Jumat, 30 November 2012

Pythagorean trigonometric identity

Proof based on right-angle triangles

Any similar triangles have the property that if we select the same angle in all them, the ratio of the two sides defining the angle is the same regardless of which similar triangle is selected, regardless of its actual size: the ratios depend upon the three angles, not the lengths of the sides. Thus for either of the similar right triangles in the figure, the ratio of its horizontal side to its hypotenuse is the same, namely cos θ.
The elementary definitions of the sine and cosine functions in terms of the sides of a right triangle are:
\sin \theta = \frac{\mathrm{opposite}}{\mathrm{hypotenuse}}= \frac{b}{c}
\cos \theta = \frac{\mathrm{adjacent}}{\mathrm{hypotenuse}} = \frac{a}{c} \ .
The Pythagorean identity follows by squaring both definitions above, and adding; the left-hand side of the identity then becomes
\frac{\mathrm{opposite}^2 + \mathrm{adjacent}^2}{\mathrm{hypotenuse}^2}
which by the Pythagorean theorem is equal to 1. Note, however, that this definition is valid only for angles between 0 and π/2 radians (not inclusive) and therefore this argument does not prove the identity for all angles. Values of 0 and π/2 are trivially proven by direct evaluation of sin and cos at those angles.
To complete the proof, the identities found at Trigonometric symmetry, shifts, and periodicity may be employed. By the periodicity identities we can say if the formula is true for −π < θ ≤ π then it is true for all real θ. Next we prove the range π/2 < θ ≤ π, to do this we let t = θ − π/2, t will now be in the range 0 < t ≤ π/2. We can then make use of squared versions of some basic shift identities (squaring conveniently removes the minus signs):
\sin^2\theta+\cos^2\theta = \sin^2\left(t+\frac{1}{2}\pi\right) + \cos^2\left(t+\frac{1}{2}\pi\right) = \cos^2t+\sin^2t = 1.
All that remains is to prove it for −π < θ < 0; this can be done by squaring the symmetry identities to get
\sin^2\theta=\sin^2(-\theta)\text{ and }\cos^2\theta=\cos^2(-\theta).\,

Related identities

Similar right triangles illustrating the tangent and secant trigonometric functions.
The identities
1 + \tan^2 \theta = \sec^2 \theta\,
and
1 + \cot^2 \theta = \csc^2 \theta\,
are also called Pythagorean trigonometric identities.[1] If one leg of a right triangle has length 1, then the tangent of the angle adjacent to that leg is the length of the other leg, and the secant of the angle is the length of the hypotenuse.
 \tan \theta =\frac{b}{a} \ ,
and:
 \sec \theta = \frac{c}{a} \ .
In this way, this trigonometric identity involving the tangent and the secant follows from the Pythagorean theorem. The angle opposite the leg of length 1 (this angle can be labeled φ = π/2 − θ) has cotangent equal to the length of the other leg, and cosecant equal to the length of the hypotenuse. In that way, this trigonometric identity involving the cotangent and the cosecant also follows from the Pythagorean theorem.
Tabulation of derivations
Another way of thinking about the other identities is to derive them from the original identity. The following table shows how this is done by dividing each element of the original Pythagorean Identity by a common divisor.
Original Identity Divisor Divisor Equation Derived Identity Derived Identity (Alternate)
\sin^2 \theta + \cos^2 \theta = 1\!



\sin^2 \theta + \cos^2 \theta = 1\!  \cos^2 \theta \!   \frac{\sin^2 \theta}{\cos^2 \theta}  + \frac{\cos^2 \theta}{\cos^2 \theta} = \frac{1}{\cos^2 \theta}\!  \tan^2 \theta + 1 = \sec^2 \theta\!  1 + \tan^2 \theta  = \sec^2 \theta\!
\sin^2 \theta + \cos^2 \theta = 1\!  \sin^2 \theta \!   \frac{\sin^2 \theta}{\sin^2 \theta}  + \frac{\cos^2 \theta}{\sin^2 \theta} = \frac{1}{\sin^2 \theta}\!  1 + \cot^2 \theta = \csc^2 \theta\!  \cot^2 \theta + \ 1 = \csc^2 \theta\!

