What is the value of tan^2(pi/4)+sin^2(pi/6)+cos^2(pi/6) ?

The value of tan^2(pi/4)+sin^2(pi/6)+cos^2(pi/6) is 2

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tan^2(pi/4)+sin^2(pi/6)+cos^2(pi/6)

Solution

tan^{2} (\frac{π}{4}) + sin^{2} (\frac{π}{6}) + cos^{2} (\frac{π}{6})

Solution

2

Solution steps

tan^{2} (\frac{π}{4}) + sin^{2} (\frac{π}{6}) + cos^{2} (\frac{π}{6})

Use the following trivial identity:

tan (\frac{π}{4}) = 1

tan (\frac{π}{4})

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

= 1

Use the following trivial identity:

sin (\frac{π}{6}) = \frac{1}{2}

sin (\frac{π}{6})

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{1}{2}

Use the following trivial identity:

cos (\frac{π}{6}) = \frac{3}{2}

cos (\frac{π}{6})

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{3}{2}

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

Simplify

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

Combine the fractions

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

Apply rule

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

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

Add the numbers:

1 + 3 = 4

= \frac{4}{2 ^{2}}

Factor

4 : 2^{2}

Factor

4 = 2^{2}

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

Cancel the common factor:

2^{2}

= 1

= 1 + 1

Add the numbers:

1 + 1 = 2

= 2

= 2

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Frequently Asked Questions (FAQ)

What is the value of tan^2(pi/4)+sin^2(pi/6)+cos^2(pi/6) ?
The value of tan^2(pi/4)+sin^2(pi/6)+cos^2(pi/6) is 2