SOLUTION: Find {{{ sin2x }}} if {{{ secx=-4 }}} and {{{ sinx<0 }}}

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Question 289398: Find +sin2x+ if +secx=-4+ and +sinx%3C0+
Answer by Edwin McCravy(20064) About Me  (Show Source):
You can put this solution on YOUR website!
Find +sin2x+ if +secx=-4+ and +sinx%3C0+

Since the secant and the sine of x are both negative, that means
x is in the third quadrant. between 180° and 270° in degress,
or between pi and 3pi%2F2 in radians.  So therefore
2x will be between 360° and 540° which is actually the 1st and
2nd quadrants.  The sine is positive in both the 1st and 2nd
quadrants, so the answer will be posotive.

We need these identities:

1. Sin%282theta%29+=+2Sin%28theta%29Cos%28theta%29
2. Cos%5E2theta+%2B+Sin%5E2theta=1 
3. Cos%28theta%29=1%2FSec%28theta%29 
 
First we'll find Cos%28x%29 using 3.

Cos%28x%29=1%2FSec%28x%29=1%2F%28-4%29=-1%2F4

Next we'll find Sin%28x%29 using 2.

Cos%5E2x+%2B+Sin%5E2x=1

%28-1%2F4%29%5E2%2BSin%5E2x=1
 
1%2F16%2BSin%5E2x=1

Sin%5E2x=1-1%2F16

Sin%5E2x=16%2F16-1%2F16

Sin%5E2x=15%2F16

Sin%28x%29=%22%22%2B-sqrt%2815%2F16%29=%22%22%2B-sqrt%2815%29%2F4

However, since Sin%28x%29%3C0 we take the negative sign,

Sin%28x%29=-sqrt%2815%29%2F4

Now we can substitute in 1:

Sin%282x%29+=+2Sin%28x%29Cos%28x%29

Sin%282x%29+=+2%28-sqrt%2815%29%2F4%29%28-1%2F4%29

Sin%282x%29+=+2sqrt%2815%29%2F16%29

Sin%282x%29+=+sqrt%2815%29%2F8

Edwin