Verified answer

➊ cos[ 2 cos⁻¹(-√2/2)]     ← Note:   cos⁻¹ only returns angles between 0 and π

                                                     (i.e. 1st or 2nd Quadrant angles only).

                                   So, cos⁻¹(-√2/2) is an angle θ for which its cosine is

                                   negative. That puts θ in the 2nd Quadrant so that

                                   θ = 3π/4 = cos⁻¹(-√2/2)

  = cos[ 2 (3π/4)]            ← by substitution

  = 0                       ← ANSWER

➋ sin[ 2 sin⁻¹(1/-2]            ← Note:   sin⁻¹ only returns angles between

                                    -π/2 and π/2 (i.e. 4th or 1st Quadrant angles only).

                     So, sin⁻¹(1/-2) = sin⁻¹(-1/2) so that we are looking for an angle θ

                     or which its sine is negative. That puts θ in the 4th Quadrant

                     expressed as a negative angle so that sin⁻¹(-1/2) = -π/6  NOT  11π/6

  = sin[ 2 (-π/6)]            ← by substitution

  = sin(-π/3)            ← remember ... sin(-β) = -sin(β)

  = -sin(π/3)

  = -(√3)/2            ← ANSWER

➌ tan[2 sin⁻¹(12/13)]            ← Think of this as tan[2θ]

                                         Note:  sin⁻¹ only returns angles between

                                 -π/2 and π/2 (i.e. 4th or 1st Quadrant angles only).

Since 12/13 is positive and less than 1.57 (i.e. less than π/2),

sin⁻¹(12/13)  =  θ  is a 1st Quadrant angle.

So, sinθ = 12/13  and 0 < θ < π/2    ☚ ✰

cosθ = √[1 - sin²θ]            ← Positive square root only because 0 < θ < π/2

cosθ = √[1 - (12/13)²]            ← by substitution

cosθ = √[13² - 12²] / 13

cosθ = √[(13-12)(13+12)] / 13            ← factored difference of squares

cosθ = 5/13    ☚ ✰

tanθ = sinθ/cosθ = (12/13) / (5/13) = 12/5    ☚ ✰

                 2 tanθ            2(12/5)          -120

tan2θ = ————— = ——————– = ———  or  -1.0084      ← ANSWER

              1 - tan²θ         1 - (12/5)²          119

Hope this helped.

(1)√2 / 2 = 1 / √2

Let cos^(- 1) (- √2/2) = cos^(- 1) (- 1/√2) = Θ

=> cos Θ = - 1/√2 ........ (1)

Hence, cos {2 cos^(- 1) (- √2/2)} = cos 2Θ

= 2 cos²Θ - 1 ; substituting the value of cos Θ from (1)

= 2 (- 1/√2)² - 1

= 2(1/2) - 1

= 1 - 1 = 0

(2) Let sin^(- 1) (12/13) = Θ

=> sin Θ = 12/13

=> cos Θ = √(1 - sin²Θ) = √(1 - 144/1169) = √(25/169) = ± 5/13

=> tan Θ = ± 12/5

Now tan {2 sin^(- 1) (12/13)} = tan2Θ

= (2 tan Θ) / (1 - tan²Θ)

= ± 2(5/13) / ( 1 - 144/25)

= ± (10/13) / (- 119/25)

= ± (10/13)*(25/119) = ± 250/1547

3) let sin^(- 1) (1/-2) = Θ

=> sin Θ = - 1/2

=> cos Θ = √(1 - sin²Θ) = ±√(1 - 1/4) = ±√3/2

Hence, sin {2sin^(- 1) (1/-2)} = sin 2Θ = 2 sin Θ cos Θ

= 2(- 1/2)(±√3/2) = ±√3/2

what is the exponent of 2cos here if it is -1 then is ans. is √2

the ans. goes like this

cos[ 2cos^-1(-√2/2)]

=cos[2/cos(-√2/2)] since cos^-1 = 1/cos

=2*(-√2/2) since cos and cos get cancelled

=√2