Question 615521
Since it's easier to type, I'm going to use A and B instead of alpha and beta.<br>
cos(A+B) = cos(A)cos(B) - sin(A)sin(B)
We've been given cos(A) and sin(B). We need sin(A) and cos(B).<br>
We can find sin(A) from cos(A) using:
{{{sin^2(A) = 1 - cos^2(A)}}}
Substituting in the given value for cos(A) we get:
{{{sin^2(A) = 1 - (1/2)^2}}}
which simplifies as follows:
{{{sin^2(A) = 1 - 1/4}}}
{{{sin^2(A) = 4/4 - 1/4}}}
{{{sin^2(A) = 3/4}}}
Now we will find the square root of each side. Since we are told that A is between {{{-pi/2}}} 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:
{{{sin(A) = -sqrt(3/4)}}}
which simplifies:
{{{sin(A) = -sqrt(3)/sqrt(4)}}}
{{{sin(A) = -sqrt(3)/2}}}<br>
We can use a similar process to find cos(B):
{{{cos^2(B) = 1-sin^2(B)}}}
{{{cos^2(B) = 1-(1/3)^2}}}
{{{cos^2(B) = 1-(1/9)}}}
{{{cos^2(B) = 9/9-(1/9)}}}
{{{cos^2(B) = 8/9}}}
Since B is between 0 and {{{pi/2}}} it must terminate in the 1st quadrant. cos is positive in the first quadrant so we will use the positive square root:
{{{cos(B) = sqrt(8/9)}}}
{{{cos(B) = sqrt(8)/sqrt(9)}}}
{{{cos(B) = (sqrt(4*2))/3}}}
{{{cos(B) = (sqrt(4)*sqrt(2))/3}}}
{{{cos(B) = (2*sqrt(2))/3}}}<br>
Now that we have all the values we need, we can go back to:
cos(A+B) = cos(A)cos(B) - sin(A)sin(B)
and substitute in the values we have:
{{{cos(A+B) = (1/2)((2*sqrt(2))/3) - (-sqrt(3)/2)(1/3)}}}
which simplifies as follows:
{{{cos(A+B) = sqrt(2)/3 - (-sqrt(3)/6)}}}
{{{cos(A+B) = sqrt(2)/3 + sqrt(3)/6}}}
This may an acceptable answer. If not, then add the terms together:
{{{cos(A+B) = (2sqrt(2))/6 + sqrt(3)/6}}}
{{{cos(A+B) = (2sqrt(2) + sqrt(3))/6}}}