 cos (3 6^{\circ}) sin (1 8^{\circ}) = sin (5 4^{\circ}) - sin (1 8^{\circ})

Show that:

2 cos (3 6^{\circ}) sin (1 8^{\circ}) = \frac{1}{2}

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

sin (7 2^{\circ}) = 2 sin (3 6^{\circ}) cos (3 6^{\circ})

sin (7 2^{\circ}) sin (3 6^{\circ}) = 4 sin (3 6^{\circ}) sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

Divide both sides by

sin (3 6^{\circ})

sin (7 2^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

Use the following identity:

sin (x) = cos (9 0^{\circ} - x)

sin (7 2^{\circ}) = cos (9 0^{\circ} - 7 2^{\circ})

cos (9 0^{\circ} - 7 2^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

cos (1 8^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

Divide both sides by

cos (1 8^{\circ})

1 = 4 sin (1 8^{\circ}) cos (3 6^{\circ})

Divide both sides by

2

\frac{1}{2} = 2 sin (1 8^{\circ}) cos (3 6^{\circ})

Substitute

\frac{1}{2} = 2 sin (1 8^{\circ}) cos (3 6^{\circ})

\frac{1}{2} = sin (5 4^{\circ}) - sin (1 8^{\circ})

sin (5 4^{\circ}) = cos (9 0^{\circ} - 5 4^{\circ})

\frac{1}{2} = cos (9 0^{\circ} - 5 4^{\circ}) - sin (1 8^{\circ})

\frac{1}{2} = cos (3 6^{\circ}) - sin (1 8^{\circ})

Show that:

cos (3 6^{\circ}) + sin (1 8^{\circ}) = \frac{5}{4}

Use the factorization rule:

a^{2} - b^{2} = (a + b) (a - b)

a = cos (3 6^{\circ}) + sin (1 8^{\circ})

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - (cos (3 6^{\circ}) - sin (1 8^{\circ}))^{2} = ((cos (3 6^{\circ}) + sin (1 8^{\circ})) + (cos (3 6^{\circ}) - sin (1 8^{\circ}))) ((cos (3 6^{\circ}) + sin (1 8^{\circ})) - (cos (3 6^{\circ}) - sin (1 8^{\circ})))

Refine

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - (cos (3 6^{\circ}) - sin (1 8^{\circ}))^{2} = 2 (2 cos (3 6^{\circ}) sin (1 8^{\circ}))

Show that:

2 cos (3 6^{\circ}) sin (1 8^{\circ}) = \frac{1}{2}

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

sin (7 2^{\circ}) = 2 sin (3 6^{\circ}) cos (3 6^{\circ})

sin (7 2^{\circ}) sin (3 6^{\circ}) = 4 sin (3 6^{\circ}) sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

Divide both sides by

sin (3 6^{\circ})

sin (7 2^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

Use the following identity:

sin (x) = cos (9 0^{\circ} - x)

sin (7 2^{\circ}) = cos (9 0^{\circ} - 7 2^{\circ})

cos (9 0^{\circ} - 7 2^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

cos (1 8^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

Divide both sides by

cos (1 8^{\circ})

1 = 4 sin (1 8^{\circ}) cos (3 6^{\circ})

Divide both sides by

2

\frac{1}{2} = 2 sin (1 8^{\circ}) cos (3 6^{\circ})

Substitute

2 cos (3 6^{\circ}) sin (1 8^{\circ}) = \frac{1}{2}

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - (cos (3 6^{\circ}) - sin (1 8^{\circ}))^{2} = 1

Substitute

cos (3 6^{\circ}) - sin (1 8^{\circ}) = \frac{1}{2}

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - (\frac{1}{2})^{2} = 1

Refine

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - \frac{1}{4} = 1

Add

\frac{1}{4}

to both sides

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - \frac{1}{4} + \frac{1}{4} = 1 + \frac{1}{4}

Refine

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} = \frac{5}{4}

Take the square root of both sides

cos (3 6^{\circ}) + sin (1 8^{\circ}) = \pm \frac{5}{4}

cos (3 6^{\circ})

cannot be negative

sin (1 8^{\circ})

cannot be negative

cos (3 6^{\circ}) + sin (1 8^{\circ}) = \frac{5}{4}

Add the following equations

cos (3 6^{\circ}) + sin (1 8^{\circ}) = \frac{5}{2}

((cos (3 6^{\circ}) + sin (1 8^{\circ})) + (cos (3 6^{\circ}) - sin (1 8^{\circ}))) = (\frac{5}{2} + \frac{1}{2})

Refine

cos (3 6^{\circ}) = \frac{5 + 1}{4}

= \frac{5 + 1}{4}

= \frac{\frac{2 5 - 5}{4}}{\frac{5 + 1}{4}}

Simplify

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

\frac{\frac{2 5 - 5}{4}}{\frac{5 + 1}{4}}

Divide fractions:

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

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

Cancel the common factor:

4

= \frac{2 5 - 5}{5 + 1}

Rationalize

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

\frac{2 5 - 5}{5 + 1}

Multiply by the conjugate

\frac{5 - 1}{5 - 1}

= \frac{2 5 - 5 ( 5 - 1 )}{( 5 + 1 ) ( 5 - 1 )}

(5 + 1) (5 - 1) = 4

(5 + 1) (5 - 1)

