Show that sin(A+B)=sinAcosB+sinBcosA...
diagram can be use if necessary
To prove the identity sin(A+B) = sinAcosB + sinBcosA, you can use the trigonometric addition formula for sine. This formula states:
sin(A + B) = sinAcosB + cosAsinB
Now let's examine how to derive this formula using a diagram.
Consider a circle with radius 1. Let O be its center, and let A and B be two points on the circumference of the circle. Assume that angle AOB is A and angle ABO is B.
Now draw a perpendicular line from point A to line OB, and denote the intersection as P. The length of OP is sinB, and the length of AP is sinA.
Also, draw a line perpendicular to OB from point B, and denote the intersection as Q. The length of OQ is cosA, and the length of BQ is cosB.
Using the assumption that the radius of the circle is 1, we can see that AP represents sinA, OP represents sinB, BQ represents cosB, and OQ represents cosA.
Now, draw a line parallel to line OP from point A, intersecting OB at point R. Similarly, draw a line parallel to line BQ from point B, intersecting OB at point S. The distance between points P and R is cosB, and the distance between points Q and S is sinA.
Consider the triangle OPR. The length of OP is sinB, and the length of OR is cosB. By the Pythagorean theorem, we have:
(OP)^2 + (OR)^2 = (PR)^2
(sinB)^2 + (cosB)^2 = (PR)^2
sin^2 B + cos^2 B = (PR)^2
1 = (PR)^2
Therefore, PR (which is equal to AP) also represents sinAcosB.
Similarly, consider the triangle OSQ. The length of OQ is cosA, and the length of OS is sinA. By the Pythagorean theorem, we have:
(OQ)^2 + (OS)^2 = (QS)^2
(cosA)^2 + (sinA)^2 = (QS)^2
cos^2 A + sin^2 A = (QS)^2
1 = (QS)^2
Therefore, QS (which is equal to BQ) also represents sinBcosA.
Hence, we have proven that sin(A+B) = sinAcosB + sinBcosA using the geometric illustration explained above.