<p>I am confused with only this certain topic in the chapter about trigonometry. The directions/tips in my textbook are surprisingly very vague (every other instructions are very clear, thorough, and understandable).</p>
<p>Write sin(tan^-1*u*) as an algebraic expression containing u.</p>
<p>Let Θ = tan^-1*u* so that tanΘ = u, -π/2 < Θ < π/2, -∞ < u < ∞. As a result, we know that secΘ > 0. Then</p>
<p>sin(tan^-1*u*) = sinΘ = sinΘ x cosΘ/cosΘ = tanΘcosΘ = tanΘ/secΘ = tanΘ / √1+tan^2Θ = u / √1+u^2
........................... ↑ ........................ ↑ ............................... ↑
.....Multiply by 1: cosΘ / cosΘ ...... sinΘ / cosΘ = tanΘ ............sec^2Θ = 1 + tan^2Θ, secΘ > 0</p>
<p>I am befuddled...</p>
<p>Why do “we know that secΘ > 0” if “tanΘ = u, -π/2 < Θ < π/2, -∞ < u < ∞?”
& why use “cosΘ/cosΘ” as “1” to multiply to “sinΘ?”</p>
<p>
It’s just a way of converting the sin to a tan. Basically, you’re doing this:
We know tan = sin/cos
=> sin = tan * cos</p>
<p>sin Θ = tan Θ * cos Θ
= tan Θ * (1/sec Θ)
= tan Θ / √(1 + tan² Θ)
We know u = tan Θ</p>
<p>=> sin Θ = u/√(1+u²)
=> sin(atan u) = u/√(1+u²)</p>
<p>
-π/2 < Θ < π/2 => Θ is in quadrant I or IV. We know cos is positive here, therefore sec is also positive here.</p>
<p>Shiz. I love you. Seriously. Thank you so much, HarveyLewis! My brain now is no longer aching :)</p>
<p>Jesus I havn’t done this kind of **** in forever.</p>
<p>neither have I. That’s why they made wolfram alpha</p>
<p>Usually I would draw a right triangle to help me on this type of problem.</p>
<p>"That’s why they made wolfram alpha " → It’s true.</p>
<p>TRUFFLIEPUFF: No problem.</p>
<p>[noparse]We apologize for the horrible pun.[/noparse]</p>