Evaluate each expression without a calculator and write your answer in radians.?

Question

1. sin (2 cos^-1 1/5) 

2. csc (tan^-1 3/4) 

3. cos (sin^-1 3/5 +cos ^-1 5/13) 

4. sec [cos^-1 (√2/2)+ sin^-1 (-1)] 

5. sin (sin^-1 1/2 + tan ^-1 (-3)) 

6. cos (tan^-1 5/12- tan^-1 3/4))

Answer 1

• 1). sin(2cos^{- 1} (1/5)).

Let, 2cos^{- 1} (1/5) = t

Then, sin(2cos^{- 1} (1/5)) = sin (t).

cos^{- 1} (1/5) = t/2

⇒ cos(t/2) = 1/5.

Half angle formula : cos(A/2) = ± √[(1 + cos A)/2].

± √[(1 + cos t)/2] = 1/5

Squaring on both sides.

(± √[(1 + cos t)/2])² = (1/5)²

(1 + cos t)/2 = 1/25

1 + cos t = 2/25

cos t = 2/25 - 1 = - 23/25.

⇒ sin t = √(1 - (23/25)²) = √(625 - 529)/25 = √96/25 = 0.3919.

Put, t = 2cos^{- 1} (1/5).

sin(2cos^{- 1} (1/5)) = 0.3919.

Therefore, sin(2cos^{- 1} (1/5)) = 0.3919.

• 2). 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.

Therefore,  csc(tan^{- 1}(3/4)) = 5/3.

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

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

Then, sin x = 3/5  and  cos y = 5/13

If sin x = 3/5, then cos x = √(1 - sin² x) = √(1 - (3/5)²) = 4/5.

If cos y = 5/13, then sin y = √(1 - cos² y) = √(1 - (5/13)²) = 12/13.

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

Substitute cirresponding values.

cos(x + y) = (4/5)(5/13) - (3/5)(12/13)

= (20 - 36)/(65)

= - 16/65.

Therefore, cos[sin^{- 1}(3/5) + cos^{- 1}(5/13)] = - 16/65.

Answer 2

• 4). sec[cos^{- 1}(√2/2) + sin^{- 1}(- 1)].

sec[cos^{- 1}(1/√2) + sin^{- 1}(- 1)].

Let cos^{- 1}(1/√2) = x  and  sin^{- 1}(- 1) = y.

Then, cos x = 1/√2  and  sin y = - 1

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

If sin y = - 1, then cos y = √(1 - sin² y) = √(1 - (- 1)²) = 0.

Using reciprocal identity : sec x = 1/cos x.

sec[cos^{- 1}(1/√2) + sin^{- 1}(- 1)] = 1/{ cos[cos^{- 1}(1/√2) + sin^{- 1}(- 1)] }

= 1/cos (x + y).

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

Substitute corresponding values.

cos(x + y) = (1/√2)(0) - (1/√2)(- 1)

= 0 + 1/√2

= 1/√2.

⇒1/cos (x + y) = 1/(1/√2) = √2.

sec[cos^{- 1}(√2/2) + sin^{- 1}(- 1)] = √2.

Therefore, sec[cos^{- 1}(√2/2) + sin^{- 1}(- 1)] = √2.

• 5).sin[sin^{- 1}(1/2) + tan^{- 1}(- 3)].

Let sin^{- 1}(1/2) = x  and  tan^{- 1}(- 3) = y.

Then, sin x = 1/2  and  tan y = - 3

If sin x = 1/2, then cos x = √(1 - sin² x) = √(1 - (1/2)²) = √3/2.

tan y = opposite/adjacent = - 3/1.

Phythagoras theorem : hypotenuse² = adjacent² + opposite² .

hypotenuse² = 1² + (- 3)²

hypotenuse² = 1 + 9 = 10

hypotenuse = √10.

⇒ cos y = adjacent/hypotenuse = 1/√10, and

    sin y = opposite/hypotenuse = - 3/√10.

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

Substitute corresponding values.

sin(x + y) = (1/2)(1/√10) + (√3/2)(- 3/√10)

= (1 - 3√3)/(2√10).

Therefore, sin[sin^{- 1}(1/2) + tan^{- 1}(- 3)] = (1 - 3√3)/(2√10).

• 6). cos[tan^{- 1}(5/12) - tan^{- 1}(3/4)].

Let tan^{- 1}(5/12) = x  and  tan^{- 1}(3/4) = y.

Then, tan x = 5/12  and  tan y = 3/4

• tan x = opposite/adjacent = 5/12.

Phythagoras theorem : hypotenuse² = adjacent² + opposite² .

hypotenuse² = 5² + 12²

hypotenuse² = 25 + 144 = 169

hypotenuse = 13.

⇒ cos x = adjacent/hypotenuse = 12/13, and

    sin x = opposite/hypotenuse = 5/13.

• tan y = opposite/adjacent = 3/4.

Phythagoras theorem : hypotenuse² = adjacent² + opposite² .

hypotenuse² = 3² + 4²

hypotenuse² = 9 + 16 = 25

hypotenuse = 5.

⇒ cos y = adjacent/hypotenuse = 4/5, and

    sin y = opposite/hypotenuse = 3/5.

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

Substitute cirresponding values.

cos(x + y) = (12/13)(4/5) + (5/13)(3/5)

= (48 + 15)/(65)

= 63/65.

Therefore, cos[tan^{- 1}(5/12) - tan^{- 1}(3/4)] = 63/65.