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use the formulas to find this
ok u can put 3sinx-4sin^3x rather than sin3x (3sinx-4sin^3x)/(sinx)(cosx) now take sinx common from above equation sinx(3-4sin^2x)/sinx cosx =(3-4sin^2x)/cosx
now you can put sin^2x=1-cos^2x (3-4(1-cos^2x)/cosx (3-4+4cos^2x)/cosx -1+4cos^2x/cosx -1(1-4cos^2x)/cosx
as u know cos2x=1-2cos^2x and cos4x=1-4cos^2x now put 1-4cos^2x as cos4x
so part a is incorrect
Thank you for pointing it out step by step! I have a really hard time figuring out which identities to use and how to simplify them. I hope you don't mind me asking, what step was secant changed?
are u asking about second equation ? and no problem it's my pleasure to help u .
Yes please, I'd love for help with the second equation!
ok lemme check
ok sin3x=3sinx-4cos^3x cos3x=4cos^3x-3cosx
we will solve both separately first then we'll multiply them ok u can put 3sinx-4sin^3x rather than sin3x (3sinx-4sin^3x-sinx) now take sinx common from above equation sinx(3-4sin^2x-1) =sinx(2-4sin^2x)
and for cos3x+cosx as cos3x=4cos^3x-3cosx so 4cos^3x-3cosx+cosx now take cosx common cosx(4cos^2x-3+1)=cosx(4cos^2x-2)
now combine both equations sinx(2-4sin^2x)/cosx(4cosx^2-2)
take 2 from above and 2 from below 2(sinx)(1-2sin^2x)/2(cosx)(2cos^2-1) =sinx/cosx=tanx and yes it is true
Thank you! The second one was true but apparently the first one was also true, not false. I appreciate the time it took to go through all the steps, though! It makes understanding the concept a lot easier.
may be first one can true .. perhaps i did some mistake in solving this
It might be the secant part in (sin3x/sin x cos x)=4 cos x-sec x ? I'm not very good at trig identities, though....
ok try this urself by using the formulas which i told u . u can do this now just don't skip any step ok
\[3sinx-4\sin^3/\sin x \cos x =4 \cos x-\sec x\] \[(3-4\sin^2x)/cosx=4 \cos x - \sec x\] \[(3-4(1-\cos^2x)/cosx=4\cos x-\sec x\] \[-1+4\cos^2x/cosx\] \[ -1-4\cos^2x/\cos=4\cos x-\sec x\] \[4\cos^2x-1/\cos=4\cos^2x-1/\cos\] I tried my best c': it's just my best is sub par....