Question

Proving trig equations
(sin x)(tan x cos x - cot x cos x) = 1 - 2 cos^2 x

Answer 1

On the left hand side of the equation use:
\[\tan x=\frac{\sin x}{\cos x}\]
and: \[\cot x=\frac1{\tan x}=\frac{\cos x}{\sin x}\]

Answer 2

\((sin x)(\dfrac{sinx}{cosx}cosx-\dfrac{cosx}{sinx}cosx)=1-2soc^2x\)

Answer 3

uh try to fix what is inside the second parenthesis, see how you can manipulate it

Answer 4

good, now you can simplify this part
\[\frac{\sin x}{\cos x}{\cos x}\]

Answer 5

Wouldn't \(\dfrac{sinx}{cosx}\) and \(\dfrac{cosx}{sinx}\) cancel?

Answer 6

if you're multiplying
and it does not cancel, it gives you 1

Answer 7

Oh ok

Answer 8

\((sin x)(1-\dfrac{cosx}{sinx}cosx)=1-2cos^2x\)

Answer 9

whoa whoa whoa how is that possible?

Answer 10

\[(\sin x)\left(\frac{\sin x}{\cos x}{\cos x}-\frac{\cos x}{\sin x}\cos x\right)=1-2\cos^2x\\
(\sin x)\left(\frac{\sin x}{\cancel{\cos x}}{\cancel{\cos x}}-\frac{\cos x}{\sin x}\cos x\right)=1-2\cos^2x\\\vdots\]

Answer 11

Oh so \((sin x)(sinx-\dfrac{cosx}{sinx}cosx)=1-2cos^2x\) right?

Answer 12

you're not done yet!

Answer 13

that's better, now distribute the (sin x) into the big brackets

Answer 14

Ok, \((sinx^2-\dfrac{cosx}{sinx}cosx~sinx)=1-2cos^2x\). Then \(\dfrac{cosx}{sinx}cosx~sinx\) cancels to become cosx*cosx?

Answer 15

yeah and cos x * cos x = cos^2 x

Answer 16

Ok, so now it is \(sin^2x-cos^2x=1-2cos^2x\)?

Answer 17

good

Answer 18

now what do you get if you add cos^2x to both sides?

Answer 19

\(\sf sin^2x=1\)

Answer 20

not quite

Answer 21

Oh, no, \(\sf sin^2x=1-cos^2x\)

Answer 22

can you rearrange this to the one trig identity you can remember?

Answer 23

\(\sf sin^2x+cos^2x=1\)?

Answer 24

yep, !

Answer 25

:D

Answer 26

wait but when we have to prove both sides
we can solve only one side
mean to say cannot move left to right or right to left anything right ??

Answer 27

Why is that a condition of a proof?

Answer 28

:D nvm my teacher said that

Answer 29

Nnesha wht ur teacher said?

Answer 30

my teacher also used to say that

Answer 31

your both teachers are good
they are guiding you how to write a neat looking proof :)

Answer 32

What is wrong with using equations?

Answer 33

nothing
it is an equation from beginning to end
LAUGHING OUT LOUD

Answer 34

nothing wrong
adding stuff to both sides and messing with both sides looks messy.. thats all..

Answer 35

it doesnt change much when u use both sides to prove ,

the proof remains true

Answer 36

^^^ lol yeah
thanks
to all of you
but btw if i do same thing she probably gonna take offff ppppoints :D:(

Answer 37

first 5 minutes of this video
https://www.youtube.com/watch?v=NuGDkmwEObM

Answer 38

it says "well organized".

Answer 39

it says below strategy doesnt work always but i can't think of a simple example where it could fail at the moment :

A = B
..=..
..=..
1 = 1

Answer 40

im sure this shouldn't matter for almost all the proofs in trig

Answer 41

okay thanks :)

Answer 42

@UnkleRhaukus good work :):)

Answer 43

\[LHS =(\sin )\left(\tan{\cos }-\cot\cos \right)\\
\qquad=(\sin )\left(\frac{\sin }{\cos }{\cos }-\frac{\cos }{\sin }\cos \right)\\
\qquad=\sin^2-\cos^2\]
Now using the the identity \[\sin^2+\cos^2=1\\\qquad\quad\sin^2=1-\cos^2\]
\[LHS = (1-\cos^2)-\cos^2\\\qquad=1-2\cos^2\\\qquad=RHS\]

(Even without the x's) this looks more messy to me.

Answer 44

Thank you for the help @UnkleRhaukus

Answer 45

I think so, both are messy in some sense and they are okay for trig proofs.
Having said that, the video i attached earlier is talking about a proof like this i guess :

\[\begin{align} &\sin^2x + \cos^2x = 1 \\~\\
&\sin^2x -\cos^2 x = 1-2\cos^2x\\~\\
& \sin x \left[\tan x \cos x - \cot x \cos x \right] = 1-2\cos^2x\\~\\
\end{align}\]

Here we have started with a known identity and arrived at the desired result. Ofcourse we had to do some bottom up scratch work

Answer 46

i like it