how do I find sin 5ᴨ/8, cos 5ᴨ/8 and tan 5ᴨ/8 using trig identities

- anonymous

how do I find sin 5ᴨ/8, cos 5ᴨ/8 and tan 5ᴨ/8 using trig identities

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- anonymous

What is ᴨ

- anonymous

use formula for sin4x

- anonymous

the problem doesn't say...just says to use trigonometric identities to find the values

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## More answers

- anonymous

But what is "ᴨ" is it theta or something else. What is it?

- anonymous

expand sin4x, put x as 5pi/8 and equate the expa nsion of sin4x to 1

- anonymous

Is it pi?

- anonymous

yes

- anonymous

Sin ((270-45)/2)=sin (5(pi)/8)

- anonymous

Do you understand this step?

- anonymous

Do you understand the step "Sin ((270-45)/2)=sin (5(pi)/8)"

- anonymous

not really, but i think you got the 270 from 3pi/2 and the 45 from pi/4....

- anonymous

Well, I am trying to express the thing in degree format. (Sorry for using jumbled form). By the term I mean 180 degree

- anonymous

By the term pi mean 180 degree

- anonymous

\[Sin ((3\pi/2-\pi/4)/2)=sin (5\pi/8)\]

- anonymous

Now does that makes sense?

- anonymous

yes

- anonymous

*make

- anonymous

Now can you do the rest, or I will have to show you every steps.

- anonymous

so, sin (225/2)=sin (112.5)?
you might have to show me the steps.

- anonymous

or is it sin (5pi/4 divided by 2)

- anonymous

Yes, thats true, but thats not helpful here.
\[Sin ((3\pi/2-\pi/4)/2)=sin (5\pi/8)\]
\[=>sin (5\pi/8)=\sqrt(0.5(1-cos(3\pi/2-\pi/4))\]

- anonymous

Now you got to use addition formula for cos to solve the part with cos, and then it will be done

- anonymous

I am totally lost .....this does not look at all familiar.....

- anonymous

are you familiar with double angle formula for trig?

- anonymous

i'm sure we went over them....but nothing coming to mind

- anonymous

Then you can't expect to get this thing within your understandability range

- anonymous

that's why I am having trouble....thought if someone walked me through it, it would make sense

- anonymous

If you wish I can tutor you (in exchange of some nominal payment) :)

- anonymous

Seems like you are not interested. So may I leave?

- anonymous

didn't know i had to pay for help

- anonymous

No no not that

- anonymous

you can get free help here for ever

- anonymous

But I don't think they will be very helpful

- anonymous

So incase you need personal help

- anonymous

I mean through audio chatting and virtual black board

- anonymous

I can't teach you everything here. This place is too inconvenient to type even a simple equation

- anonymous

I mean no one can

- anonymous

thanks for the offer, but rather just stay here

- anonymous

Thats great. No problem

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