Question

Verify each trigonometric equation by substituting identities to match the right hand side of the equation to the left hand side of the equation.

cot x sec4x = cot x + 2 tan x + tan3x

Answer 1

yikes

Answer 2

Lol, oh no.

Answer 3

is it \(\tan^3(x)\) (i hope) or \(\tan(3x)\) (i hope not)

Answer 4

one minute.

Answer 5

1st

Answer 6

whew

Answer 7

Lol

Answer 8

ok lets work on the right only, it is mostly algebra
and to make it easier, lets replace cosine by \(a\) and sine by \(b\)

Answer 9

you have with those replacements
\[\frac{a}{b}+\frac{2b}{a}+\frac{b^2}{a^3}\] to add those fractions we need the common denominator which is \(a^3b\)

Answer 10

oops typo there \[\frac{a}{b}+\frac{2b}{a}+\frac{b^3}{a^3}\]

Answer 11

add just like always
\[\frac{a^4+2a^2b^2+b^4}{a^3b}\]

Answer 12

you with me so far?

Answer 13

yea, so

Answer 14

don't you subtract your denominator & numerator.

Answer 15

you said add them?

Answer 16

no it is addition only

Answer 17

a^7 + 2a^5b^3 + b^5

Answer 18

lets go slow

Answer 19

something like that?

Answer 20

you good to here \[\frac{a}{b}+\frac{2b}{a}+\frac{b^3}{a^3}\]?

Answer 21

this is where you say "yes, i get that part"

Answer 22

yea

Answer 23

ok
now the denomiators are \(a,b,a^3\) the least common multiply of those three is \(a^3b\)

Answer 24

ok?

Answer 25

yes.

Answer 26

\[\frac{a}{b}+\frac{2b}{a}+\frac{b^3}{a^3}\]
\[=\frac{a}{b}\times \frac{a^3}{a^3}+\frac{2b}{a}\times \frac{a^2b}{a^2b}+\frac{b^3}{a^3}\times \frac{b}{b}\]\[=\frac{a^4+2a^2b^2+b^4}{a^3b}\]

Answer 27

one more step
the numerator is a perfect square
it is \[(a^2+b^2)^2\]

Answer 28

now back to trig
since \(a=\cos(x), b =\sin(x)\)you have
\[\frac{(\cos^2(x)+\sin^2(x))^2}{\cos^3(x)\sin(x)}\]

Answer 29

and we recall that
\[\sin^2(x)+\cos^2(x)=1\] so the numerator is only a one \[\frac{(\cos^2(x)+\sin^2(x))^2}{\cos^3(x)\sin(x)}=\frac{1}{\cos^3(x)\sin(x)}\]

Answer 30

From the left hand side: sec^2 = 1+tan^2, hence, sec^4 = (1+tan^2)^2therefore, sec^4 = 1+ 2 tan^2 + tan^4 replace allcot ( 1+ 2 tan^2 + tan^4) distribute cot inside the bracket, cancel more by using cot = 1tanto get the right hand side

Answer 31

i was working from the right to the left
you can work the other way if you like but i find it more confusing

Answer 32

besides it pretty much says to work from right to left if i read it correctly

Answer 33

Oh right, ok. continue :)

Answer 34

\[\frac{1}{\cos^3(x)\sin(x)}=\frac{\cos(x)}{\cos^4(x)\sin(x)}=\frac{1}{\cos^4(x)}\times \frac{\cos(x)}{\sin(x)}=\sec^4(x)\cot(x)\]

Answer 35

whew that was exhauting
now i have to go have a drink

Answer 36

That's it?

Answer 37

that is it yes

Answer 38

My teacher's insane...

Answer 39

teaching math will do that for you

Answer 40

go back over the steps, make sure you understand them
all steps are there
i have to run, good luck

Answer 41

Thank you :)

Answer 42

I will...

Answer 43

btw convince yourself that it is 95% algebra, very little trig
yw

Answer 44

Ok :)