PROVE:cos(A+B)= cosAcosB- sinAsinB.?

Answer 1

Use supplemental angles to rewrite cos(A + B) as a sine function. Then use the proof of the angle sum sine formula (provided in my reference).

Answer 2

Try this: write down the power series for cos(a+x)
expanded about the point a:

cos(a+x) = cos(a) - x*sin(a)/ 1! - x^2cos(a)/2! - x^3sin(a)/3! - ...

Now collect terms in the series according to whether they contain a sine or a cosine. This gives:

cos(a+x) = cos(a)[ 1 - x^2/2 + x^4/4 - x^6/6 + ... ]
- sin(a)[ x - x^3 /3! + x^5/5! - x^7/7! + ... ]
But the quantities inside the brackets are the cosine and sine of x, so we have the trigonometric identity:

cos(a+b) = cos(a)cos(b) - sin(a)sin(b)

BUT if you want a simple geometric proof, check out the source. It would be difficult to describe using text but it is basically just setting up a triangle in the appropriate way and the identity simply falls out of it. Briefly you draw a right triangle with the right angle at the bottom left. You then draw a line from the upper vertex with a length of one down to the bottom side. from this point a line is drawn up to the hypoteneuse that is perpendicular to to it. This forms a new right triangle inside the original one with a hypoteneuse of 1. Now label the right most angle of the triangle as ALPHA and make BETA the angle at the top in the small right triangle that is formed when the unit line is drawn. The side of the original triangle on the right is just the cos(BETA) and sin(BETA) is the segment along the bottom side of the triangle. Now, the bottom angle of the new right triangle is just ALPHA+BETA. Anyway, you get the idea as to how much trouble it is to put into words. But once you label all the sides, etc just do:

sin(ALPHA)/cos(ALPHA) = cos(BETA)/[sin(BETA) + cos(ALPHA+BETA)/sin(ALPHA)]

And there you are.

Check out the source, piece of cake.