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