Question 922568
I think you're having trouble "reading" these functions
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It helps if you can visualize the plot of {{{ f(t) }}} and {{{ t }}}
for example, 
{{{ f(t) = sin( t - 6pi ) }}}
{{{ 6pi }}} is called the phase angle
If you set {{{ t = 0 }}} then you have:
{{{ f(0) = sin( -6*pi ) }}}
{{{ -6*pi = -3*( 2*pi ) }}}
This is a multiple of {{{ 2*pi }}}, and since
{{{ sin( 2*pi ) = 0 }}}, then
{{{ sin( -6*pi ) = 0 }}}
So, I have:
{{{ f(0) = 0 }}}
Here's the plot:
{{{ graph( 400, 400, -10, 10, -1.5, 1.5, sin( x - 6*pi )) }}}
As you can see, there is no difference between this
function and {{{ sin(t) }}}
Now I can say:
{{{ csc(t) = 4 }}}
{{{ sin(t) = 1/4 }}}
{{{ sin( t - 6*pi ) = 1/4 }}}
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You are correct that the angle whose csc is {{{ 4 }}}
and whose tan is > 0 is in the 1st quadrant
{{{ sin(t) = 1/4 }}}
If {{{ a = 1 }}}
{{{ c = 4 }}} , then
{{{ b = sqrt( 4^2 - 1^2 ) }}}
{{{ b = sqrt( 15 ) }}}
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Now note that {{{ b/c = cos(t) }}}
{{{ c/b = sec(t) }}}
Hope this helps a little -you just have to keep
testing yourself on what you are vague on