SOLUTION: If sin(x)=-5/13 and x is in quadrant III, with 0° ≤ x < 360°, find the exact values of the expressions without solving for x. sin(x/2) cos(x/2) tan(x/2)

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Question 1205032: If sin(x)=-5/13 and x is in quadrant III, with 0° ≤ x < 360°, find the exact values of the expressions without solving for x.
sin(x/2)
cos(x/2)
tan(x/2)

Found 2 solutions by Edwin McCravy, ikleyn:
Answer by Edwin McCravy(20060)   (Show Source): You can put this solution on YOUR website!
If sin(x)=-5/13 and x is in quadrant III,...
So for angle x, we have a 5-12-13 right triangle in quadrant III,
where x=-12, y=-5, and r=+13.

There is often a conflict of notation when x is used for an angle, and also
for values of the adjacent side of the defining right triangle. There is a
problem here. But I think you won't get confused.  Teachers aren't always
careful to point out this conflict, which happens a lot. (just like the problem
of how to say "the sign of the sine". J )

So cos(x)=x/r=(-12)/(+13)=-12/13 and tan(x)=y/x=(-5)/(-12)=+5/12


 so x/2 is in quadrant II (the upper half of quadrant II).

So sin(x/2) is positive, cos(x/2) is negative, and tan(x/2) is negative.

So we just use the half-angle formulas:

, 

We know to use the + because x/2 is in quadrant II, where sine is positive.



We know to use the - because x/2 is in quadrant II, where cosine is negative.



Now we could use a formula for tan(x/2), but now all we need is 



Edwin



Answer by ikleyn(52814)   (Show Source): You can put this solution on YOUR website!
.
If sin(x)=-5/13 and x is in quadrant III, with 0° ≤ x < 360°, find the exact values of the expressions without solving for x.
sin(x/2)
cos(x/2)
tan(x/2)
~~~~~~~~~~~~~~~~~~~~

As the problem is worded,  printed and presented in the post,  it has a shocking deficiency.

One part says that  x  is in quadrant III,  and then next part says  0° ≤ x < 360°,
which is  INCONSISTENT  writing  (making a reader tremble in horror).

There are two ways to fix it and to present it correctly/consistently.

One way is to omit this inequality  0° ≤ x < 360°.

Other way is to write  180° ≤ x < 270°.


As it is seen from the post,  the math composer is quite illiterate in writing  Math.


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        I am writing these lines in order for to prevent this formulation
        to go out in the Internet for spreading/distribution in this form
                            as it is presented in the post.

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With friendly greetings.




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