Question

proving identities. please help

Answer 1

what's the question?

Answer 2

no. 3

Answer 3

@robtobey

Answer 4

\(\color{#000000}{ \displaystyle \frac{\sin 3x}{\sin x}+\frac{\cos 3x}{\cos x} =2 }\)

\(\color{#000000}{ \displaystyle \frac{\cos x\sin 3x}{\sin x\cos x}+\frac{\sin x\cos 3x}{\sin x\cos x} =2 }\)

\(\color{#000000}{ \displaystyle \frac{\cos x\sin 3x+\sin x\cos 3x}{\sin x\cos x} =2 }\)

\(\color{#000000}{ \displaystyle \cos x\sin 3x+\sin x\cos 3x=2\sin x\cos x }\)

\(\color{#000000}{ \displaystyle \sin (3x+x)=2\sin x\cos x }\)

\(\color{#000000}{ \displaystyle \sin (4x)=\sin (2x) }\)

Answer 5

So whether or not #1 is true, should not be a pain to figure.

Answer 6

i alreay got no.1 im trying to figure out no. 3 :( @SolomonZelman

Answer 7

and the given in no. 1 is minus instead of plus...

Answer 8

oh, then #1 is true.

Answer 9

there will be a minus in there, so sin(3x-x)=sin(2x), basically...

Answer 10

I need to ask you something about the problems tho

Answer 11

what is it?. and i tried no. 3 but then again i got confused at the last part

Answer 12

In general when it says "verify" it is assumed that I can only use one side, and when it says "prove" it is assumed that I can use both sides.

So, can I use both sides, or should I use one side only?

Answer 13

(well, this in general, is hat I heard, but I wouldn't bet my life on what these words exactly imply regarding the permitted procedures)

Answer 14

my teacher says we can use both sides or one side.. so yeah u can use any. :)

Answer 15

but i tried proving it with one side but i got stucked up manipulating \[\sin 3x\]

Answer 16

\(\color{#000000}{ \displaystyle \frac{\sin (3t)}{\sin t \cos t}=4\cos t-\sec t }\)

\(\color{#000000}{ \displaystyle \frac{\sin 2t\cos t+\sin t\cos 2t}{\sin t \cos t}=4\cos t-\sec t }\)

\(\color{#000000}{ \displaystyle \frac{\sin 2t}{\sin t }+\frac{\cos 2t}{ \cos t}=4\cos t-\sec t }\)

\(\color{#000000}{ \displaystyle \frac{2\sin t\cos t}{\sin t }+\frac{\cos^2t-\sin^2t}{ \cos t}=4\cos t-\sec t }\)

\(\color{#000000}{ \displaystyle 2\cos t+\cos t-\frac{\sin^2t}{ \cos t}=4\cos t-\sec t }\)

\(\color{#000000}{ \displaystyle -\frac{\sin^2t}{ \cos t}=\cos t-\sec t }\)

Answer 17

can you go from here?

Answer 18

(Combine fractions on the right .... )

Answer 19

ok wait. i don't get it :(

Answer 20

everything or the specific steps?

Answer 21

the specific steps.. like on the last part and the equation before that

Answer 22

\(\color{#000000}{ \displaystyle \sin(\color{red}{x}+\color{blue}{y})=\sin(\color{red}{x})\cos(\color{blue}{y}) +\sin(\color{blue}{y})\cos(\color{red}{x}) }\)
\(\color{#000000}{ \displaystyle \sin(\color{red}{x}-\color{blue}{y})=\sin(\color{red}{x})\cos(\color{blue}{y}) -\sin(\color{blue}{y})\cos(\color{red}{x}) }\)

\(\color{#000000}{ \displaystyle \cos(\color{red}{x}+\color{blue}{y})=\cos(\color{red}{x})\cos(\color{blue}{y}) -\sin(\color{red}{x})\sin(\color{blue}{y}) }\)
\(\color{#000000}{ \displaystyle \cos(\color{red}{x}-\color{blue}{y})=\cos(\color{red}{x})\cos(\color{blue}{y}) +\sin(\color{red}{x})\sin(\color{blue}{y}) }\)

these are the properties I was using basically.....

Answer 23

how can i get the -sin^2t ... i already understand that sin^2t= cos^2t-1

Answer 24

oh wait i already got it

Answer 25

\[\frac{ -(1-\cos ^{2}t) }{ cost t } = -\frac{ 1 }{ \cos t } + \frac{ \cos ^{2} t }{ cost }\]
\[= -\sec t = \cos t\]
cost t - sec t

Answer 26

is this it???

Answer 27

\(\color{#000000}{ \displaystyle \frac{\sin (3t)}{\sin t \cos t}=4\cos t-\sec t }\)

\(\color{#ff0000}{ {\rm Since,}~~~\displaystyle \sin(x+y)=\sin x \cos y+\sin y \cos x }\)
\(\color{#ff0000}{ {\rm therefore,}~~~\displaystyle \sin(3x)=\sin(2x+x)= \sin 2x \cos x+\sin x \cos 2x }\)

\(\color{#000000}{ \displaystyle \frac{\sin 2t\cos t+\sin t\cos 2t}{\sin t \cos t}=4\cos t-\sec t }\)

\(\color{#ff0000}{ {\rm Splitting~the~fraction~, }~~~ }\)
\(\color{#ff0000}{ {\rm and~canceling~out~what~cancels~in~each~fraction. } }\)

\(\color{#000000}{ \displaystyle \frac{\sin 2t}{\sin t }+\frac{\cos 2t}{ \cos t}=4\cos t-\sec t }\)

\(\color{#ff0000}{ {\rm Since,}~~~\displaystyle \sin(x+y)=\sin x \cos y+\sin y \cos x }\)
\(\color{#ff0000}{ {\rm therefore,}~~~\displaystyle \sin(2x)=\sin(x+x)= \sin x \cos x+\sin x \cos x=2\sin x \cos x }\)

\(\color{#ff0000}{ {\rm and~since,}~~~\displaystyle \cos(x+y)=\cos x \cos y-\sin y \sin x }\)
\(\color{#ff0000}{ {\rm therefore,}~~~\displaystyle \cos(2x)= \cos(x+x)=\cos x \cos x-\sin x \sin x \\ =\cos^2x-\sin^2x }\)

\(\color{#000000}{ \displaystyle \frac{2\sin t\cos t}{\sin t }+\frac{\cos^2t-\sin^2t}{ \cos t}=4\cos t-\sec t }\)


\(\color{#ff0000}{ {\rm Algebra ~~~....~, }~~~ }\)

\(\color{#000000}{ \displaystyle 2\cos t+\cos t-\frac{\sin^2t}{ \cos t}=4\cos t-\sec t }\)

\(\color{#000000}{ \displaystyle -\frac{\sin^2t}{ \cos t}=\cos t-\sec t }\)

Answer 28

I am lagging with so many codes. Takes for ever for this box to respond((

Answer 29

thankyou very much for taking time to write the equations and explain them to me:)

Answer 30

but is my last part correct?

Answer 31

yes, it is.

Answer 32

and then the obvious equivalence that you wrote .....
so it is indeed an indentity.

Answer 33

identity **

Answer 34

yeaay! okaay2. thaanks again solomon!! :))

Answer 35

Not a problem!