Question 1165906


a)  

{{{tan(-210)}}}

Use the following property :

 {{{tan(-x)=-tan (x)}}}

{{{tan(-210)=-tan(210)}}}

{{{tan(210)=tan(180)+tan(30)}}} ....since {{{tan(180)=0}}}

{{{tan(210)=tan(30)}}}

{{{tan(30)=-1/sqrt(3)}}}

=> {{{tan(-210)=-1/sqrt(3)}}}


b)   

{{{sin(300)}}} degrees

Use the following property :

{{{sin(x)=cos(90-x)}}}

=> {{{sin(300)=cos(90-300)}}}=> {{{sin(300)=cos(-210)}}}

then {{{cos(-210)=-cos(210)}}}

 and {{{-cos(210)=-cos(180+30) }}}

Using the summation identity :

{{{cos(x+ y )= cos (x ) cos (y )- sin (x ) sin (y )}}}

{{{cos(180+ 30 )= cos (180 ) cos (30 )- sin (180 ) sin (30 )}}}

since 
{{{cos(180)=-1}}} 
{{{cos(30)=sqrt(3)/2}}} 
{{{sin (180 )=0}}}
{{{sin (30 )=1/2}}}

we have

{{{cos(180+ 30 )= (-1)(sqrt(3)/2)- 0*(1/2)}}}

{{{cos(180+ 30 )= -sqrt(3)/2)}}}

=>{{{sin(300)= -sqrt(3)/2)}}}