Question

express cos theta (sec theata-cos theata) in terms of sin theta.

Answer 1

@iPwnBunnies

Answer 2

@satellite73

Answer 3

Err, as in JUST sin? cos(x) = sin(90 - x)

Answer 4

sec(x) = 1/cos(x) = 1/sin(90-x)

Answer 5

i dnt gt wht ur doin @iPwnBunnies

Answer 6

I'm rewriting cos(x) and sec(x) in terms of sin(x)...
you'll have to rewrite the expression

Answer 7

cos(x) = sin(90 - x)
Isn't cos(60) = sin(90-60) = 1/2?

Answer 8

where did u gt 90-x?

Answer 9

@iPwnBunnies

Answer 10

Well, I just proved it with one example.

Answer 11

cos(60) = sin(90 - 60) = sin(30)

Answer 12

so is sin30 the answer?

Answer 13

@iPwnBunnies

Answer 14

@mathmale plzzzzzzz

Answer 15

"express cos theta (sec theata-cos theata) in terms of sin theta"\[\cos \theta (\sec \theta - \cos \theta ) ~In~terms~of~\sin \theta\]

Answer 16

First, Lyubas, multiply cos theta by sec theta. Think carefully first! How is sec theta definied?

Answer 17

sec=cos?

Answer 18

????

Answer 19

do i have to convert the whole thing into sin?

Answer 20

I encourage you to make up a table of the basic trig functions for later reference. You'll need to know those functions.

1
Note that the secant is defined as sec theta = -----------
cos theta

Please try again, Lyubas. Multiply cos theta by sec theta. Cancel wherever possible.

Answer 21

do i have to convert the whole thing into sin? Yes, but that's the next-to-last step. Hold on that.

Answer 22

so now i have cos theta(1/cos theata)-cis theta

Answer 23

cos theta(1/cos theata-cos theta )

Answer 24

is tht right so far??

Answer 25

Yes.

What is the value of \[\frac{ \cos \theta }{ \cos \theta }?\]

Answer 26

1?

Answer 27

What is the value of \[(\cos \theta)*(\cos \theta)~?\]

Answer 28

\[\cos ^{2}\]\[\cos ^{2} \]

Answer 29

cos^2 theta

Answer 30

(cos theta) / (cos theta) =1, and yes, (cos theta)(cos theta) = (cos theta)^2.

Can you think of an identity for (1-(cos theta)^2) ?

Answer 31

cos(a-b)=cosAcosB+sinAsinB??

Answer 32

\[Because~ \sin ^{2}\theta+\cos ^{2}\theta = 1, 1-\cos ^{2}\theta=?\]

Answer 33

ugh nvm im nt getting this thnks anyways.

Answer 34

sorry! But we're almost there. Mind sticking with this until we're done?

Answer 35

sure

Answer 36

I'm going to use x in place of theta. It's so much easier to type.

cos x (sec x - cos x) = 1 - (cos x)^2. Can you agree with that?

Answer 37

yes but how did u no to divide cos x/cos x and multiply cos x times cos x?

Answer 38

cos x (sec x - cos x) = 1 - (cos x)^2 involves multiplication only.

remember, lyubas, that we discussed cos x sec x being equal to (cos x)/(cos x). This follows the important identity that

1
sec x = ----------
cos x

Answer 39

Note, too, that 1 - (cos x)^2 = (sin x)^2. that stems from another important identity:\[\sin ^{2}\theta+\cos ^{2}\theta = 1\]

Answer 40

Please, lyubas, copy down these identities; you will need them later. We can add to your list of identities as we work on various trig problems.

Answer 41

i shuld copy both of these identities?
1
sec x = ---------- and sin^2theta +cos^2theta=1
cos x

Answer 42

Yes. Copy them down for later reference.

1
sec x = ------- and sin^2theta +cos^2theta=1
cos x

Answer 43

ok what do i do next?

Answer 44

You end up with 1 - (cos x) ^2, which is equal to (sin x)^2, which is the answer.
You were supposed to rewrite cos x (sec x - cos x) in terms of sin x, and we have now done exactly that, so you are finished with this problem.

Answer 45

1 - (cos x) ^2, which is equal to (sin x)^2 is the answer right?

Answer 46

Yes, and that's all y ou have to do! But you might want to write that as \[\sin ^{2}\theta\]because the original problem was written in terms of theta. More power to you. I need to get off the Internet soon, but will look forward to working with you again soon!

Good night!

Answer 47

thnks sry for driving u crzy

Answer 48

have agood night