Question 984820
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First, even though *[tex \Large \sin{x}] and *[tex \Large \sin{y}] are given, we still need to know *[tex \Large \cos{x}] and *[tex \Large \cos{y}] in order to calculate *[tex \Large \sin(x\ +\ y)]

If we know that *[tex \Large \sin\varphi\ =\ a], then we can determine the value of cosine by the following process:

Given:                  *[tex \LARGE \sin\varphi\ =\ a]
Square both sides:      *[tex \LARGE \sin^2\varphi\ =\ a^2]
Apply the Pythagorean
  Identity              *[tex \LARGE 1\ -\ \cos^2\varphi\ =\ a^2]
A little algebra        *[tex \LARGE \cos^2\varphi\ =\ 1\ -\ a^2]
Take the root           *[tex \LARGE \cos\varphi\ =\ \pm\sqrt{1\ -\ a^2}]

Since your problem does not restrict the value of *[tex \Large x] or *[tex \Large y] to a specific quadrant, the value of cosine, when calculated on the basis of the value of sine, is ambiguous.  We should look out for more than one solution to your problem.

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin{x}\ =\ 0.8]
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos{x}\ =\ \pm\sqrt{1\ - \ 0.64}\ =\ \pm 0.6]
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin{y}\ =\ 0.6]
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos{y}\ =\ \pm\sqrt{1\ - \ 0.36}\ =\ \pm 0.8]

Now we have all the information required to perform the needed calculation.

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin(x\ +\ y)\ =\ \sin{x}\cos{y}\ +\ \cos{x}\sin{y}]
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin(x\ +\ y)\ =\ (0.8)(\pm 0.8)\ +\ (\pm 0.6)(0.6)]
So
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin(x\ +\ y)\ =\ 0.64\ +\ 0.36\ =\ 1]
or
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin(x\ +\ y)\ =\ -0.64\ -\ 0.36\ =\ -1]
Hence
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ y\ =\ \frac{\pi}{2}]
or
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ y\ =\ \frac{3\pi}{2}]
</pre>
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \