MATH 1113 PreCalculus

Day #30, More Complicated Identities

Cosine and Sine of the sum and difference of two angles:

Let's look at two different ways of calculating length of the chord representing the difference of angles B and A in the Geogebra applet, below.

Geogebra Applet

If you FOIL out the squares in the equation, you get:
cos²(B) + cos²(A) - 2*cos(B)cos(A) + sin²(B) + sin²(A) - 2*sin(B)sin(A) =
cos²(B-A) - 2cos(B-A) + 1 + sin²(B-A)
 

Then, if you apply the Pythagorean Identity to all of the sin²s and cos²s, it simplifies to:

2 - 2cos(B)cos(A) - 2sin(B)sin(A) = 2 - 2cos(B-A)

Switching sides and canceling 2's all over, you get:

cos(B-A) = cos(B)cos(A) + sin(B)sin(A)

which is the angle difference formula for cosines.

Then the angle sum formula for cosines comes directly from the oddness of sine and the evenness of cosine:

cos(B+A) = cos(B)cos(A) - sin(B)sin(A)

Applying the identity: sin(B+A) = cos((B+A)-pi/2) = cos(B+(A-pi/2)) =

cos(B)cos(A-pi/2) - sin(B)sin(A-pi/2) =

cos(B)sin(A) - sin(B)*(-1)sin(pi/2-A) (since sine is odd) =

cos(B)sin(A) + sin(B)cos(A) (using the cofunction identity).

So:

sin(B+A) = sin(B)cos(A) + cos(B)sin(A)

And again employing the evenness of cosine and oddness of sine, we get:

sin(B-A) = sin(B)cos(A) - cos(B)sin(A)

Again, using the definition of tangent and a little trickiness, you can end up with:

tan(B+A) = (tan(B)+tan(A))/(1-tan(B)tan(A))

tan(B - A) = (tan(B) - tan(A))/(1+tan(B)tan(A))

These formulas should not be memorized, but you should remember that they exist, and be able to look them up and apply them.

Example:  Find the exact value for the cosine of 75^o.

75^o = 30^o + 45^o.  Therefore:

cos(75^o) = cos(30^o+45^o) = cos(30)cos(45) - sin(30)sin(45) =

[sqrt(3)/2]*[sqrt(2)/2] - [1/2]*[sqrt(2)/2] = sqrt(2)*[sqrt(3) - 1]/4

Next, we'll see two important special cases, the Double and Half-Angle formulas.

Page Last Modified: 25 March, 2007

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