Question

WHO IS GOOD AT PRECALCULUS

Answer 1

Well, I'd say there are plenty of people who can help out. Ask your question and get assisted.

Answer 2

Well, u can ask questions and let me know what U want to know..... :D

Answer 3

Please ask your question.

Answer 4

http://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf
look at the "product to sum" formulas

Answer 5

oops posted the wrong one hold on.

Answer 6

so again, use the product to sum formulas on the link
it's the third one in that section of four formulas

Answer 7

\[\sin x\cos\frac\pi7-\sin\frac\pi7\cos x\]apply\[\sin\alpha\cos\beta=\frac12[\sin(\alpha+\beta)+\sin(\alpha-\beta)]\]

Answer 8

then do you plug them in

Answer 9

yes

Answer 10

so sinαcosβ=1/2[sin(pi/7+pi/7)+sin(pi/7−pi/7)]

Answer 11

those should have x's in there

Answer 12

hmmmmm

Answer 13

\[\alpha=x\]in the first one

Answer 14

oh ok so what difference does it make?

Answer 15

\[\alpha=x\]\[\beta=\frac\pi7\]in the first term
plug that in and see what you get

Answer 16

so is the answer A and C ?

Answer 17

You don't get free answers like that on here, you are given help to figure it out for yourself.

Answer 18

no thats what i got?

Answer 19

i will post my work hold on

Answer 20

oh I see what you are saying, sorry

Answer 21

I haven't checked myself, hold on...

Answer 22

= 1/2 [sin(x+pi/7) + sin(x-pi/7)]

Answer 23

I am getting a different answer
let me double check

Answer 24

right that is the first term
and what s the next?

Answer 25

i dont understand your question

Answer 26

you want to change\[\sin x\cos\frac\pi7-\sin\frac\pi7\cos x\]with the formula\[\sin\alpha\cos\beta=\frac12[\sin(\alpha+\beta)+\sin(\alpha-\beta)]\]notice this formula only covers each *pair* sin*cos, so you need to use it twice...

Answer 27

that just confused me didnt we already do that?

Answer 28

we have only done it for \[\sin x\cos\frac\pi7\]we have not yet subtracted the \[\sin\frac\pi7\cos x\]part

Answer 29

so do we plug that in for x?

Answer 30

that question doesn't make much sense...
plug in for x ? we are going to find alpha and beta again
what are they this time?

Answer 31

alpha is x and beta is - sin pi/7cos x

Answer 32

no, compare\[\sin\frac\pi7\cos x\]\[\sin\alpha\cos\beta\]then\[\alpha=?\]\[\beta=?\]

Answer 33

α= pi/7
β=x

Answer 34

yes, now use the product to sum formula to change that to the other form

Answer 35

\[\sin\alpha\cos\beta=\frac12[\sin(\alpha+\beta)+\sin(\alpha-\beta)]\]

Answer 36

sin pi/7 cos x= 1/2 [sin(pi/7+x)+sin(pi/7−x)]

Answer 37

good
now put the two together!
sinx cos pi/7-sin pi/7 cos x=????

Answer 38

i have no idea on this part!

Answer 39

sinx cos pi/7=1/2[sin(x+pi/7)+sin(x−pi/7)]
sin pi/7 cos x= 1/2 [sin(pi/7+x)+sin(pi/7−x)]

so what is
sinx cos pi/7-sin pi/7 cos x=?

(just subtract the terms in the second form, see what cancels)

Answer 40

cos pi/7 ?

Answer 41

\[\sin\alpha\cos\beta=\frac12[\sin(\alpha+\beta)+\sin(\alpha-\beta)]\]\[\sin x\cos\frac\pi7=\frac12[\sin(x+\frac\pi7)+\sin(x-\frac\pi7)]\]\[\sin \frac\pi7\cos x=\frac12[\sin(\frac\pi7+x)+\sin(\frac\pi7-x)]\]so\[\sin x\cos\frac\pi7-\sin \frac\pi7\cos x\]\[=\frac12[\sin(x+\frac\pi7)+\sin(x-\frac\pi7)]-\frac12[\sin(\frac\pi7+x)+\sin(\frac\pi7-x)]\]simplify

Answer 42

they all cancel?

Answer 43

which terms are the same? is\[\sin(\frac\pi7+x)=\sin(x+\frac\pi7)\]?

Answer 44

no

Answer 45

i just dont get hwo you simplify that..

Answer 46

they are the same because addition is *always* commutative\[a+b=b+a\]\(always\)**

Answer 47

so they dont cancel?

Answer 48

so we can rewrite this\[=\frac12[\cancel{\sin(x+\frac\pi7)}+\sin(x-\frac\pi7)]-\frac12[\cancel{\sin(x+\frac\pi7)}+\sin(\frac\pi7-x)]\]\[=...\]

Answer 49

1/2 2 sin (x-pi/7)

Answer 50

yes

Answer 51

which simplifies to...?

Answer 52

sin(x-pi/7)

Answer 53

yes
=?
(I must go, you can bump this problem and someone else can finish it for me)

Answer 54

so now have\[\sin(x-\frac\pi7)=-\frac{\sqrt2}2\]good luck from there!

Answer 55

thank you!

Answer 56

i guessig you find referanc eangle