<p>Here is one question that is bothering me and wasn't mentioned in the official thread. If sin(x)=.91 and tan(x)=some number I can't remember. What is cos(x)?</p>
<p>I'm not gonna explain here, for the sake of keeping it in one thread. Check the official thread.</p>
<p>It wasn't on the official thread, that I could find!</p>
<p>So ask on the official thread. That's where I addressed it, check now.</p>
<p>Let's say that number you can't remember is "a."</p>
<p>Tan(x)=Sin(x)/Cos(x)</p>
<p>Solve for Cos(x)...</p>
<p>Cos(x)=Sin(x)/Tan(x)</p>
<p>Now, since you have both values, you can plug them in and get your answer.</p>
<p>Cos(x)=(.91)/(a)</p>
<p>sin(x)^2+cos(x)^2=1</p>
<p>cos(x)^2=1-sin(x)^2</p>
<p>cos(x)^2=1-.91^2</p>
<p>cosx=.4146</p>
<p>There is one problem in your proof towerpumpkin. If cos(x)^2=0.1719, then cos(x) is equal to plus or minus .4146. Which is it? In the official thread, Bog tried to explain it and got -.4146 as the answer but his explination didn't make sense.</p>
<p>it was -.414
I'm 100% sure.</p>
<p>Just multiply tan(x) by cos(x).</p>
<p>Oh...crap...I forgot completely about that LOL (sigh this is what happens when you don't use your brain in over a month)....If you had the value of tan(x), however, then you can figure out the sign of cos(x) pretty easily (if tan is negative, cos is negative and vice versa).</p>
<p>I like alessandro's and towerpumpkin's (if he had considered the negative tangent) methods and wish I had thought of those, but I simply realized that the angle, x, had to be in the second quadrant because of the signs, so I took the x that I got from the arcsin and subtracted it from 180 (reflected it over the y-axis). Then I took the cosine of that, and that was my answer.</p>
<p>How is it that my explanation didn't make sense if you proceeded to get an answer from it? It must have been somewhat coherent then.</p>
<p>Any of them would work fine. The only reason I chose to do it my way is so you don't come up with two answers and possible make a mistake. Whenever you are given information regarding (in this case) a tangent and a sine, you should use an identity/definition that includes both of them for the sake of simplicity.</p>
<p>Alessandro: Very nicely done. I admire the elegance and simplicity of your method, and I'll definitely use that if I ever come across a problem like this in the future.</p>
<p>I used something like Bog's method on the actual test, however.</p>
<p>Was -.414 answer C, and 4.14 D?</p>
<p>it was negative!!!!!!!!!</p>
<p>yea, it was negative i believe</p>
<p>from the signs of sin and tan, the quadrant in which the angle lies can be determined. the rest is trivial</p>