What is the value of csc^2((7pi)/4) ?

The value of csc^2((7pi)/4) is 2

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csc^2((7pi)/4)

csc^{2} (\frac{7 π}{4})

2

Solution steps

csc^{2} (\frac{7 π}{4})

Rewrite using trig identities:

1 + cot^{2} (\frac{3 π}{4})

csc^{2} (\frac{7 π}{4})

Use the Pythagorean identity:

csc^{2} (x) = 1 + cot^{2} (x)

= 1 + cot^{2} (\frac{7 π}{4})

cot (\frac{7 π}{4}) = cot (\frac{3 π}{4})

cot (\frac{7 π}{4})

Rewrite

\frac{7 π}{4}

π + \frac{3 π}{4}

= cot (π + \frac{3 π}{4})

Apply the periodicity of

cot

cot (x + π) = cot (x)

cot (π + \frac{3 π}{4}) = cot (\frac{3 π}{4})

= cot (\frac{3 π}{4})

= 1 + cot^{2} (\frac{3 π}{4})

= 1 + cot^{2} (\frac{3 π}{4})

Use the following trivial identity:

cot (\frac{3 π}{4}) = - 1

cot (\frac{3 π}{4})

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

= - 1

= 1 + (- 1)^{2}

Simplify

1 + (- 1)^{2} : 2

1 + (- 1)^{2}

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

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

n

is even

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

= 1^{2}

Apply rule

1^{a} = 1

= 1

= 1 + 1

Add the numbers:

1 + 1 = 2

= 2

= 2

ln (3^{2} + 4^{3} - 0^{4}) + cos (3 + 4) + 0^{2}

cos (\frac{3 π}{4} - \frac{11 π}{6})

e^{- 1} cos (1)

\frac{120}{tan ( 1 9 ^{\circ} )}

\frac{12}{tan ( 5 3 ^{\circ} )}