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Chapter 5: Problem 40

Find an identity expressing \(\sin \left(\tan ^{-1} t\right)\) as a nice function of \(t\).

Short Answer

Expert verified
The identity expressing \(\sin\left(\tan^{-1}(t)\right)\) as a function of \(t\) is: \[\sin(\tan^{-1}(t)) = \frac{t}{\sqrt{t^2 + 1}}\]

Step by step solution

01

Define a right triangle with tan

Let's create a right-angled triangle where \(\tan(\theta) = t\). To get the \(\sin(\theta)\) related to \(t\), we use the right triangle, where \(t\) is the ratio of the lengths of the opposite and adjacent sides. Let the side opposite to \(\theta\) be of length \(x\) and the side adjacent to \(\theta\) be a length of \(y\). Then the tan relation would be \[t = \frac{x}{y}\]
02

Use the Pythagorean theorem

In a right-angled triangle, the Pythagorean theorem states that the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. We will denote the hypotenuse as \(z\). Hence, \[z^2 = x^2 + y^2\] From the tan relation, we can re-write the equation: \[z^2 = (ty)^2 + y^2\] Factoring out \(y^2\), we get \[z^2 = y^2(t^2 + 1)\] and solving for \(y^2\) gives us \[y^2 = \frac {z^2}{t^2 + 1}\]
03

Find \(\sin(\theta)\)

Using the right triangle definition, we can find \(\sin(\theta)\) - it is simply the ratio of the length of the side opposite \(\theta\) (side with length x) to the length of the hypotenuse (side with length z). Thus, \[\sin(\theta) = \frac{x}{z}\]
04

Substitute and simplify

Since we know that \(x = ty\), we can substitute this into the sin equation: \[\sin(\theta) = \frac{ty}{z}\] Now, we need to express \(z\) in terms of \(y\), which we can do using the relation we found in Step 2: \[z^2 = y^2(t^2 + 1) \Longrightarrow z = y\sqrt{t^2 + 1}\] Finally, we can substitute this value of \(z\) into the sin equation: \[\sin(\theta) = \frac{ty}{y\sqrt{t^2 + 1}}\]
05

Find the identity

Now, we can cancel out the \(y\) in the numerator and denominator, which gives us: \[\sin(\tan^{-1}(t)) = \sin(\theta) = \frac{t}{\sqrt{t^2 + 1}}\] So, the identity expressing \(\sin\left(\tan^{-1}(t)\right)\) as a function of \(t\) is: \[\sin(\tan^{-1}(t)) = \frac{t}{\sqrt{t^2 + 1}}\]

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Most popular questions from this chapter

Evaluate the given quantities assuming that \(u\) and \(v\) are both in the interval \(\left(-\frac{\pi}{2}, 0\right)\) and \(\tan u=-\frac{1}{7} \quad\) and \(\quad \tan v=-\frac{1}{8}\) $$\sin (2 v)$$

Show that \(\cos 20^{\circ}\) is a zero of the polynomial \(8 x^{3}-6 x-1\) [Hint: Set \(\theta=20^{\circ}\) in the identity from the previous problem.]

Find the smallest positive number \(\theta\) such that \(e^{\tan \theta}=500\).

Find a formula for \(\cos (4 \theta)\) in terms of \(\cos \theta\).

The next two exercises emphasize that \(\sin (x-y)\) does not equal \(\sin x-\sin y .\) For \(x=5.7\) radians and \(y=2.5\) radians, evaluate each of the following: (a) \(\sin (x-y)\) (b) \(\sin x-\sin y\)
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