Proof using the unit circle

Point P(x,y) on the circle of unit radius at an obtuse angle θ > π/2
Sine function on unit circle (top) and its graph (bottom)
The unit circle centered at the origin in the Euclidean plane is defined by the equation:[2]
x^2 + y^2 = 1.\,
Given an angle θ, there is a unique point P on the unit circle at an angle θ from the x-axis, and the x- and y-coordinates of P are:[3]
x = \cos \theta \ \mathrm { and} \ y = \sin \theta \ .
Consequently, from the equation for the unit circle:
 \cos^2 \theta + \sin^2 \theta = 1 \ ,
the Pythagorean identity.
In the figure, the point P has a negative x-coordinate, and is appropriately given by x = cosθ, which is a negative number: cosθ = −cos(π−θ ). Point P has a positive y-coordinate, and sinθ = sin(π−θ ) > 0. As θ increases from zero to the full circle θ = 2π, the sine and cosine change signs in the various quadrants to keep x and y with the correct signs. The figure shows how the sign of the sine function varies as the angle changes quadrant.
Because the x- and y-axes are perpendicular, this Pythagorean identity is actually equivalent to the Pythagorean theorem for triangles with hypotenuse of length 1 (which is in turn equivalent to the full Pythagorean theorem by applying a similar-triangles argument). See unit circle for a short explanation.

Proof using power series

The trigonometric functions may also be defined using power series, namely (for x an angle measured in radians):[4][5]
\begin{align}
  \sin x &= \sum_{n = 0}^\infty \frac{(-1)^n}{(2n + 1)!} x^{2n + 1},\\
  \cos x &= \sum_{n = 0}^\infty \frac{(-1)^n}{(2n)!} x^{2n}.
\end{align}
Using the formal multiplication law for power series at Multiplication and division of power series (suitably modified to account for the form of the series here) we obtain

\begin{align}
\sin^2 x & = \sum_{i = 0}^\infty \sum_{j = 0}^\infty \frac{(-1)^i}{(2i + 1)!} \frac{(-1)^j}{(2j + 1)!} x^{(2i + 1) + (2j + 1)} \\
& = \sum_{n = 1}^\infty \left(\sum_{i = 0}^{n - 1} \frac{(-1)^{n - 1}}{(2i + 1)!(2(n - i - 1) + 1)!}\right) x^{2n} \\
& = \sum_{n = 1}^\infty \left( \sum_{i = 0}^{n - 1} {2n \choose 2i + 1} \right) \frac{(-1)^{n - 1}}{(2n)!} x^{2n},\\
\cos^2 x & = \sum_{i = 0}^\infty \sum_{j = 0}^\infty \frac{(-1)^i}{(2i)!} \frac{(-1)^j}{(2j)!} x^{(2i) + (2j)} \\
& = \sum_{n = 0}^\infty \left(\sum_{i = 0}^n \frac{(-1)^n}{(2i)!(2(n - i))!}\right) x^{2n} \\
& = \sum_{n = 0}^\infty \left( \sum_{i = 0}^n {2n \choose 2i} \right) \frac{(-1)^n}{(2n)!} x^{2n}.
\end{align}
Note that in the expression for sin2, n must be at least 1, while in the expression for cos2, the constant term is equal to 1. The remaining terms of their sum are (with common factors removed)
\sum_{i = 0}^n {2n \choose 2i} - \sum_{i = 0}^{n - 1} {2n \choose 2i + 1}
= \sum_{j = 0}^{2n} (-1)^j {2n \choose j}
= (1 - 1)^{2n}
= 0
by the binomial theorem. Consequently,
\sin^2 x + \cos^2 x = 1 \ ,
which is the Pythagorean trigonometric identity.
The Pythagorean theorem is not closely related to the Pythagorean identity when the trigonometric functions are defined in this way; instead, in combination with the theorem, the identity now shows that these power series parameterize the unit circle, which we used in the previous section. Note that this definition actually constructs the sin and cos functions in a rigorous fashion and proves that they are differentiable, so that in fact it subsumes the previous two.

