Solve for 0<@<360
cos(@+40)xcos(@-10)=0.5
@ is an angle and everything is in degrees
i don't know what to do!!
Is "x" a variable? I suspect not. Let's use "*" (or simple juxtaposition) for multiplication instead.
What trig identities have they given you? If you have the angle-sum and -difference identities, then here's the "trick" (and no, it is
not "obvious"!):
. . . . .cos(A + B) = cos(A)cos(B) - sin(A)sin(B)
. . . . .cos(A - B) = cos(A)cos(B) + sin(A)sin(B)
Add these:
. . . . .cos(A + B) + cos(A - B) = 2cos(A)cos(B)
. . . . .(1/2)[cos(A + B) + cos(A - B)] = cos(A)cos(B)
Now let A = @ + 40 and let B = @ - 10:
. . . . .(1/2)[cos(@ + 40 + @ - 10) + cos(@ + 40 - @ + 10)] = cos(@ + 40)cos(@ - 10)
. . . . .(1/2)[cos(2@ + 30) + cos(50)] = cos(@ + 40)cos(@ - 10)
Of course, we started with the right-hand side equalling 1/2 so, after substituting and then multiplying through by 2 to clear the fraction on the left-hand side, we get:
. . . . .cos(2@ + 30) + cos(50) = 1
. . . . .cos(2@ + 30) = 1 - cos(50)
Plug the right-hand side into your calculator, using the inverse-cosine key to find the decimal approximation. Then find all values of @ within the stated interval for which 2@ + 30 produce equivalent cosine values.
Whew!
