2015年7月6日 · How do you find all the solutions for #2 \sin^2 \frac{x}{4}-3 \cos \frac{x}{4} = 0# over the

2015年4月23日 · How do you verify the identity #sin4x= 4sinxcosx(1-sin^2x)#? Trigonometry Trigonometric Identities and Equations Proving Identities. 1 Answer

2015年5月28日 · I'll start from the double angle identities: cos 2theta = cos^2 theta - sin^2 theta sin 2theta = 2sin theta cos theta Then: sin 4x = 2sin 2x cos 2x =2(2 sin x cos x)(cos ^2x - sin^2 x) = 2(2sin x cos x cos^2 x - 2sinx cos x sin^2 x) =4sin x cos^3 x - 4 sin^3 x cos x

#x in {30^@, 120^@, 150^@, 240^@}# #color(white)("XXX")+- "any integer multiple of " 360^@#

2015年2月3日 · The answer is: 2. The amplitude of a periodic function is the numer that multiply the function itself. Using the double-angle formula of sinus, that says: sin2alpha=2sinalphacosalpha, we have: y=2*2sinxcosx=2sin2x. So the amplitude is 2. This is the sinus function: graph{sinx [-10, 10, -5, 5]} This is the y=sin2x function (the period becomes pi): graph{sin(2x) [-10, 10, -5, 5]} and this is the ...

2016年6月13日 · As below RHS =4sinxcosx-8sin^3xcosx =4sinxcosx(1-2sin^2x) =2sin2xcos2x=sin4x=LHS

2017年1月26日 · see below 4 sin x cos x-2sinx =2sinx(cosx-1) or 4 sin x cos x-2sinx =2(2sinxcosx)-2sinx =2sin2x-2sinx

2017年2月27日 · x = 1/12 pi, 5/12 pi, 13/12 pi and 17/12 pi 4sin x cos x = 1 2 ( 2 sin x cos x) =1, where 2sin x cos x = sin 2 x 2 sin 2x = 1 sin 2x = 1/2 2x =sin^(-1)(1/2) 2x = 1/6 ...

2016年11月20日 · How do you simplify #(-4cosxsinx+2cos2x)^2+(2cos2x+4sinxcosx)^2#? Trigonometry Trigonometric Identities and Equations Double Angle Identities

2016年11月28日 · We can use here the algebraic identity. #a^3-b^3=(a-b)(a^2+b^2+ab)# Using this #(cos^3-sin^3x)/(cosx-sinx)# = #((cosx-sinx)(cos^2x+sin^2x+sinxcosx))/(cosx-sinx)#