Question

Are
(sin3x/sin x cos x)=4 cos x-sec x
and
(sin 3x-sin x)/(cos 3x+cos x)= tan x
identities?

Answer 1

use the formulas to find this

Answer 2

sin3x=3sinx-4sin^3

Answer 3

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

Answer 4

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

Answer 5

as u know cos2x=1-2cos^2x
and cos4x=1-4cos^2x
now put 1-4cos^2x as cos4x

Answer 6

-cos4x/cosx

Answer 7

so part a is incorrect

Answer 8

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?

Answer 9

are u asking about second equation ? and no problem it's my pleasure to help u .

Answer 10

Yes please, I'd love for help with the second equation!

Answer 11

ok lemme check

Answer 12

ok sin3x=3sinx-4cos^3x
cos3x=4cos^3x-3cosx

Answer 13

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)

Answer 14

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)

Answer 15

now combine both equations
sinx(2-4sin^2x)/cosx(4cosx^2-2)

Answer 16

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

Answer 17

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.

Answer 18

may be first one can true .. perhaps i did some mistake in solving this

Answer 19

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....

Answer 20

ok try this urself by using the formulas which i told u . u can do this now just don't skip any step ok

Answer 21

\[3sinx-4\sin^3/\sin x \cos x =4 \cos x-\sec x\]
\[(3-4\sin^2x)/cosx=4 \cos x - \sec x\]
\[(3-4(1-\cos^2x)/cosx=4\cos x-\sec x\]
\[-1+4\cos^2x/cosx\]
\[ -1-4\cos^2x/\cos=4\cos x-\sec x\]
\[4\cos^2x-1/\cos=4\cos^2x-1/\cos\]
I tried my best c': it's just my best is sub par....