Give the exact value of each expression?

Question

1. cos165⁰
2. sin 11π/12
3. tan465°

Answer 1

1) cos165^o

= cos(120^o + 45^o)

= cos(120^o) cos(45^o) - sin(120^o)sin(45^o)  [Since cos(A + B) = cosAcosB - sinAsinB]

= cos(120^o)(1/√2) - sin(120^o)(1/√2)

= 1/√2 [ cos(120^o) - sin(120^o) ]

= 1/√2 [ cos(90^o + 30^o) - sin(90^o + 30^o)]

= 1/√2 [ {cos(90^o)cos(30^o) - sin(90^o)sin(30^o)} - sin(90^o + 30^o)] [Since cos(A + B) = cosAcosB - sinAsinB]

= 1/√2 [ {(0)(√3/2) - (1)(1/2)} - sin(90^o + 30^o)]

= 1/√2 [ 0 - 1/2 - sin(90^o + 30^o)]

= 1/√2 [ -1/2 - sin(90^o + 30^o)]

= 1/√2 [ -1/2 - {sin(90^o)cos(30^o) + cos(90^o)sin(30^o)}]   [Since sin(A + B) = sinAcosB + cosAsinB]

= 1/√2 [ -1/2 - {(1)(√3/2) + (0)(1/2)}]

= 1/√2 [ -1/2 - {√3/2 + 0}]

= 1/√2 [ -1/2 - √3/2]

= 1/√2 [ (-1 - √3) / 2]

= -1 - √3 / 2√2

= -(1+√3) / 2√2

Therefore cos165^o = -(1+√3) / 2√2.

Answer 2

2) sin 11π/12

= sin 11(180)/12

= sin (1980)/12

= sin 165^o

= sin(120^o + 45^o)

= sin(120^o) cos(45^o) + cos(120^o)sin(45^o)  [Since sin(A + B) = sinAcosB + cosAsinB]

= sin(120^o) (1/√2) + cos(120^o)(1/√2)

= (1/√2) [ sin(120^o) + cos(120^o) ]

= (1/√2) [ sin(90^o + 30^o) + cos(90^o + 30^o) ]

= (1/√2) [ {sin(90^o) cos(30^o) + cos(90^o)sin(30^o)} + cos(90^o + 30^o) ] [Since sin(A + B) = sinAcosB + cosAsinB]

= (1/√2) [ {(1) (√3/2) + (0)(1/2)} + cos(90^o + 30^o) ]

= (1/√2) [ {√3/2 + 0} + cos(90^o + 30^o) ]

= (1/√2) [ √3/2 + cos(90^o + 30^o) ]

= (1/√2) [ √3/2 + cos(90^o) cos(30^o) - sin(90^o)sin(30^o) ] [Since cos(A + B) = cosAcosB - sinAsinB]

= (1/√2) [ √3/2 + (0)(√3/2) - (1)(1/2) ]

= (1/√2) [ √3/2 + 0 - 1/2 ]

= (1/√2) [ √3/2 - 1/2 ]

= (1/√2) [ (√3 - 1)/2 ]

= (√3 - 1) / 2√2

Therefore sin11π/12 = (√3 - 1) / 2√2.

 

Answer 3

3) tan465^o

= tan(360^o +105^o)

= tan(2π + 105^o)

= tan(105^o)                                                                         [Since tan(2π + α) = tan α ]

= sin(105^o) / cos(105^o)

= sin (60^o + 45^o) / cos(60^o + 45^o)

= {sin(60^o)cos(45^o) + cos(60^o)sin(45^o)} / cos(60^o + 45^o)    [Since sin(A + B) = sinAcosB + cosAsinB]

= {cos(60^o)cos(45^o) - sin(60^o)sin(45^o)} / cos(60^o + 45^o)    [Since cos(A + B) = cosAcosB - sinAsinB]

= {(√3/2)(1/√2) + (1/2)(1/√2)} / cos(60^o + 45^o)

= {√3 / 2√2 + 1/2√2} / cos(60^o + 45^o)

= {(√3 + 1)/2√2} / cos(60^o + 45^o)

= {(√3 + 1) / 2√2} / {cos(60^o)cos(45^o) - sin(60^o)sin(45^o)}    [Since cos(A + B) = cosAcosB - sinAsinB]

= {(√3 + 1) / 2√2} / {(1/2)(1/√2) - (√3/2)(1/√2)}

= {(√3 + 1) / 2√2} / {1 / 2√2 - √3 / 2√2)}

= {(√3 + 1) / 2√2} / {(1 - √3) / 2√2}

= (1 + √3) / (1 - √3)

Multiply numerator and denominator by (√3 + 1).

= (1 + √3)(1 + √3) / (1 - √3)(1 + √3)

= (1 + √3)² / {(1)² - (√3)²}

= (1 + 2√3 + 3) / (1 - 3)

= (4 + 2√3) / (- 2)

= 2(2 + √3) / (- 2)

= - 2 - √3.

Therefore tan 465^o = - 2 - √3.