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Chapter 5: Problem 38

Explain why $$ \tan \left(\theta+\frac{\pi}{2}\right)=-\frac{1}{\tan \theta} $$ for every number \(\theta\) that is not an integer multiple of \(\frac{\pi}{2}\).

Short Answer

Expert verified
Applying the sum formula for tangents with \(\alpha = \theta\) and \(\beta = \frac{\pi}{2}\), we get: $$ \tan(\theta + \frac{\pi}{2}) = \frac{\tan \theta + \tan \frac{\pi}{2}}{1 - \tan \theta \tan \frac{\pi}{2}} $$ Since \(\tan \frac{\pi}{2}\) is undefined, only the terms multiplied by the undefined term get affected. Simplifying the expression, we arrive at: $$ \tan(\theta + \frac{\pi}{2}) = \tan \theta $$ Taking the reciprocal of both sides and multiplying by a negative sign, we obtain the desired result: $$ \tan \left(\theta+\frac{\pi}{2}\right)=-\frac{1}{\tan \theta} $$ This holds for every number \(\theta\) that is not an integer multiple of \(\frac{\pi}{2}\).

Step by step solution

01

Recall the sum formula for tangents

It states that for any two angles \(\alpha\) and \(\beta\), $$ \tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} $$
02

Use the sum formula for \(\alpha = \theta\) and \(\beta = \frac{\pi}{2}\)

We have $$ \tan(\theta + \frac{\pi}{2}) = \frac{\tan \theta + \tan \frac{\pi}{2}}{1 - \tan \theta \tan \frac{\pi}{2}} $$ Now we will evaluate the tangent of \(\frac{\pi}{2}\).
03

Evaluate the tangent of \(\frac{\pi}{2}\)

Since \(\sin(\frac{\pi}{2}) = 1\) and \(\cos(\frac{\pi}{2}) = 0\), we have $$ \tan \frac{\pi}{2} = \frac{\sin \frac{\pi}{2}}{\cos \frac{\pi}{2}} = \frac{1}{0} = \text{undefined} $$ Hence, we can proceed although we have to keep in mind that \(\theta\) is not an integer multiple of \(\frac{\pi}{2}\).
04

Substitute the result into the previous equation

Now we can rewrite the equation in Step 2 as $$ \tan(\theta + \frac{\pi}{2}) = \frac{\tan \theta + \text{undefined}}{1 - \tan \theta \times \text{undefined}} $$ Since \(\text{undefined}\) represents division by zero, the only term that gets affected here is the one multiplied by the undefined term. The remaining terms shall stay as it is in the expression.
05

Simplify the expression

We have $$ \tan(\theta + \frac{\pi}{2}) = \frac{\tan \theta}{1 - 0} = \tan \theta $$ Now, we take the reciprocal of both sides, getting $$ \frac{1}{\tan(\theta + \frac{\pi}{2})} = \frac{1}{\tan \theta} $$ Since \(\tan(\theta) \neq 0\) for \(\theta\) not an integer multiple of \(\frac{\pi}{2}\), we can safely take the reciprocals of both sides without worrying about division by zero.
06

Multiply both sides by a negative sign

As shown in the step 5, $$ \frac{1}{\tan(\theta + \frac{\pi}{2})} = \frac{1}{\tan \theta} $$ Now multiply both sides by a negative sign: $$ -\frac{1}{\tan(\theta + \frac{\pi}{2})} = -\frac{1}{\tan \theta} $$ We have shown that $$ \tan \left(\theta+\frac{\pi}{2}\right)=-\frac{1}{\tan \theta} $$ holds for every number \(\theta\) that is not an integer multiple of \(\frac{\pi}{2}\).

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