O orca432 May 2010 2 0 May 27, 2010 #1 Hey, I have a few questions I just can't seem to get down, looking for some help here... 1) sin2x = 2csc2x-tanx 1-cos2x 2) cos(x+y)cos(x-y)=(cosx)^2 +(cosy)^2 -1 3) (cosx)^6 + (sinx)^6 = 1-3(sinx)^2 + 3(sinx)^4 Thanks in advance!

Hey, I have a few questions I just can't seem to get down, looking for some help here... 1) sin2x = 2csc2x-tanx 1-cos2x 2) cos(x+y)cos(x-y)=(cosx)^2 +(cosy)^2 -1 3) (cosx)^6 + (sinx)^6 = 1-3(sinx)^2 + 3(sinx)^4 Thanks in advance!

harish21 Feb 2010 1,036 386 Dirty South May 27, 2010 #2 orca432 said: Hey, I have a few questions I just can't seem to get down, looking for some help here... 1) sin2x = 2csc2x-tanx 1-cos2x 2) cos(x+y)cos(x-y)=(cosx)^2 +(cosy)^2 -1 3) (cosx)^6 + (sinx)^6 = 1-3(sinx)^2 + 3(sinx)^4 Thanks in advance! Click to expand... You need to show you work too.. (2) \(\displaystyle cos(x+y)cos(x-y) = (cosx.cosy-sinx.siny)(cosx.cosy+sinx.siny)\) \(\displaystyle = (cos^2x.cos^2y) - (sin^2x.sin^2y)\) \(\displaystyle = (cos^2x.cos^2y) - [(1-cos^2x)(1-cos^2y)]\) multiply and cancel to complete the proof Reactions: orca432

orca432 said: Hey, I have a few questions I just can't seem to get down, looking for some help here... 1) sin2x = 2csc2x-tanx 1-cos2x 2) cos(x+y)cos(x-y)=(cosx)^2 +(cosy)^2 -1 3) (cosx)^6 + (sinx)^6 = 1-3(sinx)^2 + 3(sinx)^4 Thanks in advance! Click to expand... You need to show you work too.. (2) \(\displaystyle cos(x+y)cos(x-y) = (cosx.cosy-sinx.siny)(cosx.cosy+sinx.siny)\) \(\displaystyle = (cos^2x.cos^2y) - (sin^2x.sin^2y)\) \(\displaystyle = (cos^2x.cos^2y) - [(1-cos^2x)(1-cos^2y)]\) multiply and cancel to complete the proof

harish21 Feb 2010 1,036 386 Dirty South May 27, 2010 #3 for (3) \(\displaystyle (cosx)^6+(sinx)^6\) \(\displaystyle = (cos^{2}x)^3 + (sin^{2}x)^3\) since \(\displaystyle (a^3+b^3) = (a+b)(a^2-ab+b^2)\), you can write: \(\displaystyle = (cos^{2}x+sin^{2}x)(cos^{4}x-cos^{2}x. sin^{2}x+sin^{4}x)\) \(\displaystyle = 1 (cos^{4}x-cos^{2}x. sin^{2}x+sin^{4}x)\) convert all cosines into sines and your proof will be complete. Reactions: orca432

for (3) \(\displaystyle (cosx)^6+(sinx)^6\) \(\displaystyle = (cos^{2}x)^3 + (sin^{2}x)^3\) since \(\displaystyle (a^3+b^3) = (a+b)(a^2-ab+b^2)\), you can write: \(\displaystyle = (cos^{2}x+sin^{2}x)(cos^{4}x-cos^{2}x. sin^{2}x+sin^{4}x)\) \(\displaystyle = 1 (cos^{4}x-cos^{2}x. sin^{2}x+sin^{4}x)\) convert all cosines into sines and your proof will be complete.

O orca432 May 2010 2 0 May 27, 2010 #4 harish21 said: for (3) \(\displaystyle (cosx)^6+(sinx)^6\) \(\displaystyle = (cos^{2}x)^3 + (sin^{2}x)^3\) since \(\displaystyle (a^3+b^3) = (a+b)(a^2-ab+b^2)\), you can write: \(\displaystyle = (cos^{2}x+sin^{2}x)(cos^{4}x-cos^{2}x. sin^{2}x+sin^{4}x)\) \(\displaystyle = 1 (cos^{4}x-cos^{2}x. sin^{2}x+sin^{4}x)\) convert all cosines into sines and your proof will be complete. Click to expand... When I converted them, I got \(\displaystyle 1-(sinx)^2+(sinx)^4\), anyone care to help me?

harish21 said: for (3) \(\displaystyle (cosx)^6+(sinx)^6\) \(\displaystyle = (cos^{2}x)^3 + (sin^{2}x)^3\) since \(\displaystyle (a^3+b^3) = (a+b)(a^2-ab+b^2)\), you can write: \(\displaystyle = (cos^{2}x+sin^{2}x)(cos^{4}x-cos^{2}x. sin^{2}x+sin^{4}x)\) \(\displaystyle = 1 (cos^{4}x-cos^{2}x. sin^{2}x+sin^{4}x)\) convert all cosines into sines and your proof will be complete. Click to expand... When I converted them, I got \(\displaystyle 1-(sinx)^2+(sinx)^4\), anyone care to help me?

harish21 Feb 2010 1,036 386 Dirty South May 27, 2010 #5 orca432 said: When I converted them, I got \(\displaystyle 1-(sinx)^2+(sinx)^4\), anyone care to help me? Click to expand... \(\displaystyle cos^{4}x = (cos^{2}x)^2= (1-sin^{2}x)^2=1-2sin^{2}x+sin^{4}x\) substitute this in what you had earlier on \(\displaystyle 1 . (cos^{4}x-cos^{2}x. sin^{2}x+sin^{4}x)\) \(\displaystyle =1-2sin^{2}x+sin^{4}x-[(1-sin^{2}x). sin^{2}x]+sin^{4}x\) finish it...

orca432 said: When I converted them, I got \(\displaystyle 1-(sinx)^2+(sinx)^4\), anyone care to help me? Click to expand... \(\displaystyle cos^{4}x = (cos^{2}x)^2= (1-sin^{2}x)^2=1-2sin^{2}x+sin^{4}x\) substitute this in what you had earlier on \(\displaystyle 1 . (cos^{4}x-cos^{2}x. sin^{2}x+sin^{4}x)\) \(\displaystyle =1-2sin^{2}x+sin^{4}x-[(1-sin^{2}x). sin^{2}x]+sin^{4}x\) finish it...

pickslides MHF Helper Sep 2008 5,237 1,625 Melbourne May 27, 2010 #6 Here's a little more. \(\displaystyle \cos^{4}x-\cos^{2}x \sin^{2}x+\sin^{4}x\) \(\displaystyle \cos^{2}x\cos^{2}x-\cos^{2}x \sin^{2}x+\sin^{4}x\) \(\displaystyle (1-\sin^2{x})(1-\sin^2{x})-(1-\sin^2{x}) \sin^{2}x+\sin^{4}x\) Now expand this.

Here's a little more. \(\displaystyle \cos^{4}x-\cos^{2}x \sin^{2}x+\sin^{4}x\) \(\displaystyle \cos^{2}x\cos^{2}x-\cos^{2}x \sin^{2}x+\sin^{4}x\) \(\displaystyle (1-\sin^2{x})(1-\sin^2{x})-(1-\sin^2{x}) \sin^{2}x+\sin^{4}x\) Now expand this.