Ducky Havok Posted October 10, 2005 Share Posted October 10, 2005 I just took the SAT Math 2 subject test Saturday, and the very last question was the only that I couldn't figure out, so I'm kind of curious how to do it. The question read if [math]\sin{\theta}=x^2[/math], then what does [math]\sin{2\theta}[/math] equal in terms of x? I've never been very good at trig, so I'm kind of interested in how to do it (and if it's something glaringly obvious I'm gonna feel like an idiot). Link to comment Share on other sites More sharing options...
573v3 Posted October 10, 2005 Share Posted October 10, 2005 This is a standard formula: Sin (2a) = 2 sin (a) cos (a) Lets use it for your problem (I replaced theta with t for easy typing, hope it isn’t to confusing with t-formulas) Sin (2t) = 2 sin (t) cos (t) of which we can replace Sin (t) with x^2 so: Sin (2t) = 2 x^2 cos (t) somehow I sense there’s a mush more elegant solution then this, but technically, this would be a valid answer. And your signature does ask for simplicity Link to comment Share on other sites More sharing options...
Ducky Havok Posted October 10, 2005 Author Share Posted October 10, 2005 The only problem is they wanted it completely in terms of x, so no theta's at all. Stupid multiple choice questions. The answer I ended up picking was something like [math]2x^2\sqrt{1-x^4}[/math], so I guess now all that I'm missing is a simplification for cos(t). I really wish my highschool had offered trig, but the closest I got was geometry with a Puertorican teacher who we couldn't understand. Link to comment Share on other sites More sharing options...
ecoli Posted October 10, 2005 Share Posted October 10, 2005 cos (t) = 1 - sin (t) you could have plugged x^2 back in from the original equation. so Sin (2t) = 2 x^2 (1 - x^2) Link to comment Share on other sites More sharing options...
573v3 Posted October 10, 2005 Share Posted October 10, 2005 It's been a long time since I dealth with this, but aren't you confusing: cos (t) = 1 - sin (t) with: cos (t)^2 = 1 - sin (t)^2 Link to comment Share on other sites More sharing options...
Ducky Havok Posted October 10, 2005 Author Share Posted October 10, 2005 How's cos(t)=1-sin(t)? Example: [math]\cos{\frac{\pi}{6}}=\frac{\sqrt{3}}{2}[/math], but [math]1-\sin{\frac{\pi}{6}}=.5[/math] cos(t)=sin(pi/2-t), but I don't see how that's much help Link to comment Share on other sites More sharing options...
Ducky Havok Posted October 10, 2005 Author Share Posted October 10, 2005 Oh, I got it thanks to 573v3! cos(t)=sqrt(1-sin²(t)) so cos(t)=sqrt(1-x^4), meaning I was right! I'm a damn good guesser Link to comment Share on other sites More sharing options...
573v3 Posted October 10, 2005 Share Posted October 10, 2005 damn, you beat me to it, and there i was wasting my time ttrying it with: Sin (2a) = [2 tg (a) ] / [1 + tg^2 (a)] Sin (2t) = {2 sin (t) / cos (t) } / {1 + [sin^2 (t) / cos^2 (t)]} Sin (2t) = {4x^2 / cos (t) } / { 1 + [x^4 / cos^2 (t) ]} Sin (2t) = {4x^2 / cos (t) } + {4x^2 / cos (t) }/ {x^4 / cos^2 (t)} Sin (2t) = {4x^2 / cos (t) } + {4x^2 / cos (t) }{cos^2 (t) / x^4 } Sin (2t) = {4x^2 / cos (t) } + {4 cos(t) / x^2 } Sin (2t) = 4{ [x^4 + cos^2(t) ] / [x^2 cos (t) ]} Didn't amount to much Glad you found the answer though Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now