What is the value of 3tan^2(480) ?

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3tan^2(480)

3 tan^{2} (48 0^{\circ})

9

Solution steps

3 tan^{2} (48 0^{\circ})

tan (48 0^{\circ}) = tan (12 0^{\circ})

tan (48 0^{\circ})

Rewrite

48 0^{\circ}

18 0^{\circ} \cdot 2 + 12 0^{\circ}

= tan (18 0^{\circ} 2 + 12 0^{\circ})

Apply the periodicity of

tan

tan (x + 18 0^{\circ} \cdot k) = tan (x)

tan (18 0^{\circ} \cdot 2 + 12 0^{\circ}) = tan (12 0^{\circ})

= tan (12 0^{\circ})

= 3 tan^{2} (12 0^{\circ})

Rewrite using trig identities:

tan^{2} (12 0^{\circ}) = sec^{2} (12 0^{\circ}) - 1

tan^{2} (12 0^{\circ})

Use the Pythagorean identity:

tan^{2} (x) + 1 = sec^{2} (x)

tan^{2} (x) = sec^{2} (x) - 1

= sec^{2} (12 0^{\circ}) - 1

= 3 (sec^{2} (12 0^{\circ}) - 1)

Use the following trivial identity:

sec (12 0^{\circ}) = - 2

sec (12 0^{\circ})

sec (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} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} 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} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

= - 2

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

Simplify

3 ((- 2)^{2} - 1) : 9

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

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

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

= 3 (2^{2} - 1)

2^{2} = 4

= 3 (4 - 1)

Subtract the numbers:

4 - 1 = 3

= 3 \cdot 3

Multiply the numbers:

3 \cdot 3 = 9

= 9

= 9

cos (3.4)

\frac{tan ( 3 0 ^{\circ} ) + tan ( 6 0 ^{\circ} )}{1 - tan ( 3 0 ^{\circ} ) tan ( 6 0 ^{\circ} )}

sin (arccos (\frac{2}{3}) - arctan (\frac{1}{2}))

cos (7 5^{\circ}) cos (1 5^{\circ}) - sin (7 5^{\circ}) sin (1 5^{\circ})

\frac{22}{tan ( 1 2 ^{\circ} )}