What is the value of cos(2arccos(-1)) ?

The value of cos(2arccos(-1)) is 1

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cos(2arccos(-1))

cos (2 arccos (- 1))

1

Solution steps

cos (2 arccos (- 1))

Rewrite using trig identities:

2 cos^{2} (arccos (- 1)) - 1

cos (2 arccos (- 1))

Use the Double Angle identity:

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

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

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

Rewrite using trig identities:

cos (arccos (- 1)) = - 1

Use the following identity:

cos (arccos (x)) = x

= - 1

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

Simplify

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

2 (- 1)^{2} - 1

2 (- 1)^{2} = 2

2 (- 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

= 2 \cdot 1

Multiply the numbers:

2 \cdot 1 = 2

= 2

= 2 - 1

Subtract the numbers:

2 - 1 = 1

= 1

= 1

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