Proof using the differential equation

Sine and Cosine can be defined as the two solutions to the differential equation:[6]
y'' + y = 0\,
satisfying respectively y(0) = 0, y′(0) = 1 and y(0) = 1, y′(0) = 0. It follows from the theory of ordinary differential equations that the first solution, sine, has the second, cosine, as its derivative, and it follows from this that the derivative of cosine is the negative of the sine. The identity is equivalent to the assertion that the function
z = \sin^2 x + \cos^2 x\,
is constant and equal to 1. Differentiating using the chain rule gives:
 \frac{d}{dx} z = 2 \sin x \ \cos x + 2 \cos x \ (-\sin x) = 0 \ ,
so z is constant. A calculation confirms that z(0) = 1, and z is a constant so z = 1 for all x, so the Pythagorean identity is established.
A similar proof can be completed using power series as above to establish that the sine has as its derivative the cosine, and the cosine has as its derivative the negative sine. In fact, the definitions by ordinary differential equation and by power series lead to similar derivations of most identities.
This proof of the identity has no direct connection with Euclid's demonstration of the Pythagorean theorem.
sumber : http://en.wikipedia.org/wiki/Pythagorean_trigonometric_identity

RUMUS-RUMUS BANGUN DATAR DAN BANGUN RUANG

RUMUS BANGUN DATAR
a.     Persegi
Bangun persegi memiliki 4 buah simetri putar dan 4 buah simetri lipat.
Rumus :
·         Keliling : 4 x s
·         Luas : s x s (s2)
S = sisi
b.     Persegi panjang
Bangun persegi  panjang memiliki 2 buah simetri putar dan 2 buah simetri lipat.
Rumus :
·         Keliling : 2 x (p+l)
·         Luas : p x l
P= panjang
L= lebar
c.      Segitiga
1.      Segitiga sama kaki
Bangun segitiga sama kaki memiliki 1 buah simetri putar dan 1 buah simetri lipat.
2.      Segitiga sama sisi
Bangun segitiga sama sisi memiliki 3 buah simetri putar dan 3 buah simetri lipat.
3.      Segitiga siku-siku
Bangun segitiga siku-siku tidak memiliki simetri lipat dan memiliki 1 buah simetri putar.
4.      Segitiga sembarang
Bangun segitiga sembarang tidak memiliki simetri lipat dan memiliki 1 buah simetri putar.
Rumus :
·         Keliling : AB+BC+AC
·         Luas : ½  x a x t
a = alas
t= tinggi
d.     Jajargenjang
Bangun jajargenjang memiliki 2 buah simetri putar dan tidak memiliki simetri putar.
Rumus :
·         Keliling: AB+BC+CD+AD
·         Luas: a x t
a=alas
t=tinggi
e.     Trapesium
1.      Trapesium  sembarang
Bangun trapesium sembarang memiliki  1 buah simetri putar dan tidak memiliki simetri lipat.
2.      Trapesium sama kaki
Bangun trapesium sama kaki  memiliki 1 buah simetri putar dan 1 buah simetri lipat.
3.      Trapesium siku-siku
Bangun trapesium siku-siku memiliki 1 buah simetri putar dan tidak memiliki simetri lipat.
Rumus :
·         Keliling : AB+BC+CD+DA
·         Luas: ½ x jumlah sisi sejajar x tinggi
f.        Layang-layang
Bangun layang-layang memiliki 1 simetri putar dan 1 simetri lipat
Rumus:
·         Keliling: 2(AB+BC)
·         Luas: ½ x d1 x d2
d = diagonal
g.     Belah ketupat
Bangun belah ketupat memiliki 2 buah simetri lipat dan 2 buah simetri putar.
Rumus :
·         Keliling : 4 x s
·         Luas: ½ x d1 x d2
d = diagonal
B.    RUMUS BANGUN RUANG
a.     Kubus
Rumus:
·         Luas permukaan: 6 x s2 =6s2
·         Volume: s x s x s= s3
b.     Balok
Rumus:
·         Luas permukaan: 2{(p x l)+(p x t)+(l x t)}
·         Volume: p x l x t
c.      Limas
Rumus:
·         Luas permukaan: La + jumlah luas segitiga pada bidang tegak
·         Volume : 1/3 x La x t
La=luas alas
t= tinggi
d.     Prisma
Rumus:
·         Luas permukaan : (2 x La)+(K x t)
·         Volume: La x t
La= luas alas
K= keliling alas
t= tinggi
e.     Tabung
Rumus:
·         Luas permukaan: 2 π r (r+t)
·         Luas selimut: 2 π r t
·         Volume : π r2 t
π= 22/7 atu 3,14
r= jari-jari alas
t= tinggi tabung
f.       Kerucut
Rumus:
·         Luas permukaan: π r (r+s)
·         Luas selimut: π r s
·         Volume: 1/3 π r2 t
r= jari-jari lingkaran alas
s= panjang garis pelukis kerucut
t= tinggi kerucut
g.     Bola
Rumus :
·         Luas permukaan: 4 π r2
·         Volume: 4/3 π r3
r= jari-jari bola

