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anonymous
 one year ago
Are
(sin3x/sin x cos x)=4 cos xsec x
and
(sin 3xsin x)/(cos 3x+cos x)= tan x
identities?
anonymous
 one year ago
Are (sin3x/sin x cos x)=4 cos xsec x and (sin 3xsin x)/(cos 3x+cos x)= tan x identities?

This Question is Closed

sohailiftikhar
 one year ago
Best ResponseYou've already chosen the best response.4use the formulas to find this

sohailiftikhar
 one year ago
Best ResponseYou've already chosen the best response.4sin3x=3sinx4sin^3

sohailiftikhar
 one year ago
Best ResponseYou've already chosen the best response.4ok u can put 3sinx4sin^3x rather than sin3x (3sinx4sin^3x)/(sinx)(cosx) now take sinx common from above equation sinx(34sin^2x)/sinx cosx =(34sin^2x)/cosx

sohailiftikhar
 one year ago
Best ResponseYou've already chosen the best response.4now you can put sin^2x=1cos^2x (34(1cos^2x)/cosx (34+4cos^2x)/cosx 1+4cos^2x/cosx 1(14cos^2x)/cosx

sohailiftikhar
 one year ago
Best ResponseYou've already chosen the best response.4as u know cos2x=12cos^2x and cos4x=14cos^2x now put 14cos^2x as cos4x

sohailiftikhar
 one year ago
Best ResponseYou've already chosen the best response.4so part a is incorrect

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank 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?

sohailiftikhar
 one year ago
Best ResponseYou've already chosen the best response.4are u asking about second equation ? and no problem it's my pleasure to help u .

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes please, I'd love for help with the second equation!

sohailiftikhar
 one year ago
Best ResponseYou've already chosen the best response.4ok sin3x=3sinx4cos^3x cos3x=4cos^3x3cosx

sohailiftikhar
 one year ago
Best ResponseYou've already chosen the best response.4we will solve both separately first then we'll multiply them ok u can put 3sinx4sin^3x rather than sin3x (3sinx4sin^3xsinx) now take sinx common from above equation sinx(34sin^2x1) =sinx(24sin^2x)

sohailiftikhar
 one year ago
Best ResponseYou've already chosen the best response.4and for cos3x+cosx as cos3x=4cos^3x3cosx so 4cos^3x3cosx+cosx now take cosx common cosx(4cos^2x3+1)=cosx(4cos^2x2)

sohailiftikhar
 one year ago
Best ResponseYou've already chosen the best response.4now combine both equations sinx(24sin^2x)/cosx(4cosx^22)

sohailiftikhar
 one year ago
Best ResponseYou've already chosen the best response.4take 2 from above and 2 from below 2(sinx)(12sin^2x)/2(cosx)(2cos^21) =sinx/cosx=tanx and yes it is true

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank 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.

sohailiftikhar
 one year ago
Best ResponseYou've already chosen the best response.4may be first one can true .. perhaps i did some mistake in solving this

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It might be the secant part in (sin3x/sin x cos x)=4 cos xsec x ? I'm not very good at trig identities, though....

sohailiftikhar
 one year ago
Best ResponseYou've already chosen the best response.4ok try this urself by using the formulas which i told u . u can do this now just don't skip any step ok

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[3sinx4\sin^3/\sin x \cos x =4 \cos x\sec x\] \[(34\sin^2x)/cosx=4 \cos x  \sec x\] \[(34(1\cos^2x)/cosx=4\cos x\sec x\] \[1+4\cos^2x/cosx\] \[ 14\cos^2x/\cos=4\cos x\sec x\] \[4\cos^2x1/\cos=4\cos^2x1/\cos\] I tried my best c': it's just my best is sub par....
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