Which sum or difference identity would you use to verify that cos (180° - q) = -cos q? a.) sin (a -b) = sin a cos b – cos a sin b b.) cos (a -b) = cos a cos b – sin a sin b c.) cos (a -b) = cos a cosb + sin a sin b d.) sin (a + b) = sin a cos b + cos a sin b

The correct answer is Option C

Step-by-step explanation:

In this question first we use the sum and difference formula :

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

Let a = 180 and b = q, then

cos (180-q) = cos (180) cos (q) + sin (180) sin (q)

Now the sin of 180 degree is 0. Since 0 multiplying  with anything leaves 0, this makes sin (180) sin (q) = 0

cos (180-q) = cos (180) cos(q)

Now the cos of 180 degree is -1.

cos(180-q) = -1 cos(q), or cos (180-q) = -cos(q)

last option

step-by-step explanation:

given that δdef is right and isoceles,

assuming that side df is the hypotenuse means that de an ef are the other 2 sides that intersect at 90° and are of equal length.

because they are of equal length, we can say that they are congruent

because they intersect at 90°, their slopes must be opposite reciprocals.

only the last option satisfies these conditions.

idk

step-by-step explanation:

use that smart brain