Trig! inverse equations?

Question

I need help solving this sheet. 

I need steps also so I know how to do it for my exam. thank uou

Answer 1

• 1). sin^{- 1} ( - √3/2).

sin(sin^-1(x)) = x   for every x in the interval [-1, 1].

sin^-1(sin(x)) = x   for every x in the interval [ - π/2, π/2].

= sin^{- 1} ( sin (- π/3))

= - π/3.

∴ sin^{- 1} ( - √3/2) = - π/3.

• 2). sin(arccos(- 1/3)).

= sin(cos^{- 1} (- 1/3))

Apply formula : sin(cos^{- 1} (x)) = √(1 - x²).

⇒ sin(cos^{- 1} (- 1/3))

= √(1 - (- 1/3)²)

= √(1 - (1/9))

= √((9 - 1)/9)

= √(8/9)

= (2√2)/3.

∴ sin(cos^{- 1} (- 1/3)) = (2√2)/3.

• 4). sin^{- 1} ( sin (330)).

330⁰ in radians is 11π/6.

⇒ sin^{- 1} ( sin (330)) = sin^{- 1} ( sin (11π/6))

∴ sin^{- 1} ( sin (330)) = 11π/6.

• 5). cos^{- 1} ( cos (7π/6)).

cos(cos^-1(x)) = x     for every x in the interval [-1, 1]

cos^-1(cos(x)) = x     for every x in the interval [0, π].

∴ cos^{- 1} ( cos (7π/6)) = 7π/6.

• 6). csc(tan^{- 1}(3/4)).

Apply formula : csc(tan^{- 1} (x)) = √(1 + x²)/x.

⇒ csc(tan^{- 1}(3/4)) = √(1 + (3/4)²)/(3/4).

= 4√(1 + (9/16))/3

= 4√((16 + 9)/16)/3

= 4√(25/16)/3

= 5/3.

∴ csc(tan^{- 1}(3/4)) = 5/3.

• 7). sin[cos^{- 1}(3/5) - tan^{- 1}(7/13)].

Let cos^{- 1}(3/5) = x  and  tan^{- 1}(7/13) = y.

Then, cos x = 3/5  and  tan y = 7/13

If cos x = 3/5, then sin x = √(1 - cos² x) = √(1 - (3/5)²) = 4/5.

tan y = opposite/adjacent = 7/13.

Phythagoras theorem : hyp² = adj² + opp² .

hyp² = 7² + 13²

hyp² = 49 + 169 = 218

hyp = √218.

⇒ cos y = adjacent/hypotenuse = 13/√218, and

    sin y = opposite/hypotenuse = 7/√218.

Apply formula : sin(x - y) = sin x cos y - cos x sin y.

Substitute cirresponding values.

sin(x - y) = (4/5)(13/√218) - (3/5)(7/√218)

= (52 - 21)/(5√218)

= 31/√218.

∴ sin[cos^{- 1}(3/5) - tan^{- 1}(7/13)] = 31/√218.

Answer 2

8).

• a).csc(sin^{- 1}(1/x)).

Using reciprocal identities : csc x = 1/sin x.

csc(sin^{- 1}(1/x)) = 1/[sin(sin^{- 1}(1/x))].

= 1/(1/x)

= x.

∴ csc(sin^{- 1}(1/x)) = x.

• b). csc(tan^{- 1}(√(9 - x²)/x)).

Use the formula : √(1 - x²)/x = tan(cos^{- 1} x).

√(9 - x²)/x = [√(1 - (x/3)²)/(x/3)] = tan(cos^{- 1} (x/3)).

⇒ csc(tan^{- 1}(√(9 - x²)/x)) = csc[ tan^{- 1}(tan(cos^{- 1} (x/3))) ]

= csc(cos^{- 1} (x/3))

= 1/sin(cos^{- 1} (x/3))

Use formula : sin(cos^{- 1} (x)) = √(1 - x²).

= 1/√(1 - (x/3)²)

= 3/√(9 - x²).

∴  csc(tan^{- 1}(√(9 - x²)/x)) = 3/√(9 - x²).