Question

Find the six trig function values of the angle 240*Show all work, do not use calculator

Answer 1

Given:

`240^0`

To get sin 240 degrees:

240 degrees falls in the third quadrant.

In the third quadrant, only tangent is positive. Hence, sin 240 will be negative.

`sin240^0=sin(180+60)`

Using the trigonometric identity;

`sin(x+y)=sinx\text{ }cosy+cosx\text{ }siny`

Hence,

`\begin{gathered} sin(180+60)=sin180cos60+cos180sin60 \\ sin180=0 \\ cos60=(1)/(2) \\ cos180=-1 \\ sin60=(√(3))/(2) \\ \\ Thus, \\ sin180cos60+cos180sin60=0((1)/(2))+(-1)((√(3))/(2)) \\ sin180cos60+cos180sin60=0-(√(3))/(2) \\ sin180cos60+cos180sin60=-(√(3))/(2) \\ \\ Hence, \\ sin240^0=-(√(3))/(2) \end{gathered}`

To get cos 240 degrees:

240 degrees falls in the third quadrant.

In the third quadrant, only tangent is positive. Hence, cos 240 will be negative.

`cos240^0=cos(180+60)`

Using the trigonometric identity;

`cos(x+y)=cosx\text{ }cosy-sinx\text{ }siny`

Hence,

`\begin{gathered} cos(180+60)=cos180cos60-sin180sin60 \\ sin180=0 \\ cos60=(1)/(2) \\ cos180=-1 \\ sin60=(√(3))/(2) \\ \\ Thus, \\ cos180cos60-sin180sin60=-1((1)/(2))-0((√(3))/(2)) \\ cos180cos60-sin180sin60=-(1)/(2)-0 \\ cos180cos60-sin180sin60=-(1)/(2) \\ \\ Hence, \\ cos240^0=-(1)/(2) \end{gathered}`

To get tan 240 degrees:

240 degrees falls in the third quadrant.

In the third quadrant, only tangent is positive. Hence, tan 240 will be positive.

`tan240^0=tan(180+60)`

Using the trigonometric identity;

`tan(180+x)=tan\text{ }x`

Hence,

`\begin{gathered} tan(180+60)=tan60 \\ tan60=√(3) \\ \\ Hence, \\ tan240^0=√(3) \end{gathered}`

To get cosec 240 degrees:

`\begin{gathered} cosec\text{ }x=(1)/(sinx) \\ csc240=(1)/(sin240) \\ sin240=-(√(3))/(2) \\ \\ Hence, \\ csc240=(1)/((-√(3))/(2)) \\ csc240=-(2)/(√(3)) \\ \\ Rationalizing\text{ the denominator;} \\ csc240=-(2)/(√(3))*(√(3))/(√(3)) \\ \\ Thus, \\ csc240^0=-(2√(3))/(3) \end{gathered}`

To get sec 240 degrees:

`\begin{gathered} sec\text{ }x=(1)/(cosx) \\ sec240=(1)/(cos240) \\ cos240=-(1)/(2) \\ \\ Hence, \\ sec240=(1)/((-1)/(2)) \\ sec240=-2 \\ \\ Thus, \\ sec240^0=-2 \end{gathered}`

To get cot 240 degrees:

`\begin{gathered} cot\text{ }x=\frac{1}{tan\text{ }x} \\ cot240=(1)/(tan240) \\ tan240=√(3) \\ \\ Hence, \\ cot240=(1)/(√(3)) \\ \\ Rationalizing\text{ the denominator;} \\ cot240=(1)/(√(3))*(√(3))/(√(3)) \\ \\ Thus, \\ cot240^0=(√(3))/(3) \end{gathered}`