<p>Sorry to ask for help, this was a problem on an assignment quiz of ours due, well, earlier today, and it's still driving me crazy. We all gave up. According to my math teacher, it's hard but we should've gotten it. Anyway, it's a proof, manipulate one side or the other but not both. x is theta.</p>
<p>sinx cos2x sin3x = 1/4(1-cos2x+cos4x-cos6x)</p>
<p>39 views, 0 replies. Help, anyone?</p>
<p>use partial fractions</p>
<p>what math is this?</p>
<p>it's trig, use your identities, (double angle, sum, etc)</p>
<p>Have either of you gone through all the steps, or are you just suggesting? I know it's manipulating trig identities 'cause the unit IS trig identities/functions/equations, but you could go various ways with them. If you have gone through all the steps, would you care to post the whole process? I've gone through it (lazily) several different ways without seeing how they would lead to the answer. Oh well. I'll try again now.</p>
<p>I don't really believe there's an elegant solution. Here's my long-winded attempt:</p>
<p>4 sinx cos2x sin3x = 1-cos2x+cos4x-cos6x
RHS = 1-cos2x+cos4x-cos6x = 1-cos2x+cos(2x+2x)-cos(2x+4x) = 1-cos2x+cos2xcos2x-sin2xsin2x-cos2xcos4x+sin2xsin4x = 1-cos2x+cos2xcosx-sin2xsin2x-cos2x(cos2xcos2x-sin2xsin2x)+sin2x(sin2xcos2x+cos2xsin2x) = 1-cos2x+cos2xcosx-sin2xsin2x+cos2xcos2xcos2x-co2xsin2xsin2x+sin2xsin2xcos2x+sin2xsin2xcos2x = 1-cos2x+cos²2x+cos³2x-sin²2x+sin²2xcos2x=1-cos2x+cos²2x+cos³2x-(1-cos²2x)+(1-cos²2x)cos2x=1-cos2x+cos²2x+cos³2x-1+cos²2x+cos2x-cos²2x=cos³2x+cos²2x=cos²2x(cos2x+1)=(2cos²x-1)(2cos²x)=4(cos²x)²-2cos²x=4(cos²x)²-2(1-2sin²x)=4(cos²x)²-2+4sin²x=4(1-sin²x)²-2+4sin²x=4(1-2sin²x+(sin²x)²)-2+4sin²x=4-8sin²x+(sin²x)²+2+4sin²x=(sin²x)²-4sin²x+2</p>
<p>I don't know where I went wrong, but I guess this could help...</p>
<p>There are three formulas you need:
cos(2x)=1-sin(x)^2
cos(u)-cos(v)=-2<em>sin[(u+v)/2]</em>sin[(u-v)/2]
sin(u)+sin(v)=2<em>sin[(u+v)/2]</em>cos[(u-v)/2]</p>
<p>Modify equation to (for your work though, keep 1/4 on RHS):4sin(x)<em>cos(2x)</em>sin(3x)=1-cos(2x)+cos(4x)-cos(6x)</p>
<p>Pay attention to brackets, these will be used to group things and show equality between expressions.</p>
<p>1-cos(2x)+cos(4x)-cos(6x)=[1-cos(2x)]+[cos(4x)-cos(6x)]=[2sin(x)^2]+[2<em>sin(x)</em>sin(5x)]=[2<em>sin(x)]</em>[sin(x)+sin(5x)]=[2<em>sin(x)]</em>[2<em>sin(3x)</em>cos(-2x)]=4<em>sin(x)</em>sin(3x)*cos(2x)</p>
<p>Therefore, sin(x)<em>cos(2x)</em>sin(3x)=1/4*[1-cos(2x)+cos(4x)-cos(6x)]</p>
<p>QED</p>
<p>averagemathgeek, where did you get those formulas?</p>
<p>the formulas I have are
sin2x=2sinxcosx
cos2x=(cos^2)x-(sin^2)x
cos2x=1-2(sin^2)x
cos2x=2(cos^2)x-1</p>
<p>sin(a+b) = sinacosb+cosasinb
sin(a-b) = sinacosb-cosasinb
cos(a+b) = cosacosb-sinasinb
cos(a-b) = cosacosb+sinasinb</p>
<p>Those formulas are generally in the back covers of precalculus and calculus texts. They can be verified by the formulas you are familiar with (as well as derived).</p>