Find the sum of these two trinomials.
(4a^3 – 5a^2 + 6a) + (a^3 + 2a - 3)

(4a^3 – 5a^2 + 6a) + (a^3 + 2a - 3)

Remove parentheses:

4a^3 – 5a^2 + 6a + a^3 + 2a - 3

Group like terms:

4a^3+a^3-5a^2+6a+2a-3

Add similar elements 4a^3+a^3:

4a^3+a^3= 5a^3

5a^3 -5a^2+6a+2a-3

Add similar elements 6a+2a:

6a+2a= 8a

5a^3-5a^2+8a-3

Your Answer Is 5a^3-5a^2+8a-3

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, because the cosine and sine are complements

Step-by-step explanation:

Sine and cosine are trigonometric complementary functions. If we are talking about a right triangle we will have the following:

(1)

(2)

Let's prove it:

(3)

(4)

(5)

(6)