Question

prove 1+sinx=2cos^2x

Answer 1

it is not true! if x=0 then the left side is
\[\large 1+\sin0=1+0=1 \]
and the right side
\[\large 2\cos^20=2(1)^2=2 \]

Answer 2

this is great! how do we know for sure?

Answer 3

what i wrote is counterexample. it definitely shows it is not true.

Answer 4

*is a counterexample.

Answer 5

yes, you are right. would we be able to prove it by taking the arc of it?

Answer 6

to prove it is not true?

Answer 7

no, to prove it is true

Answer 8

but it is not true.

if it were true, using inverse trig functions usually is not easy 'cause they don't have any useful properties.

Answer 9

okay! i understand now. this is very helpful thank you so much. im definitely becoming a fan. are you a math teacher? i also have other questions

Answer 10

glad to help

Answer 11

solve for x : tan^2x-tanx-2=0

Answer 12

u have
\[\large \tan^2x-\tan x-2=0 \]
use the substitution \(u=\tan x\) so the equation becomes a common quadratic equation.

Answer 13

okay, so what would that give us?

Answer 14

\[\large u^2-u-2=0 \]

Answer 15

yes but that does not solve anything where would be go from here

Answer 16

did u solve it?

Answer 17

what did u get?

Answer 18

u^2-u=2

Answer 19

i mean find the roots, using the quadratic formula or factoring it.

Answer 20

u(u-1)

Answer 21

do u know the quadratic formula?

Answer 22

yes -b + or - the sqaure root of b^2-4ac over 2a
but i already factored it so what is the need for the formula?

Answer 23

u didn't factor it correctly

Answer 24

\[\large 0=u^2-u-2=(u-2)(u+1)\]

Answer 25

so you would take each one and set it equal to 0 giving u=2 and u=-1

Answer 26

yes

Answer 27

or tanx= 2 and tan x=-1

Answer 28

yes

Answer 29

tanx of 2 is extraneous but tanx=-1 can be written as 3pi/4 and 7pi/4

Answer 30

that is what i had as my original answer i wasnt sure if it is right though

Answer 31

yes, the solutions r right

Answer 32

but
\[\large 2=u=\tan x \]
would give you x=63.43º (using a calculator)

Answer 33

yes but the answer should translate to radian measure

Answer 34

in radians it would be x=1.107 (using a calculator)

Answer 35

so it would be close to pi/3

Answer 36

yes. close to pi/3.

but if u have to use tables (not allowed to use calculators) or elegant answers such as \(3\pi/4\), then u=2 can be "dismissed". You should ask your teacher.

Answer 37

yes making it extraneous. perfect. you are great

Answer 38

i hope i am not bothering you this homework is very hard. i have a couple of more questions

Answer 39

i am fine. go ahead.

Answer 40

solve for x: 2sinxcosx=sin2x

Answer 41

would we rearrange the equation to 2sinxcosx-sinx2x=0

Answer 42

it is
\[\large 2\sin x\cos x-\sin x\cdot 2x=0 \]
?

Answer 43

no the original problem is \[2sinxcosx=\sin2x\]

Answer 44

this is an identity (always true)

Answer 45

yes but it is asking to solve for x not prove

Answer 46

the identity is 2sinAcosA=Sin2A

Answer 47

yeah, but u don't solve an identity. any x will solve it.

Answer 48

yes you are right!

Answer 49

verify: \[(2+\tan ^{2}x \div \sec ^{2}x) -1\]

Answer 50

aren't u missing an equal sign?

Answer 51

no it says to verify

Answer 52

but it is not a valid proposition. there's nothing to verify, at least from my point of view.

maybe there's a typo in the question.

Answer 53

no its like you have to use trigonometric identities to prove it, or, to simply. lets say it is to simplify

Answer 54

ok
\[\large \frac{2+\tan^2x}{\sec^2x}-1=\frac{2+\tan^2x-\sec^2x}{\sec^2x} \]

Answer 55

now since
\[\large \tan^2x+1=\sec^2x \]

Answer 56

then
\[\large =\frac{2-1}{\sec^2x}=\frac{1}{\sec^2x}=\cos^2x\]

Answer 57

and that's it, I think

Answer 58

ok! i have a few more similiar to this one simplify \[\tan ^{2}x (\sec ^{2}x \div \csc ^{2}x)\]

Answer 59

use the definitions of sec and csc
\[\large \sec x=1/\cos x\qquad\csc x=1/\sin x \]

Answer 60

sorry, but here it is late. I gotta get some rest.

Answer 61

see u some other time.

Answer 62

thank you very much for your help

Answer 63

u r welcome. bye