document.write( "Question 615521: Given cosα = 1/2, -pi/2 < α < 0; sinβ=1/3, 0 < β < pi/2. Find cos(α + β) \n" ); document.write( "
Algebra.Com's Answer #387227 by jsmallt9(3758)\"\" \"About 
You can put this solution on YOUR website!
Since it's easier to type, I'm going to use A and B instead of alpha and beta.

\n" ); document.write( "cos(A+B) = cos(A)cos(B) - sin(A)sin(B)
\n" ); document.write( "We've been given cos(A) and sin(B). We need sin(A) and cos(B).

\n" ); document.write( "We can find sin(A) from cos(A) using:
\n" ); document.write( "\"sin%5E2%28A%29+=+1+-+cos%5E2%28A%29\"
\n" ); document.write( "Substituting in the given value for cos(A) we get:
\n" ); document.write( "\"sin%5E2%28A%29+=+1+-+%281%2F2%29%5E2\"
\n" ); document.write( "which simplifies as follows:
\n" ); document.write( "\"sin%5E2%28A%29+=+1+-+1%2F4\"
\n" ); document.write( "\"sin%5E2%28A%29+=+4%2F4+-+1%2F4\"
\n" ); document.write( "\"sin%5E2%28A%29+=+3%2F4\"
\n" ); document.write( "Now we will find the square root of each side. Since we are told that A is between \"-pi%2F2\" and 0 it must terminate in the 4th quadrant. Since sin is negative in the 4th quadrant, we know to use the negative square root:
\n" ); document.write( "\"sin%28A%29+=+-sqrt%283%2F4%29\"
\n" ); document.write( "which simplifies:
\n" ); document.write( "\"sin%28A%29+=+-sqrt%283%29%2Fsqrt%284%29\"
\n" ); document.write( "\"sin%28A%29+=+-sqrt%283%29%2F2\"

\n" ); document.write( "We can use a similar process to find cos(B):
\n" ); document.write( "\"cos%5E2%28B%29+=+1-sin%5E2%28B%29\"
\n" ); document.write( "\"cos%5E2%28B%29+=+1-%281%2F3%29%5E2\"
\n" ); document.write( "\"cos%5E2%28B%29+=+1-%281%2F9%29\"
\n" ); document.write( "\"cos%5E2%28B%29+=+9%2F9-%281%2F9%29\"
\n" ); document.write( "\"cos%5E2%28B%29+=+8%2F9\"
\n" ); document.write( "Since B is between 0 and \"pi%2F2\" it must terminate in the 1st quadrant. cos is positive in the first quadrant so we will use the positive square root:
\n" ); document.write( "\"cos%28B%29+=+sqrt%288%2F9%29\"
\n" ); document.write( "\"cos%28B%29+=+sqrt%288%29%2Fsqrt%289%29\"
\n" ); document.write( "\"cos%28B%29+=+%28sqrt%284%2A2%29%29%2F3\"
\n" ); document.write( "\"cos%28B%29+=+%28sqrt%284%29%2Asqrt%282%29%29%2F3\"
\n" ); document.write( "\"cos%28B%29+=+%282%2Asqrt%282%29%29%2F3\"

\n" ); document.write( "Now that we have all the values we need, we can go back to:
\n" ); document.write( "cos(A+B) = cos(A)cos(B) - sin(A)sin(B)
\n" ); document.write( "and substitute in the values we have:
\n" ); document.write( "
\n" ); document.write( "which simplifies as follows:
\n" ); document.write( "\"cos%28A%2BB%29+=+sqrt%282%29%2F3+-+%28-sqrt%283%29%2F6%29\"
\n" ); document.write( "\"cos%28A%2BB%29+=+sqrt%282%29%2F3+%2B+sqrt%283%29%2F6\"
\n" ); document.write( "This may an acceptable answer. If not, then add the terms together:
\n" ); document.write( "\"cos%28A%2BB%29+=+%282sqrt%282%29%29%2F6+%2B+sqrt%283%29%2F6\"
\n" ); document.write( "\"cos%28A%2BB%29+=+%282sqrt%282%29+%2B+sqrt%283%29%29%2F6\" \n" ); document.write( "
\n" );