1.

`cosx=-1/3`, x in quadrant III. Find the values of other five trignometric functions

Answer» `therefore cosx=-1/3`
`therefore 2cos^(2)x/2-1=-1/3`
or `2cos^(2)x/2=1-1/3=2/3`,
or `cos^(2)x/2=1/3`
or `cosx/2=+-1/sqrt(3)=+-sqrt(3)/3`
Again, `cosx=-1/3`
`rArr 1-2sin^(2)x/2=-1/3` or `2sin^(2)x/2=1+1/3=4/3`
or `sin^(2)x/2=2/3`,
or `sinx/2=+sqrt(2)/(3)`,
or `sinx/2=+-sqrt(6)/3`
`therefore` x lies in III quadrant.
`rArr` The value of x will be lie between `pi` and `(3pi)/(2)`
i.e., `pi lt x lt (3pi)/(2)`
Then the value of `x/2` will lie between `pi/2` and `(3pi)/(4)` i.e., `pi/2 lt x/2 lt (3pi)/(4)`
Then `x/2`, lies in 2nd quadrant in which `sinx/2` is positive while `cosx/2` and `tanx/2` are negative.
`therefore sinx/2=sqrt(6)/(3)`, then `cosx/2=-sqrt(3)/3`
and `tanx/2=(sqrt(6)//3)/(-sqrt(3)//3)=-sqrt(2)`
Therefore, `sinx/2=sqrt(6)/(3), cosx/2=-sqrt(3)/3` and `tanx/2=-sqrt(2)`


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