# How would you calculate: cos(a+b)*cos(a+c)

jgoulden | Certified Educator

We'll use the same cos identity twice:

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

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

Multiplying...

(cos a cos b - sin a sin b)(cos a cos c - sin a sin c)

cos^2 a cos b cos c - cos a sin a cos c sin b - cos a sin a cos b sin c + sin^2 a sin b sin c

cos^2 a cos b cos c + sin^2 a sin b sin c - cos a sin a (sin b cos c + cos b sin c )

Note that (sin b cos c + cos b sin c) = sin (b+c) so we can simplify this a bit as

cos^2 a cos b cos c + sin^2 a sin b sin c - cos a sin a sin(b+c)

neela | Student

### cos(a+b)*cos(a+c).

The expression is in a compact form already and to calculate cos(a+b)*cos(a+c) , we can  give values in the very expression and find out. If we expand there would be more number of terms and calculation would be over a number of terms resulting purposeless longer calculation.

But cos(a+b) * cos (a+c) = (1/2)[ cos (a+b+a+c)/2 + cos [a+b-(a+c))]/2

= (1/2) { cos[a+(b+c)/2]+cos[(b-c)/2 }. Using this you can calulate as the sum of two terms calulate the value of expressions.

Please do not expand into many terms for calculation of the cosine  value of the expression, which is making the sum more complex and tedius to calculate the value.