Trig

[Ducky Havok]

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).

[573v3]

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

[Ducky Havok]

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.

[ecoli]

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)

[573v3]

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

[Ducky Havok]

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

[Ducky Havok]

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

[573v3]

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