Question

simplify in cosine and sine terms : (cot t + csc t) (tan t - sin t)

Answer 1

=cot t*tant - sint*cott +csct*tant - sint*csct

Remember that cott = 1/(tant)= cost/sint
csct = (1/sint)
tant = sint/cost
Thus,
=1 -cost + 1/cost - 1
= 1/cost - cost
(This can be simplified further using the squares of cost and sint...would you like me to continue?)

Answer 2

foil the left side first and don't touch the right side

Answer 3

cotθtanθ-cotθsinθ+cscθtanθ-cscθsinθ

Answer 4

(1/tanθ tanθ)-(cosθ/sinθ sinθ)+(1/sinθ sinθ/cosθ)-(1/sinθ sinθ

Answer 5

) at the end

Answer 6

so far so good?

Answer 7

yea thats what i did but what did you get at the end ?

Answer 8

after combining on left side i got -cosθ+secθ which is QED to secθ-cosθ

Answer 9

i got \[\frac{ \sin^2 \theta }{ \cos \theta }\] on both sides

Answer 10

ok so
(1/tanθ tanθ)=1

Answer 11

(cosθ/sinθ sinθ)=cosθ

Answer 12

(1/sinθ sinθ/cosθ)=1/cosθ

Answer 13

(1/sinθ sinθ)=-1

Answer 14

1-cosθ+1/cosθ-1

Answer 15

I think there's a simpler way. AFter you got sect - cost, just multiply by 1, namely cost/cost you get (1-cos^2(t))/cost which by the identity, is sin^2(t) / cost

Answer 16

we have the same answer but you have to put it in sine and cosine terms so i think its how i did it

Answer 17

the 1 cross each other out so you end up with -cosθ+1/cosθ which is -cosθ+secθ

Answer 18

why do we have to put it into sin and cos?

Answer 19

because thats what he has been doing in class so I think thats how he wants it

Answer 20

The direction says so?

Answer 21

the directions just stated verify identity

Answer 22

actually: simplify in cosine and sine terms : (cot t + csc t) (tan t - sin t)

Answer 23

we have to verify the identity (cot theta + csc theta) (tan theta - sin theta) = sec theta - cos theta

Answer 24

yup that what the direction say

Answer 25

so simplyfying everything to cosine and sine terms i get sin^2 theta / cos theta on both sides

Answer 26

@skybunnygirl secant is implicitly in terms of sine and cosine. it is NOT sine and cosine itself.

Answer 27

so i have to change sec to cos all the time?

Answer 28

yes according to the book is one thing but what he has done in class its all sines and cosines so for the test i think its best to go with his directions

Answer 29

no. You asked why it has to be in terms of sine and cosine. Because the directions said so was my response. That is why sect - cost was not the appropriate answer. Ordinariy, if it asked for trig, you couldve said sint*tant which is the shortest answer but it asks specifically for sint and cost

Answer 30

yea because after simplying everything you get 1- cos theta + (1/cos theta) -1 so the 1s cancel and you are left with (1-cos^2 theta) / cos theta so according to the pythagorean theorem the numenator is sin^2 theta so the final answer is sin^2 theta / cos theta

Answer 31

thanks for your help @khoala4pham

Answer 32

@skybunnygirl were you able to get the same thing?

Answer 33

not yet

Answer 34

if you cancel out the 1, where is the 1 in your answer comes from?

Answer 35

\[\cot \theta \tan \theta -\cot \theta \sin \theta - cdc \theta \tan \theta - \csc \theta \sin \theta \] then you put it in sin and cos terms so you get : \[\frac{ \cos \theta \sin \theta }{ \sin \theta \cos \theta } - \frac{ \cos \theta \sin \theta }{ \sin \theta } +\frac{ \sin \theta }{ \cos \theta \sin \theta } - \frac{ \sin \theta }{ \sin \theta }\] so then you get: \[1 - \frac{ \cos \theta \sin \theta }{ \sin \theta } + \frac{ \sin \theta }{ \cos \theta \sin \theta } - 1\] then you add the fractions but cancel out the sin first on both fractions and cancel the 1s so you get :\[\frac{ \cos \theta }{ 1 } + \frac{ 1 }{ \cos \theta } and you get: \frac{ 1- \cos^2 \theta }{ \cos \theta }\] and finally according to the theorem you get: \[\frac{ \sin^2 \theta }{ \cos \theta}\]

Answer 36

on the other side you have \[\sec \theta - \cos \theta \] which is the same as \[\frac{ 1 }{ \cos \theta } - \frac{ \cos \theta }{ 1 }\] so you subtract them and you get \[\frac{ 1- \cos^2 \theta }{ \cos \theta }\] which also gives you \[\frac{ \sin ^2 \theta }{ \cos \theta }\] QED

Answer 37

@skybunnygirl better?

Answer 38

looking at it step by step

Answer 39

got it, I just had to work it out step by step so I understand what you saying. thanks a million. so from now on i have to get them all in cos and sin format in the final answer per the professor?

Answer 40

yes I will say so just so you dont get any points taken off with something as simple as just following what he does in the class you know?

Answer 41

what a pain. now i can redo some of the homework i did, ugh

Answer 42

next time just look at his notes and what we write in class to see how he works out his problems out ans the type of answers he expects also dont forget that if you get an answer like \[\frac{ 2 }{ \sqrt{29} } \] make sure you multiply numerator and denominator by sqrt of 29 to get the root on top if you dont do this he said he would mark it wrong so the answer should be \[\frac{ 2 \sqrt{29} }{ 29}\]

Answer 43

yes i learned that with my last teacher so that is an easy rule

Answer 44

are you gonna have all these answers done by tuesday?

Answer 45

ok good and never put negatives in denominator

Answer 46

anyway you can recheck mine for me?

Answer 47

yup that he and my previous teacher drilled in well

Answer 48

yes so we can compare :) i am always there early because i have another class from 4-5 so after 5 i am available in case you go early

Answer 49

i can be early as i'm coming from home. i can meet you a bit after 5 and go over some of the stuff, that would be very helpful

Answer 50

ok sound good text me when you get there

Answer 51

perfect, thanks for your help!