Use the identity cos(ø/2) = √[(1+cos(ø))/2]; then you get
sinø = √[(1+cosø)/2] Square both sides
sin^2(ø) = (1+cosø)/2
2sin^2(ø) = 1+cosø; Expand sin^2(ø) = 1 - cos^2(ø)
2 - 2*cos^2(ø) = 1 + cosø
2*cos^(ø) + cosø -1 = 0
Substitute x = cosø
2x^2 + x - 1 = 0 factor this (2x - 1)*(x + 1); so x = -1 or x = .5
so cosø = -1 or cosø = .5
ø = (2n+1)π and ø = π/3+2nπ
where n = 0,1,2...
2006-11-18 13:24:10
·
answer #1
·
answered by gp4rts 7
·
0⤊
1⤋
sin θ = cos (θ/2)
θ=60°
sin 60°=√3 /2
cos 30°=√3 /2
2006-11-18 14:26:31
·
answer #2
·
answered by yupchagee 7
·
1⤊
0⤋
sin θ = cos(θ/2)
2sin(θ/2)cos(θ/2) = cos(θ/2)
cos(θ/2) = 0 or sin(θ/2) = 1/2
All the solutions are:
θ = pi + 2npi
θ = pi/3 + 4npi
θ = 5pi/3 + 4npi
(n arbitrary integer)
2006-11-18 13:25:32
·
answer #3
·
answered by Anonymous
·
1⤊
0⤋
Hi. Try θ = 60
2006-11-18 13:12:56
·
answer #4
·
answered by Cirric 7
·
1⤊
0⤋
Well, I am not sure how you are doing Trig. I can figure this out easily using identity triangles 1 / root 3 /2 to be exact.
Do you define your sine and cosine as opposite over hypotenuse and adjacent over hypotenuse? What grade are we talking about here? I hope to goodness you aren't using the half angle formula.
2006-11-18 13:20:55
·
answer #5
·
answered by derkaiser93 4
·
0⤊
0⤋
contained in the above observe issue, the rectangle has each and each and every area of 30 and 21 contraptions. The rectangle's corners are being folded mutually, which could be pictured or portrayed visually as each and each and every area of the rectangle coinciding to grant 0.5 of its section hence yielding a triangle. bear in mind that a rectangle and parellelogram have an similar section. in view that that is authentic, that ought to intend the inverse courting with triangles because if those polygons have similar section, then 0.5 of their section might want to be equivalent to a triangle hence, making the folded corners a triangular asset, which concurrently implies the rule of thumb of particular triangles refferred to or said virtually as good triangles. because you're given 2 area lengths of the triangle and requested to locate the crease, that would want to entail some variety of rule defining the courting between triangular aspects with observe of to this style of triangle, it really is purely the right triangle. hence, the rule of thumb necessary to locate the crease or supposedly the hypotenuse is the Pythagorean Theorem. J.C
2016-11-29 06:29:30
·
answer #6
·
answered by ? 4
·
0⤊
0⤋
The formula sin(2t) = 2sin(t)cos(t) can be employed here.
Taking half of each of these arguments gives :
sin(t) = 2sin(t/2)cos(t/2)
Your problem then becomes :
sin(t) = cos(t/2) = 2sin(t/2)cos(t/2)
Now divide through by cos(t/2) to give :
1 = 2sin(t/2)
Therefore, sin(t/2) = 1/2
So, t/2 = 30º and t = 60º as the principal value.
2006-11-18 13:18:03
·
answer #7
·
answered by falzoon 7
·
1⤊
0⤋
if you have a cool calc, graph each side and see where they cross. Actually, it's probably 60 degrees and 30 degrees, in radians.
2006-11-18 13:09:53
·
answer #8
·
answered by Anonymous
·
0⤊
0⤋
180 deg
(or PI rad)
2006-11-18 13:25:12
·
answer #9
·
answered by chemicalimbalance000 4
·
1⤊
0⤋
you are smarter than I, I have tried to solve this for you, but can't, Sorry.
2006-11-18 13:12:11
·
answer #10
·
answered by St♥rmy Skye 6
·
0⤊
0⤋