Question

Trigonometry Help

Prove the Identity

(sinx - cosx)/(secx - cscx) = (sin2x)/2

I changed to cos-sin stuff and got:

(sinx - cosx)/((1/cosx)-(1/sinx))

but here i get stuck once again, idk where to go from here..

Answer 1

multiply the numerator and denominator by \(\sin x \cos x\)

Answer 2

multiply the numerator and denominator by \(\sin x \cos x\)

Answer 3

but gane, where is the sinx cosx even from? i just see sinx-cosx

Answer 4

so the sinxcosx is from the right side of the equation, i am just using it on the left?

Answer 5

nope, lets work it again from scratch

Answer 6

we take left hand side,
massage it such that we get the right hand side

Answer 7

secx - cscx = 1/c - 1/s = (s - c)/sc

Answer 8

so we start with left hand side :

\[\large \dfrac{\sin x - \cos x }{\sec x - \csc x} \]

Answer 9

your earlier first step was correct, change sec and csc to cos and sin in the denominator

Answer 10

alright and we change everything to sin/cos

Answer 11

yes

Answer 12

\[\large \dfrac{\sin x - \cos x }{\sec x - \csc x} = \dfrac{\sin x - \cos x }{\frac{1}{\cos x}- \frac{1}{\sin x}} \]

Answer 13

Now stare at the denominator fractions

Answer 14

can you combine them by getting a common denominator ?

Answer 15

yes

Answer 16

whats the LCM of \(\cos x \), \(\sin x\) ?

Answer 17

well we want to get it to sinxcosx correct?

Answer 18

yes

Answer 19

so it would be..\[\frac{ sinx-cox }{ sinxcosx }\]

Answer 20

Correct! thats the denominator

Answer 21

alright alright, next we want to get rid of the denominator's denominator?

Answer 22

well more like we dont want the denominator to be a fraction

Answer 23

\[\large \dfrac{\sin x - \cos x }{\sec x - \csc x} = \dfrac{\sin x - \cos x }{\frac{1}{\cos x}- \frac{1}{\sin x}}\]

\[ \large = \dfrac{\sin x - \cos x}{\frac{\sin x - \cos x}{\sin x \cos x}} \]

Answer 24

yes we don't want fractions

Answer 25

can we cancel sinx-cosx ?

Answer 26

alright, im not really too sure how to do that, im not good with fractions.

Answer 27

\[\large \dfrac{\sin x - \cos x }{\sec x - \csc x} = \dfrac{\sin x - \cos x }{\frac{1}{\cos x}- \frac{1}{\sin x}}\]

\[ \large = \dfrac{\sin x - \cos x}{\frac{\sin x - \cos x}{\sin x \cos x}} \]

\[ \large = \dfrac{1}{\frac{1}{\sin x \cos x}} \]

Answer 28

oh um..yes? since its in the numerator too?

Answer 29

you got it !

Answer 30

now multiply numerator and denominator to get rid of the fraction in the bottom

Answer 31

\[ \large = \dfrac{1}{\frac{1}{\sin x \cos x}} \]


\[ \large = \dfrac{\color{red}{\sin x \cos x} }{\color{red}{\sin x \cos x} \frac{1}{\sin x \cos x}} \]

Answer 32

in the denominator sinxcosx cancels out, leaving you with just sinxcosx in the numerator

Answer 33

\[ \large = \dfrac{1}{\frac{1}{\sin x \cos x}} \]


\[ \large = \dfrac{\color{red}{\sin x \cos x} }{\color{red}{\sin x \cos x} \frac{1}{\sin x \cos x}} \]

\[\large = \color{red}{\sin x \cos x}\]

Answer 34

ooh ok since it would just be sinxcosx/1 it makes it just sinxcosx
i see now

Answer 35

but how does that become sin2x/2..that makes no sense to me

Answer 36

yes next you need to recall the identity : \(\sin 2x = 2 \sin x \cos x\)

Answer 37

ok i see that, under the Double Angle Formulas

Answer 38

yes thats the name ! using that identity, conclude that \(\large \color{red}{\sin x \cos x = \dfrac{\sin 2x}{2}}\)

Answer 39

im sorry but i just dont understand the last part, with the sin 2x = 2sinx cosx.
i just dont understand how that makes sinxcosx into sin2x/2

Answer 40

we have this identity : \(\large \sin 2x = 2\sin x \cos x\)

Answer 41

divide 2 both sides, what do you get ?

Answer 42

oh lol i see now

Answer 43

good :)

\(\large \sin x \cos x = \dfrac{2}{2} \sin x \cos x = \dfrac{1}{2} \sin (2x)\)

Answer 44

ugh everyone on here is better than my professor at explaining everything lol. Thank you so much ganeshie8

Answer 45

huh @nsellers23 u wont find better than ganesh in explaining :P

Answer 46

haha i dont doubt that one bit ikram

Answer 47

lol i am flattered ! ty for the good words :) good luck !!

Answer 48

just 5 more problems to go, ill see you guys in a bit if i cant get them