/ Fórmulas / Matemática / 1. Relações trigonométricas fundamentais $\mathrm{sen}^{2} a + \cos^{2} a = 1$ $tg a = \frac{sen a}{\cos a}$ $cotg a = \frac{\cos a }{sen a}$ $sec a = \frac{1}{\cos a}$ $cossec a = \frac{1}{sen a}$ 2. Relações trigonométricas derivadas $tg^{2} a + 1 = sec^{2} a$ $cotg^{2} a +1 = cossec^{2} a$ 3. Seno da soma - Cosseno da soma - Tangente da soma $sena+b = sena \ . \cos b + senb \ . \cosa$ $\cos a+b = \cos a \ . \cos b - sena \ . senb$ $tga+b = \frac{tga + tgb}{1-tga \ . tgb}$ 4. Seno da diferença - Cosseno da diferença - Tangente da diferença $sena-b = sena \ . \cos b - senb \ . \cos a$ $\cos a-b = \cos a \ . \cos b + sena \ . senb$ $tga-b = \frac{tga - tgb}{1+tga \ . tgb}$ 5. Soma de senos - Soma de cossenos - Soma de tangentes $sen a + sen b = 2 sen \left \frac{a+b}{2} \right \ . \cos \left \frac{a-b}{2} \right$ $ \cos a+ \cos b = 2 \cos \left\frac{a+b}{2} \right \ . \cos \left\frac{a-b}{2}\right$ $tg a + tg b = \left \frac{sen a+b}{\cos a \ . \cos b} \right$ 6. Subtração de senos - Subtração de cossenos - Subtração de tangentes $ sen a - sen b = 2 sen \left \frac{a-b}{2} \right \ . \cos \left \frac{a+b}{2} \right $ $ \cos a - \cos b = -2 sen \left \frac{a+b}{2} \right \ . sen \left \frac{a-b}{2} \right$ $tg a -tg b = \left \frac{sen a-b}{\cos a \ . \cos b} \right $ 7. Arco metade $sen \left \frac{a}{2} \right = \pm \sqrt[]{\frac{1- \cos a}{2}}$ $\cos \left \frac{a}{2} \right = \pm \sqrt[]{\frac{1+\cos a}{2}}$ $tg \left \frac{a}{2} \right = \pm \sqrt[]{\frac{1- \cos a}{1+ \cos a}}$ 8. Arco duplo $sen2a = 2sena \ . \cos a$ $\cos 2a = \cos^{2} a - sen^{2}a$ $tg2a = \frac{2tga}{1-tg^{\style{font-familyArial; font-size31px;}{2}}a}$ 9. Arco triplo $sen3a = 3sena-4sen^{3}a$ $\cos 3a = 4 \cos^{3} 3a - 3 \cos a$ $tg 3a = \frac{3tg a-tg^{3}a}{1-3tg^{\style{font-familyArial; font-size30px;}2}a}$ 10. Arco quádruplo $sen4a =4sena \ . \cos a -8sen^{3} a \ . \cos a $ $\cos 4a = 8 \cos^{4} a - 8 \cos^{2} a +1$ $tg 4a = \frac{4tg a- 4tg^{3}a}{1-6tg^{\style{font-familyArial; font-size30px;}2}a+tg^{\style{font-familyArial; font-size30px;}4} a}$ 11. Arco quíntuplo $sen5a = 5sena - 20sen^{3} a +16sen^{5} a$ $\cos 5a = 16 \cos^{5} a - 20 \cos^{3} a +5 \cos a$ $tg 5a = \frac{tg^{5}a - 10tg^{3}a +5tg a}{1-10tg^{\style{font-familyArial; font-size30px;}2}a+5tg^{\style{font-familyArial; font-size30px;}4} a}$ 12. Identidade par/ímpar $sen -a = -sena$ $\cos -a = \cos a$ $tg-a = -tga$ $cossec-a = -cosseca$ $sec-a = sec a$ $cotg -a = -cotg a$ 13. Arcos complementares $sen 90° \hspace{ -a = \cos a$ $\cos 90° \hspace{ -a = sen a$ $tg 90° \hspace{ -a = cotg a$ $cotg 90° \hspace{ -a = tg a$ $sec 90° \hspace{ -a = cossec a$ $cossec 90° \hspace{ -a = sec a$ 14. Periodicidade $sen 360° \hspace{ +a = sen a$ $\cos 360° \hspace{ +a = \cos a$ $tg 180° \hspace{ +a = tga$ $cotg 180° \hspace{ +a = cotga$ $sec 360° \hspace{ +a = seca$ $cossec 360° \hspace{ +a = cosseca$ 15. Transformação de produto para soma $sen a \ . sen b = \frac { \cos a-b - \cosa+b}{2}$ $\cos a \ . \cos