Question #6f53b

1 Answer
Feb 16, 2018

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\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:

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"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)

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"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 !!"

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"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 .