rumus sin a sin b
2 Rumus Sinus Jumlah dan Selisih Dua Sudut Perhatikan rumus berikut ini. \(\sin (A + B)\) = \(\cos \left\{\frac{\pi}{2} - (A + B)\right\}\) = \(\cos (\frac{\pi}{2
Rumus trigonometri dua sudut - sin a+b = sin a cos b + cos a sin b sin a-b = sin a cos b - cos a sin b cos a+b = cos a cos b - sin a sin b cos a-b = cos a cos b + sin a sin b sina+b= sin a cos b + cos a sin b cosa+b= cos a cos b - sin a sin b sina-b= sin a cos b - cos a sin b cosa-b= cos a cos b + sin a sin b - + - + sina+b + sina-b= 2 sin a cos b cosa+b + cosa-b= 2 cos a cos b sin a + sin b= 2 sin 1/2a+b cos 1/2a-b cos a + cos b= 2 cos 1/2a+b cos 1/2a-b sina+b= sin a cos b + cos a sin b cosa+b= cos a cos b - sin a sin b sina-b= sin a cos b - cos a sin b cosa-b= cos a cos b + sin a sin b - _ - _ sin a+b - sin a-b= 2 cos a sin b cosa+b - cos a-b= -2 sin a sin b sin a - sin b= 2 cos 1/2a+b sin 1/2a-b cosa-b - cos a+b= 2 sin a sin b cos a - cos b= -2 sin 1/2a+b sin 1/2a-b cos b - cos a= 2 sin 1/2a+b sin 1/2a-b Identitas Trigonometri - sin^2 x + cos^2 x = 1 ====>> r cos a^2 + r sin a^2= r^2 berdasarkan rumus pers O -> a^2 + b^2 = c^2 r^2 cos^2 a + r^2 sin^2 a= r^2 selain itu 2a=a+a r^2 cos^2 a + sin^2 a=r^2 cos^2 a + sin^2 a=1 sin 2x= 2 sin x cos x ====>> sina+a= sin a cos a + cos a sin a sin x= 2 sin 1/2x cos 1/2x = 2 sin a cos a cos 2x= cos^2 x - sin^2 x cos x= cos^2 1/2x - sin^2 1/2x = cos^2 x -1- cos^2 X dst''' = 2 cos^2 x - 1 =1- sin^2 x - sin^2 x = 1- 2 sin^2 x ====>>cos a+a= cos a cos a - sin a sin a =cos^2 a - sin^2 a tan 2x= sin 2x - cos 2x = 2 sin x cos x - cos^2 x - sin^2 x = 2 sin x cos x 1 - X - cos^2 x - sin^2 x cos^2 x = 2 tan x - 1- tan^2 x Aturan sinus dan cosinus - a b c a^2= b^ - 2bc cos A -=-=- b^2= a^ - 2ac cos B sin a sin b sin c c^2= a^ - 2ab cos C Bagaimana bisa menemukan rumus itu? Asumsi awal; berasal dari segitigalihat buku latihan Luas segitiga menggunakan aturan trigonometry - L= 1/2ab sin C L= 1/2ac sin B L= 1/2bc sin A
suciwidia1.Rumus Sinus Sudut Ganda Dengan menggunakan rumus sin (A + B), untuk A = B maka diperoleh: sin 2A = sin (A + B) = sin A cos A + cos A sin A = 2 sin A cos A 2 Rumus Cosinus Sudut Ganda Dengan menggunakan rumus cos (A + B), untuk A = B maka diperoleh: cos 2A = cos (A + A) The Law of Sines or Sine Rule is very useful for solving triangles a sin A = b sin B = c sin C It works for any triangle a, b and c are sides. A, B and C are angles. Side a faces angle A, side b faces angle B and side c faces angle C. And it says that When we divide side a by the sine of angle A it is equal to side b divided by the sine of angle B, and also equal to side c divided by the sine of angle C Sure ... ? Well, let's do the calculations for a triangle I prepared earlier a sin A = 8 sin = 8 = b sin B = 5 sin = 5 = c sin C = 9 sin = 9 = The answers are almost the same! They would be exactly the same if we used perfect accuracy. So now you can see that a sin A = b sin B = c sin C Is This Magic? Not really, look at this general triangle and imagine it is two right-angled triangles sharing the side h The sine of an angle is the opposite divided by the hypotenuse, so a sinB and b sinA both equal h, so we get a sinB = b sinA Which can be rearranged to a sin A = b sin B We can follow similar steps to include c/sinC How Do We Use It? Let us see an example Example Calculate side "c" Law of Sinesa/sin A = b/sin B = c/sin C Put in the values we knowa/sin A = 7/sin35° = c/sin105° Ignore a/sin A not useful to us7/sin35° = c/sin105° Now we use our algebra skills to rearrange and solve Swap sidesc/sin105° = 7/sin35° Multiply both sides by sin105°c = 7 / sin35° × sin105° Calculatec = 7 / × c = to 1 decimal place Finding an Unknown Angle In the previous example we found an unknown side ... ... but we can also use the Law of Sines to find an unknown angle. In this case it is best to turn the fractions upside down sin A/a instead of a/sin A, etc sin A a = sin B b = sin C c Example Calculate angle B Start withsin A / a = sin B / b = sin C / c Put in the values we knowsin A / a = sin B / = sin63° / Ignore "sin A / a"sin B / = sin63° / Multiply both sides by B = sin63°/ × Calculatesin B = Inverse SineB = sin−1 B = Sometimes There Are Two Answers ! There is one very tricky thing we have to look out for Two possible answers. Imagine we know angle A, and sides a and b. We can swing side a to left or right and come up with two possible results a small triangle and a much wider triangle Both answers are right! This only happens in the "Two Sides and an Angle not between" case, and even then not always, but we have to watch out for it. Just think "could I swing that side the other way to also make a correct answer?" Example Calculate angle R The first thing to notice is that this triangle has different labels PQR instead of ABC. But that's OK. We just use P,Q and R instead of A, B and C in The Law of Sines. Start withsin R / r = sin Q / q Put in the values we knowsin R / 41 = sin39°/28 Multiply both sides by 41sin R = sin39°/28 × 41 Calculatesin R = Inverse SineR = sin−1 R = But wait! There's another angle that also has a sine equal to The calculator won't tell you this but sin is also equal to So, how do we discover the value Easy ... take away from 180°, like this 180° − = So there are two possible answers for R and Both are possible! Each one has the 39° angle, and sides of 41 and 28. So, always check to see whether the alternative answer makes sense. ... sometimes it will like above and there are two solutions ... sometimes it won't see below and there is one solution We looked at this triangle before. As you can see, you can try swinging the " line around, but no other solution makes sense. So this has only one solution.Jikagaris tinggi h ditarik dari titik B maka diperoleh rumus L = ½ Rumus lain dari luas segitiga ABC adalah jika diketahui panjang ketiga sisinya (yakni a, b dan c). Rumus tersebut adalah. Untuk lebih jelasnya diskusikanlah contoh soal berikut ini : 01. Tentukanlah luas segitiga ABC jika diketahui sisi BC = 4 cm, AC = 7√3 cmataubisa juga kita nyatakan dalam bentuk sin A. cos 2A = (1 - sin 2 A) - sin 2 A. cos 2A = 1 - 2 sin 2 A. dengan demikian nilai sin 2A bisa kita nyatakan ke dalam beberapa rumus, yaitu: cos 2A = cos 2 A - sin 2 A. cos 2A = 2 cos 2 A - 1. cos 2A = 1 - 2sin 2 A. Rumus untuk tan A. Dengan menggunakan rumus tan (A + B) dan jika kita