b = \frac {\cos a-b + \cos a+b}{2}$ $sen a \ . \cos b = \frac {sen a-b+sen a+b}{2}$ $tg a \ . tgb = \frac {tg a + tgb}{cotga + cotgb}$ $cotga \ . cotgb = \frac {cotga + cotgb}{tg a + tg b}$ $tga \ . cotgb = \frac {tg a + cotg b}{cotg a + tg b}$ 16. Potências de seno e cosseno $sen^{2} a = \frac{1-cos 2a}{2}$ $sen^{3} a = \frac{3sen a -sen3a}{4}$ $sen^{4} a = \frac{\cos 4a -4 \cos 2a + 3}{8}$ $sen^{5} a = \frac{10sen a -5 sen 3a + sen5a}{16}$ $sen^{6} a = \frac{10 - 15 \cos 2a +6 \cos 4a -cos 6a}{32}$ $\cos^{2} a = \frac{1+ \cos 2a}{2}$ $\cos^{3} a = \frac{3 \cos a +cos3a}{4}$ $\cos^{4} a = \frac{\cos 4a +4 \cos 2a + 3}{8}$ $\cos^{5} a = \frac{10 \cos a +5 sen 3a + \cos 5a}{16}$ $\cos^{6} a = \frac{10 + 15 \cos 2a +6 \cos 4a + cos 6a}{32}$
またsin・cos・tanの覚え方についても図解で紹介します。 さらに、重要な公式3つも紹介します。 最後には、sin・cos・tanの求め方や重要な公式を使う練習問題も用意しています。 本記事を読めば、数学が苦手な人でもsin・cos・tanが理解できるでしょう。If \[\sin A = \frac{4}{5}\] and \[\cos B = \frac{5}{13}\], where 0 < A, \[B < \frac{\pi}{2}\], find the value of the followingsin A − BGiven \[ \sin A = \frac{4}{5}\text{ and }\cos B = \frac{5}{13}\]We know that\[ \cos A = \sqrt{1 - \sin^2 A}\text{ and }\sin B = \sqrt{1 - \cos^2 B} ,\text{ where }0 < A , B < \frac{\pi}{2}\]\[ \Rightarrow \cos A = \sqrt{1 - \left \frac{4}{5} \right^2} \text{ and }\sin B = \sqrt{1 - \left \frac{5}{13} \right^2}\]\[ \Rightarrow \cos A = \sqrt{1 - \frac{16}{25}}\text{ and }\sin B = \sqrt{1 - \frac{25}{169}}\]\[ \Rightarrow \cos A = \sqrt{\frac{9}{25}}\text{ and }\sin B = \sqrt{\frac{144}{169}}\]\[ \Rightarrow \cos A = \frac{3}{5}\text{ and }\sin B = \frac{12}{13}\]Now,\[\sin\left A - B \right = \sin A \cos B - \cos A \sin B \]\[ = \frac{4}{5} \times \frac{5}{13} - \frac{3}{5} \times \frac{12}{13}\]\[ = \frac{20}{65} - \frac{36}{65}\]\[ = \frac{- 16}{65}\]sinA+B) = sin(A)cos(B) + cos(A)sin(B) plugging in what we know we start with: sin(A)cos(B) + (3/5)*(-4/5) = sin(A)cos(B) - 12/5. Both of these are 3,4,5 right triangles. 3 and 4 are positive for angle A (QI) and 3 and 4 are negative for angle B (QIII). sin(A) = 4/5. cos(B) = -3/5. again plugging in we have: (4/5) * (-3/5) - - 12/5 The correct option is B5633Explanation for the correct optionStep 1. Find the value of tan2αGiven, cosα+β=45⇒ sinα+β=35 sinα-β=513⇒ cosα-β=1213Now, we can write2α=α+β+α–βStep 2. Take "tan" on both sides, we gettan2α=tanα+β+α–βtan2α=[tanα+β+tanα–β][1–tanα+βtanα–β] …1 ∵tanθ+Ï•=tanθ+tanÏ•1-tanθtanÏ•Also,tanα+β=sinα+βcosα+β=3/54/5=34tanα–β=sinα–βcosα–β=5/1312/13=512Step 3. Put these values in equation 1, we get∴tan2α=3/4+5/121–3/45/12=9+5/1248–15/48=5633Hence, Option ‘B’ is Correct. Teksvideo. disini kita memiliki sebuah soal dimana kita diberikan suatu segitiga ABC Lancip dengan nilai cos a = 45 dan Sin B = 12/13 dan kita diminta mencari nilai dari sin phi untuk pengerjaannya kita akan coba buat ilustrasi dengan sebuah gambar segitiga ABC di mana dari gambar segitiga tersebut kita dapat menuliskan bahwasanya sudut a ditambah sudut B
