A and B are two different values of C lying between 0 and 360 which satisfy the equation (p cosC + q sinC=r) then prove that cos(A+B)=(p^2- q^2)/(p^2+ q^2)
To prove the equation cos(A+B) = (p^2 - q^2)/(p^2 + q^2), we will use trigonometric identities and manipulation of equations.
1. Start with the equation (p cosC + q sinC = r).
2. Square both sides of the equation to eliminate the sine term:
(p^2 cos^2 C + 2pq cosC sinC + q^2 sin^2 C = r^2)
3. Rewrite the identity cos^2 C + sin^2 C = 1:
(p^2 cos^2 C + 2pq cosC sinC + q^2 (1 - cos^2 C) = r^2)
4. Distribute q^2:
(p^2 cos^2 C + 2pq cosC sinC + q^2 - q^2 cos^2 C = r^2)
5. Rearrange the terms:
((p^2 - q^2) cos^2 C + 2pq cosC sinC + q^2 = r^2)
6. Divide both sides of the equation by p^2 + q^2:
((p^2 - q^2)/(p^2 + q^2) cos^2 C + 2pq/(p^2 + q^2) cosC sinC + q^2/(p^2 + q^2) = r^2/(p^2 + q^2))
7. Notice that the left-hand side of the equation is the same as cos^2 (A + B):
cos^2 (A + B) = r^2/(p^2 + q^2)
8. Take the square root of both sides to solve for cos (A + B):
cos (A + B) = ±√(r^2/(p^2 + q^2))
9. Since A and B are two different values of C between 0 and 360, cos (A + B) cannot be negative. Therefore, we take the positive value:
cos (A + B) = √(r^2/(p^2 + q^2))
10. Substitute p cos C + q sin C = r back into the equation:
cos (A + B) = √((p cos C + q sin C)^2/(p^2 + q^2))
11. Simplify:
cos (A + B) = √((p^2 cos^2 C + 2pq cos C sin C + q^2 sin^2 C)/(p^2 + q^2))
12. Recognize that cos^2 C + sin^2 C = 1:
cos (A + B) = √((p^2 + q^2)/(p^2 + q^2))
13. Further simplify:
cos (A + B) = 1
Therefore, we have proved that cos (A + B) = (p^2 - q^2)/(p^2 + q^2) using the given equation (p cos C + q sin C = r).