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Monday, May 4, 2009, 10:35 PM
John asked me how to do this one...(A maths)
prove [cos^2(2x)]-[cos^2(x)]+sinxsin3x=01st term 2nd term 3rd term For the 1st term, LEAVE IT FOR NOW. For the 2nd term, Also leave it =D For the 3rd term, We leave the 1st part. the sinx part. We work on sin3x. sin3x=sin(2x+x) We apply addition formulae. We just need to seperate sin(2x+x). sin(2x+x)=sin2xcosx+cos2xsinx Double angle formula! sin2xcosx+cos2xsinx=(2sinxcosx)(cosx)+([2cos^(x)]-1](sinx) =[2sinx[cos^2(x)]+ [2sinx[cos^2(x)]-sinx not to forget, we still have that first part we left out just now. So up till now, we have: sinxsin3x=sinx([2sinx[cos^2(x)]+ [2sinx[cos^2(x)]-sinx) =sinx([4sinx[cos^2(x)]-sinx) Expand it... =(4[sin^2(x)][cos^2(x)])-[sin^2(x)] Apply double angle on (4[sin^2(x)][cos^2(x)]), because that is actually (2sinxcosx)^2 So now we have (sin2x)^2 Putting everything we have together, (cos2x)^2-[cos^2(x)]+(sin2x)^2-[sin^2(x)] Do you see it now? Apply this one>>>(sinx)^2+(cosx)^2=1 Rearranging... {(sin2x)^2+(cos2x)^2} - {[sin^2(x)]+[cos^2(x)]}=1 - 1 =0 I know this aint much, just one qn that I thought was prettyyy challenging. I'm not very good too. I got quite a few questions that I dont know how to do. But I just wanna help a little. I dont even think most of you understand >.< You might be able to see it better if you write it out =D. |