Question

prove the ff. identities
1. (sin 1/2x-cos1/2x)^2= 1-sin x
2. csc x/sec x= 1 + cot x/1 + tan x

Answer 1

u know the formula for (a+b)^2 ?

Answer 2

nope, i forgot all of them when i was in high school

Answer 3

for 1. i will use these three formulas, write them down.
\(\large (a+b)^2=a^2+2ab+b^2------>(1)\\\large2sin xcos x=sin2x----->(2)\\\large sin^2x+cos^2x=1------>(3)\)

Answer 4

ok

Answer 5

now take left side: \((sin (x/2)-cos(x/2))^2\)
expand it accordind to (1) and tell me what u get.

Answer 6

if you square sin x/2, is the answer sin^2 x/2?

Answer 7

yes.

Answer 8

then the middle term will be -2sin cos x/2?

Answer 9

-2sin(x/2)cos(x/2)

Answer 10

so sin^2 x/2- 2 sin(x/2)cos(x/2)- cos^2 x/2

Answer 11

careful with brackets and signs.
\(sin^2 (x/2)- 2 sin(x/2)cos(x/2)+ cos^2 (x/2)\)

Answer 12

how did you use that italicized thing?

Answer 13

now use (3) formula,
what will be
\(sin^2 (x/2)+ cos^2 (x/2)=??\)

Answer 14

wait.

Answer 15

`write that in \(...........\)`
like
`\(sin^2 x\)`
u get
\(sin^2x\)

Answer 16

that would be equal to 1?

Answer 17

yes! thats one :)
so u have now :
\(1−2sin(x/2)cos(x/2)\)
ok?

Answer 18

so 1-2sin(x/2)cos(x/2) is equal to 1-sin2(x/2)?

Answer 19

YES!
from the (2) formula.
since 2sinx cos x = sin 2x
2sin(x/2)cos(x/2) = sin 2(x/2) =sin x

Answer 20

thats the final answer?

Answer 21

what will happen to 1-sin x?

Answer 22

u just proved that (sin 1/2x-cos1/2x)^2= 1-sin x
should i put all steps together?

Answer 23

yes

Answer 24

\( (sin(x/2)-cos(x/2))^2=sin^2 (x/2)- 2 sin(x/2)cos(x/2)+ cos^2 (x/2)\\=1-2 sin(x/2)cos(x/2)=1-sin x\)

Answer 25

u have list of trignometric formulas?

Answer 26

i have them, but they are a lot. what specific identities? like fundamental identities?

Answer 27

i'll list them:
\(\huge sin x =1/csc x \implies csc x=1/sin x \\\huge cos x = 1/sec x \implies secx=1/cos x \\\huge tan x=1/cot x \implies cot x =1/tanx \\\huge tan x=sin x/cos x \\\huge cot x=cos x/sinx \\
\huge sin^2x+cos^2x=1 \\\huge sec^2x=1+tan^2x\\\huge cosec^2x=1+cot^2x\)

Answer 28

here u will not need last 3,but they are VERY IMPORTANT

Answer 29

what do u mean?

Answer 30

all these are important, for 2. u'll need few from first 5.

Answer 31

so lets take right side first, can u see cot x and tan x ?
what will u write those as ?

Answer 32

tan x= 1/cot x -------cot x=1/tan x

Answer 33

that will be of no use, try cosx/sinx and sinx/cos x

Answer 34

\(\huge \frac{1+cotx}{1+tan x}=\frac{1+\frac{cos x}{sin x}}{1+\frac{sinx}{cosx}}=?\)

Answer 35

1+ cosx/sin x=1 + cot x

Answer 36

and 1 +tan x= 1+ sinx/cos x

Answer 37

??

Answer 38

so 1+cot x/1+tan x= 1+ cot x/1 +tan x

Answer 39

lol!

Answer 40

what is wrong?

Answer 41

i wrote cot x as cos x/sin x and u again wrote cos x/sin x as cot x!!!
so whats the use of my writing cot x as cos x/sin x ??

Answer 42

sorry, i just realized

Answer 43

csc x/sec x= 1 + cot x/ 1+ tan x

Answer 44

ok, i will write next step, see whether u get it:
\(\huge \frac{1+\frac{cos x}{sin x}}{1+\frac{sinx}{cosx}}=\frac{(sin x +cos x)/sinx}{(sin x+cos x)/cosx}\)

Answer 45

are we working for the two sides?

Answer 46

i started from 1+cotx/1+tanx and i will reach, csc x/sec x

Answer 47

now can u see that (sin x +cos x) gets cancelled from numerator and denominator....

Answer 48

yes

Answer 49

so what remains ?

Answer 50

cos x/sin x

Answer 51

absolutely correct, but i will keep it in the form of :
\(\huge \frac{1/sin x}{1/cos x}=?\)
now see the formula list and tell me what is 1/sin x and 1/cos x ??

Answer 52

1/sinx= csc x and 1/cos x= sec x

Answer 53

can u write the whole steps?

Answer 54

and u are done!!
see all mt comments with large font....u will get it.

Answer 55

*my

Answer 56

can u write it?

Answer 57

ok :)
\(\huge \frac{1+\frac{cos x}{sin x}}{1+\frac{sinx}{cosx}}=\frac{(sin x +cos x)/sinx}{(sin x+cos x)/cosx}=\huge \frac{1/sin x}{1/cos x}=\frac{cscx}{secx}\)

Answer 58

thanks :)

Answer 59

welcome :)