Question

prove the following identities

cos(x-pi) = sin(x+3pi/2)


tan(2x) = 2tan(x)/1-tan^2(x) (hint switch to sine and cosine)

Answer 1

why is yours red

Answer 2

use these identities for the 1st one
\[\cos(A-B)=cosAcosB+sinAsinB\]
\[\sin(A+B)=sinAcosB+sinBcosA\]

Answer 3

Qualified helpers dont come , i just bought some owl bucks for them and they didnt com for 30 minutes and still ahvent come anyways. I ased for a qualified helper too

Answer 4

for that second one use this identity-\[\tan(A+B)=\frac{ tanA+tanB }{ 1-tanAtanB}\]
try putting A=B=x in this question :)

Answer 5

@lisamath12

Answer 6

ok so for the second one that leaves me with tan(x+x) = tanx+tanx/1-(tanx)(tanx)

Answer 7

I'm not sure where to go from there though

Answer 8

yes :) ur correct
now try to simplify further

Answer 9

I am not sure how to do that
I have all the formulas for these questions what I don't know is how to solve it after that

Answer 10

Do we have to prove one or two identities for this question?

Answer 11

cos(A-B) = Cos (A + B) + 2 Sin A Sin B

Answer 12

For the second identity,

\(\sf\Large tan(2x) = \frac{sin(2x)}{cos(2x)}\)

Answer 13

Another hint for the second one:

\(\sf\Large sin(2x) =2sin(x)cos(x)\)

and

\(\sf\Large cos(2x) =cos^2(x) -sin^2(x)\)

Answer 14

@lisamath12

You will get your answer faster if you cooperate :)

Answer 15

sorry I was busy for a short while just got back

Answer 16

ok :)

Answer 17

so now we have: 2sin(x) cos(x)/ Cos^2(x) - sin^2(x) is equal too 2tan(x)/1-tan^2(x)

Answer 18

is the next step to change the tangents on the other side too sine and cosine as well?

Answer 19

we only deal with one side at a time :)

Answer 20

so we have

\(\sf\Large \frac{2sin(x)cos(x)}{cos^2(x) -sin^2(x)}\)

divide both the numerator and denominator by cos^2(x)

Answer 21

\(\sf\Huge \frac{\frac{2sin(x)cos(x)}{cos^2(x)}}{\frac{cos^2(x) -sin^2(x)}{cos^2(x)}}=??\)

Answer 22

So lets deal with the numerator first :)

\(\sf\Large \frac{2sin(x)cos(x)}{cos^2(x)} = ?\)

Simplify it :)

Answer 23

alright so now we got 2sin(x)/cos(x)

Answer 24

and the denominator is -sin^2(x)?

Answer 25

simplify 2sin(x)/cos(x)

hint: tan(x) = sin(x)/cos(x)

Answer 26

2 tan(x)

Answer 27

exactly! :D

Answer 28

ok, now we simplify the denominator part :)

Answer 29

\(\sf\Huge \frac{cos^2(x) -sin^2(x)}{cos^2(x)}=??\)

Answer 30

Hint:

(a-b)/c = a/c - b/c

Answer 31

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

Answer 32

Not sure how you got that, but :P

\(\sf\Large \frac{cos^2(x) -sin^2(x)}{cos^2(x)}=\frac{cos^2(x)}{cos^2(x)} - \frac{sin^2(x)}{cos^2(x)} = ?\)

Answer 33

oh is it - tan(x). I had cos^2(x)/cos^2(x) equal 0 and that lefted me with -sin^2(x)/cos^2(x)

Answer 34

a/a = 1

\(\sf\Large \frac{cos^2(x)}{cos^2(x)} = 1\)

so we are left with 1 - tan^2(x)

Answer 35

And therefore we have proved the identity

Answer 36

\(\sf\Huge \frac{\frac{2sin(x)cos(x)}{cos^2(x)}}{\frac{cos^2(x) -sin^2(x)}{cos^2(x)}}=\frac{2tan(x)}{1-tan^2(x)}\)

Answer 37

ohhh, yes I see it now thank you!

Answer 38

\(\sf\Large tan(2x) = \frac{2tan(x)}{1-tan^2(x)}\)

LHS = RHS

Answer 39

Anytime! :D

Answer 40

Now we have this question, right?
cos(x-pi) = sin(x+3pi/2)

Answer 41

yes exactly

Answer 42

Hold on, I'm trying to see how to solve this one. :)

Answer 43

So, I got it now :)

Answer 44

we need to use these two formulas

cos(α – β) = cos(α)cos(β) + sin(α)sin(β)
sin(α + β) = sin(α)cos(β) + cos(α)sin(β)

Answer 45

We shall seperately open up the LHS and RHS :)

Answer 46

Or we could use this actually:

cos(α – β) = cos(α + β) + 2sin(α)sin(β)

Answer 47

So @lisamath12 lets plug it into the last formula I gave :)

Answer 48

alright so then we would have: cos(x-pi) = cos(x+pi) + 2sin(x)sin(pi)

Answer 49

and lets first expand cos(x+pi) using this formula:

cos(α + β) = cos(α)cos(β) – sin(α)sin(β)

Answer 50

cos(x+pi) = cos(x)cos(pi) - sin(x)sin(pi)

Answer 51

lets simplify that

hint:

cos(pi) = -1
sin(pi) = 0

Answer 52

alright so cos(x+pi) = cos(x)(1) - sin(x)(0)

Answer 53

cos(x)(-1) **

so simplifying that gets us -cos(x) right?

Answer 54

and let's simplify the other half

2sin(x)sin(pi)

Answer 55

remember sin(pi) = 0

Answer 56

2sin(x)(0)=0

Answer 57

correct so we got the LHS as -cos(x)

lets expand the RHS too

Answer 58

sin(x+3pi/2)

using the formula:

sin(α + β) = sin(α)cos(β) + cos(α)sin(β)

Answer 59

sin(x+3pi/2) = sin(x)cos(3pi/2) + cos(x)sin(3pi/2)

Answer 60

sin(x)(0) + cos(x)(-1)

Answer 61

and simplifying it more you get -cos(x)

Answer 62

therefore

-cos(x) = -cos(x)

and LHS = RHS

Answer 63

once again big help

Answer 64

anytime :)