# Cos(^ ( 2) \theta )+ sin(^ ( 2) \theta )= sec(\theta cos(\theta)) ?

Feb 17, 2018

$\setminus q \quad \setminus q \quad \setminus q \quad \text{The statement is True; it is a trig identity.}$

#### Explanation:

$\text{We are asked to see if the following is true:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad {\cos}^{2} \left(\setminus \theta\right) \setminus + \setminus {\sin}^{2} \left(\setminus \theta\right) \setminus = \setminus \sec \left(\setminus \theta\right) \cos \left(\setminus \theta\right) . \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \setminus \left(1\right)$

$\text{We can look at each side of this statement separately.}$

 "LHS of (1):" \qquad \qquad \qquad cos^2(\theta) \ + \ sin^2(\theta) \ = \ 1;

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \text{by Fundamental Pythagorean Identity.}$

 "RHS of (1):" \qquad \quad sec(\theta) cos(\theta) \ = \ 1/cos(\theta) cos(\theta) \ = \ 1;

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \text{by Reciprocal Identities.}$

$\text{So we conclude:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \text{LHS of (1)" \ = \ "RHS of (1).}$

$\text{Thus our statement in (1) is true, it is a trig identity:}$

$\setminus q \quad \setminus \quad \text{True:} \setminus q \quad \setminus \quad \setminus \quad {\cos}^{2} \left(\setminus \theta\right) \setminus + \setminus {\sin}^{2} \left(\setminus \theta\right) \setminus = \setminus \sec \left(\setminus \theta\right) \cos \left(\setminus \theta\right) .$