## 5.2 Verifying Trigonometric Identities

## Discovering other identities

Other trigonometric identities can be derived from the elementary identities.

For example, an identity for \(\cot A\sin A\) can be found by replacing \(\cot A\) with \(\dfrac{\cos A}{\sin A}\) and simplifing to \(\cos A\).

Thus \(\cot A\sin A = \cos A\) is an identity.

## A caveat

Never begin a proof by assuming the truth of that which you are attempting to prove.

The following is an **invalid proof** of the identity above.

\(
\begin{eqnarray*}
\cot A\sin A &=& \cos A\\[12pt]
\dfrac{\cot A\sin A}{\sin A}&=&\dfrac{\cos A}{\sin A}\\[12pt]
\cot A &=& \cot A
\end{eqnarray*}
\)

This so-called ‘proof’ begins by using the very identity it seeks to prove. The presumption is that if we begin with some statement and go through a sequence of logical inferences and arrive at a true statement, then the original statement must have been true. But this presumption is false. It is possible to begin with a false statement and yet arrive at a true statement by a series of logical inferences. The fact that the final statement is true implies nothing about whether the original statement is true or false. It is a common logical fallacy that only true statements imply true statements. But false statements can imply true statements.

For example, consider the following invalid proof that \(0=1\).

\(0 = 1\)

Multiplying both sides by \(-1\) yields

\(( -1 )( 0) = ( -1 )( 1 )\) thus

\( 0 = - 1\)

Since \(0 = 1\) and \(0 = -1\), add the two equations to get

\(0 + 0 = 1 + ( -1 )\), thus

\(0 = 0\) which is true.

Thus the original statement \(0 = 1\) must be true.

This is an example of a false statement implying a true statement. These two fallacious ‘proofs’ illustrate why you cannot prove an identity if you begin by using the identity.

## Exercise 5.2.1

Verify that \(\sin^2 A = ( 1 – \cos A )( 1 + \cos A )\).

See Solution
## Exercise 5.2.2

Verify that \(\dfrac{1+\tan A}{\sec A} = \cos A + \sin A\)

See Solution
## Exercise 5.2.3

Verify that \(\dfrac{1}{\sec A + \tan A}= \sec A – \tan A\)

See Solution
## Exercise 5.2.4

Verify that \(\cos ( \frac{\pi}{2} – A ) \sec A = \cot ( \frac{\pi}{2} – A )\)

See Solution