sumber: http://dewi-9b199701.blogspot.com/2012/03/rumus-rumus-bangun-datar-dan-bangun.html

Kamis, 22 November 2012

Rumus Luas Segitiga Lengkap dan Cepat

Paman APIQ mengamati banyak sekali siswa dan guru yang mencari ilmu tentang rumus luas segitiga. Memang luas segitiga sangat menarik. Tampaknya luas segitiga itu sulit tapi ternyata mudah. Kadang tampaknya mudah tapi sulit.
Berikut ini adalah beberapa rumus luas segitiga. Silakan teman-teman memanfaatkannya. Jika ada rumus yang belum tercantum silakan menambahkannya lagi.
1. Setengah luas segi empat
Prinsip dasar dari luas segitiga adalah setengah dari luas segi empat yang berhubungan.
L = 1/2 (p x l)
Contoh:
Sebuah persegi panjang dengan panjang = 4 dan lebar = 6 dibagi menjadi 2 bagian sama besar menurut garis diagonal. Berapa luas salah satu segitiga yang terbentuk?
Jawab:
L = 1/2 (p x l)
= 1/2 (4 x 6) = 12 (Selesai).
2. Setengah alas x tinggi
Rumus ini merupakan rumus yang paling terkenal.
L = 1/2 a.t
Contoh:
Tentukan luas segitiga siku-siku yang ukuran sisi datar = 6 dan sisi tegak = 8.
Jawab:
L = 1/2 a.t
= 1/2 (6.8) = 24 (Selesai).
3. Rumus umum setengah alas x tinggi
Menariknya, rumus setengah alas x tinggi ini berlaku umum untuk semua segitiga. Persoalannya, kadang-kadang kita harus berusaha keras untuk menemukan alas dan tingginya.
Contoh:
Hitunglah luas seluruh segitiga yang dibentuk dengan menghubungkan titik O(0 , 0), A(1 , 4), B(3 , 0), C(5 , 4), D(10 , 0) dan alas sumbu-X.
Jawab:
L = 1/2 a.t
= 1/2 (10 x 4) = 20 (Selesai).
4. L = 1/2 a.b Sin C
Contoh:
Tentukan luas segitiga yang panjang sisi a = 4, b = 6, dan sudut C = 30 derajat.
Jawab:
L = 1/2 a.b Sin C
= 1/2 (4.6. Sin 30)
= 1/2 (4.6. 1/2)
= 6 (Selesai).
5. Luas segitiga sama sisi = 1/4 s^2 akar 3
6. Akar s(s – a)(s – b)(s – c)
7. Pengurangan terhadap segi empat berhubungan
Bagaimana menurut Anda?
Salam hangat…
(angger; agus Nggermanto: Pendiri APIQ)
sumber : http://apiqquantum.com/2011/09/08/rumus-luas-segitiga-lengkap-dan-cepat/