Question #6f53b

1 Answer
Feb 16, 2018

Answer:

# \ #

# \qquad \qquad \qquad \qquad cos( - \pi /3 ) = 1/2. #

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad sin( - \pi /3 ) = - \sqrt{3} / 2. #

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad tan( - \pi /3 ) = - \sqrt{3}. #

# \qquad \qquad \qquad \qquad csc( - \pi /3 ) = - { 2 \sqrt{3} } / 3 . #

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad sec( - \pi /3 ) = 2 / 1 = 2. #

#\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad cot( - \pi /3 ) = - { \sqrt{3} } / 3 . #

Explanation:

# \ #

# "First, note:" \qquad \qquad \qquad \qquad \qquad \qquad \pi /3 \ = \ 60^@. #

# "Now, remembering the 30-60-90 right triangle, we see:" #

# \qquad \qquad cos 60^@ = 1/2, \qquad sin 60^@ = \sqrt{3} / 2, \qquad tan 60^@ = \sqrt{3}. #

# "So to start with:" #

# \qquad cos( \pi /3 ) = 1 / 2, \qquad \ sin( \pi /3 ) = \sqrt{3} / 2, \qquad \ tan( \pi /3 ) = \sqrt{3}. \qquad \quad \ (1) #

# \ #

# "We want the values of all the 6 trig functions of:" \qquad - \pi /3. #

# "Now, one way that is easy to finish with, is by using the results" #
# "in equation (1), together with basic trig identities:" #

# 1) \qquad cos( -x ) = cos(x) \quad \rArr #

# \qquad \qquad \qquad \qquad cos( - \pi /3 ) = cos( \pi /3 ) = 1/2. #

# 2) \qquad sin( -x ) = - sin(x) \quad \rArr #

# \qquad \qquad \qquad \qquad sin( - \pi /3 ) = - sin( \pi /3 ) = - [ \sqrt{3} / 2 ] = - \sqrt{3} / 2. #

# 3) \qquad tan( -x ) = - tan(x) \quad \rArr #

# \qquad \qquad \qquad \qquad tan( - \pi /3 ) = - tan( \pi /3 ) = - [ \sqrt{3} ] = - \sqrt{3}. #

# "Now continuing, using what we have found in (1)--(3) here, and" #
# "with basic trig identities:" #

# 4")" \qquad csc( x ) = 1 / sin(x) \quad \rArr #

# \qquad \qquad \qquad \qquad csc( - \pi /3 ) = 1 / sin( - \pi /3 ) = 1 / ( - \sqrt{3} / 2 ) = - 2 / \sqrt{3} #

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ = - 2 / \sqrt{3} \cdot \sqrt{3} / \sqrt{3} \ = - { 2 \sqrt{3} } / 3 . #

# 5")" \qquad sec( x ) = 1 / cos(x) \quad \rArr #

# \qquad \qquad \qquad \qquad sec( - \pi /3 ) = 1 / cos( - \pi /3 ) = 1 / ( 1 / 2 ) = 2 / 1 = 2. #

# 6")" \qquad cot( x ) = 1 / tan(x) \quad \rArr #

# \qquad \qquad \qquad \qquad cot( - \pi /3 ) = 1 / tan( - \pi /3 ) = 1 / ( - \sqrt{3} ) = - 1 / \sqrt{3} #

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ = - 1 / \sqrt{3} \cdot \sqrt{3} / \sqrt{3} \ = - { \sqrt{3} } / 3 . #

# "Now we are finished !!" #

# \ #

# "So, summarizing:" #

# \qquad \qquad \qquad \qquad cos( - \pi /3 ) = 1/2. #

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad sin( - \pi /3 ) = - \sqrt{3} / 2. #

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad tan( - \pi /3 ) = - \sqrt{3}. #

# \qquad \qquad \qquad \qquad csc( - \pi /3 ) = - { 2 \sqrt{3} } / 3 . #

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad sec( - \pi /3 ) = 2 / 1 = 2. #

#\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad cot( - \pi /3 ) = - { \sqrt{3} } / 3 . #