Apply Difference of Two Squares Formula:

(a + b) (a - b) = a^{2} - b^{2}

a = 5, b = 1

= (5)^{2} - 1^{2}

Simplify

(5)^{2} - 1^{2} : 4

(5)^{2} - 1^{2}

Apply rule

1^{a} = 1

1^{2} = 1

= (5)^{2} - 1

(5)^{2} = 5

(5)^{2}

Apply radical rule:

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

= (5^{\frac{1}{2}})^{2}

Apply exponent rule:

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

= 5^{\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

= 5

= 5 - 1

Subtract the numbers:

5 - 1 = 4

= 4

= 4

= \frac{2 ( 5 - 1 ) 5 - 5}{4}

= \frac{2 ( 5 - 1 ) 5 - 5}{4}

= \frac{2 ( 5 - 1 ) 5 - 5}{4}

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

Simplify

(\frac{2 ( 5 - 1 ) 5 - 5}{4})^{2} : 5 - 25

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

Apply exponent rule:

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

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

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

(2 (5 - 1) 5 - 5)^{2} = (2)^{2} (5 - 5)^{2} (5 - 1)^{2}

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

(2)^{2} : 2

Apply radical rule:

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

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

Apply exponent rule:

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

= 2^{\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

= 2

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

(5 - 5)^{2} : 5 - 5

Apply radical rule:

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

= ((5 - 5)^{\frac{1}{2}})^{2}

Apply exponent rule:

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

= (5 - 5)^{\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

= 5 - 5

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

Factor

4^{2} : 2^{4}

Factor

4 = 2^{2}

= (2^{2})^{2}

Simplify

(2^{2})^{2} : 2^{4}

(2^{2})^{2}

Apply exponent rule:

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

= 2^{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= 2^{4}

= 2^{4}

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

Cancel the common factor:

2

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

(5 - 1)^{2} = 6 - 25

(5 - 1)^{2}

Apply Perfect Square Formula:

(a - b)^{2} = a^{2} - 2 ab + b^{2}

a = 5, b = 1

= (5)^{2} - 25 \cdot 1 + 1^{2}

Simplify

(5)^{2} - 25 \cdot 1 + 1^{2} : 6 - 25

(5)^{2} - 25 \cdot 1 + 1^{2}

Apply rule

1^{a} = 1

1^{2} = 1

= (5)^{2} - 2 \cdot 1 \cdot 5 + 1

(5)^{2} = 5

(5)^{2}

Apply radical rule:

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

= (5^{\frac{1}{2}})^{2}

Apply exponent rule:

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

= 5^{\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

= 5

25 \cdot 1 = 25

25 \cdot 1

Multiply the numbers:

2 \cdot 1 = 2

= 25

= 5 - 25 + 1

Add the numbers:

5 + 1 = 6

= 6 - 25

= 6 - 25

= \frac{( 6 - 2 5 ) ( 5 - 5 )}{2 ^{3}}

Factor

6 - 25 : 2 (3 - 5)

6 - 25

Rewrite as

= 2 \cdot 3 - 25

Factor out common term

2

= 2 (3 - 5)

= \frac{2 ( 3 - 5 ) ( 5 - 5 )}{2 ^{3}}

Cancel the common factor:

2

= \frac{( 3 - 5 ) ( 5 - 5 )}{2 ^{2}}

2^{2} = 4

= \frac{( 3 - 5 ) ( 5 - 5 )}{4}

Expand

(3 - 5) (5 - 5) : 20 - 85

(3 - 5) (5 - 5)

Apply FOIL method:

(a + b) (c + d) = a c + a d + b c + b d

a = 3, b = - 5, c = 5, d = - 5

= 3 \cdot 5 + 3 (- 5) + (- 5) \cdot 5 + (- 5) (- 5)

Apply minus-plus rules

+ (- a) = - a, (- a) (- b) = ab

= 3 \cdot 5 - 35 - 55 + 55

Simplify

3 \cdot 5 - 35 - 55 + 55 : 20 - 85

3 \cdot 5 - 35 - 55 + 55

Add similar elements:

- 35 - 55 = - 85

= 3 \cdot 5 - 85 + 55

Multiply the numbers:

3 \cdot 5 = 15

= 15 - 85 + 55

Apply radical rule:

a a = a

55 = 5

= 15 - 85 + 5

Add the numbers:

15 + 5 = 20

= 20 - 85

= 20 - 85

= \frac{20 - 8 5}{4}

Factor

20 - 85 : 4 (5 - 25)

20 - 85

Rewrite as

= 4 \cdot 5 - 4 \cdot 25

Factor out common term

4

= 4 (5 - 25)

= \frac{4 ( 5 - 2 5 )}{4}

Divide the numbers:

\frac{4}{4} = 1

= 5 - 25

= 5 - 25

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Frequently Asked Questions (FAQ)

What is the value of tan^2(36) ?
The value of tan^2(36) is 5-2sqrt(5)