TRIG HELP AGAIN!

[ninbam1k]

Hey guys

Find the exact value of sin(5pi/24)cos(pi/24)

thanks in advance ^^

[enwiabe]

Hey

So sin(pi/4) = sin(6pi/24) = sin(5pi/24 + pi/24) = sin(5pi/24)cos(pi/24) + sin(pi/24)cos(5pi/24)
and
sin(pi/6) = sin(4pi/24) = sin(5pi/24 - pi/24) = sin(5pi/24)cos(pi/24) - sin(pi/24)cos(5pi/24) 

Remember that sin(pi/4) = 1/srt(2) and sin(pi/6) = 1/2 an so you have:

1/sqrt(2) = sin(5pi/24)cos(pi/24) + sin(pi/24)cos(5pi/24)      eq. 1
1/2 = sin(5pi/24)cos(pi/24) - sin(pi/24)cos(5pi/24)         eq. 2

if you add equation 1 to equation 2

you get

1/sqrt(2) + 1/2 = sin(5pi/24)cos(pi/24) + sin(pi/24)cos(5pi/24) + sin(5pi/24)cos(pi/24) - sin(pi/24)cos(5pi/24)

1/sqrt(2) + 1/2 = 2sin(5pi/24)cos(pi/24)
therefore,

sin(5pi/24)cos(pi/24) = 1/2sqrt(2) + 1/4

[ninbam1k]

Thanks daniel really appreciated ^^

[enwiabe]

that's okay i'm glad you understood it sorry i didn't have time to latex it properly for you -_-

[ivanivan0002]

i need help too!

is there a way to find inverse sin (4/5) without using a calc?

[kamil9876]

No, but something tells me this is part of a bigger problem and your not doing it the best way. Can you please provide us the actual problem?

[ivanivan0002]

That was just a chunk from a question.

Simplify, in exact form:
cos(inverse sin 4/5)

[brightsky]

That was just a chunk from a question.

Simplify, in exact form:
cos(inverse sin 4/5)

Let y = sin^(-1)(4/5)

So 4/5 = sin(y) [1]

cos^2y + sin^2y = 1

So cos^2y = 1 - sin^2y

cosy = sqrt(1 - sin^2y) [2]

Sub [1] into [2] then simplify and you're done.

EDIT: On second thoughts, I agree with kamil. Drawing a triangle with sides 3, 4 and 5 and then working from that is much simpler and neater.

[ivanivan0002]

Thanks alot!

EDIT: Wouldn't know how to do the question with that. Would you mind showing?

EDIT: ooooo nvm, found it out. It was so obvious

[ivanivan0002]

Sorry for the double post but,

show that tan^-1(3) - tan^-1(1/2) = pi/4

any ideas? I've been staring at this question for quite a while :S

[kamil9876]

Yeah it's a bit tricky. It's good to interpret this geometrically.

The quantity is actually the angle between the lines and . Why?

Now draw a line segment between these two lines that is perpendicular to . A good choice would be the one that goes through . We now have a right angle triangle to which we can apply trig. You need to find the other point of the triangle by solving simultaenous equations. Then you can work our the length of the hypotenuse and hence use trig. I wish I could draw this but fail at Linux

edit: see pdf, best I could do.

[kamil9876]

actually, just realised there is a better way. Find:



By expanding with the angle addition formula. Now you should get if you do the calculations. Now because is in the first quadrant (you can argue this by saying what I said in my previous post: that it is the angle between those two lines) it follows that it must be

[TrueTears]

Sorry for the double post but,

show that tan^-1(3) - tan^-1(1/2) = pi/4

any ideas? I've been staring at this question for quite a while :S

I was gonna do kamil's way but anyways, check this out, it's related: http://vce.atarnotes.com/forum/index.php/topic,12907.msg144875.html#msg144875