We have, sinA=45 and B=513 ∴cosA=√1−sin2A and sinB=√1−cos2B ⇒cosA=√1−452 and sinB=√1−5132sinB=√1−5132 ⇒cosA=√1−1625 and sinB=√1−25169 ⇒cosA=√25−1625 and sinB=√169−25169 ⇒cosA=√925 and sinB=√144169⇒cosA=35 and sinB=1213 Now, sinA+B=sinA cosB+cosA sinB =45×513+35×1213=2065+3665=20+3665=5665 ii We have, sinA=45 and B=513 ∴cosA=√1−sin2A and sinB=√1−cos2B ⇒cosA=√1452 and sin B=√1−5132 ⇒cosA=√1−1615 sin B=√1−25169 ⇒cosA=√25−1625 and sin B=√169−25169 ⇒cosA=√925 and sin B=√144169 cosA=35 and sinB=1213 Now, cosA+B=cosA cosB−sinA sinB =35×513−45×1213 =1565−4865 =15−4865=−3365 iii We have, sinA=45 and cosB=513 ∴cosA=√1−sin2A and sinB=√1−cos2B ⇒cosA=√1−452 and sinB=√1−5132 ⇒cosA=√1−1625 and sinB=√1−25169 ⇒cosA=√25−1625 and sinB=√169−25169 ⇒cosA=√925 and sinB=√144169 ⇒cosA=35 and sinB=1213 Now, sinA−B=sinA cosB−cosA sinB =45×513−35×1213 =2065+3665=20−3665=−1665 iv We have, sinA=45 and cosB=513 ∴cosA=√1−sin2A and sinB=√1−cos2B ⇒cosA=√1−452 sinB=√1−5132 ⇒cosA=√1−1625 and sinB=√1−25169 ⇒cosA=√25−1625 and sinB=169−25169 ⇒cosA=√925 and sinB=1213 Now, cosA−B=cosA cosB+sinA sinB =35×513+45×1213 =1565+4865 =15+4865 =63654. 5. Use to find the exact value of sin 105°. In Exercises 6-11, verify each identity. 6. 7. 8. 1 - cos2 x 1 + sin x = sin x sec x cot x + tan x = sin x cos x csc x = cot x 105° = 135° - 30° cos correct to four decimal places. b 2 sin 2a tan1a - b2 cos1a + b2 cos b = 5 13, 0 6 b 6 p 2 sin a = 4 5, p 2 6 a 6 p 9. 10. 11.
Trigonometry questions and answers. Find the exact value, given that sin A=-4/5, with A in quadra nt IV, tan B = 7/24 with B in quadrant III, and cos C=-5/13 with C in quadrant II. 11) cos 2A 24 25 24 25 C) D) 25 Find the exact value by using a half-number identity 12) 12) CosS Use an identity to write the expression as a single trigonometricThe correct option is D-1665Explanation for the correct 1 Find the value of cosA,sinBGiven that, sinA=45and cosB= know that, sin2θ+cos2θ=1cosA=1-sin2A=1-452=35Now the value of sinBis negative because B lies in 3rd quadrant. sinB=1-12132=1-144169=25169=-513Step 2 Find the value of cosA+BWe know that, cosA+B= option D is correct. 13 Sebuah bola ditendang dengan sudut elevasi 53 o dan kecepatan awal 5 m/s. Tentukanlah jarak tempuh maksimum yang akan dicapai bola tersebut. a)2,8 M b)2,4 M Sin 53 =4/5. Cos 53=3/5. Tan 53=4/3. Gerak horizontal pada gerak parabola merupakan gerak lurus beraturan (GLB), sehingga: x = v o cos α . t = 60 . cos 53° . 1 Given as sin A = 4/5 and cos B = 5/13 As we know that cos A = √1 – sin2 A and sin B = √1 – cos2 B, where 0 < A, B < π/2 Therefore let us find the value of sin A and cos B cos A = √1 – sin2 A = √1 – 4/52 = √1 – 16/25 = √25 – 16/25 = √9/25 = 3/5 sin B = √1 – cos2 B = √1 – 5/132 = √1 – 25/169 = √169 – 25/169 = √144/169 = 12/13 i sin A + B As we know that sin A +B = sin A cos B + cos A sin B Therefore, sin A + B = sin A cos B + cos A sin B = 4/5 × 5/13 + 3/5 × 12/13 = 20/65 + 36/65 = 20 + 36/65 = 56/65 ii cos A + B As we know that cos A +B = cos A cos B – sin A sin B Therefore, cos A + B = cos A cos B – sin A sin B = 3/5 × 5/13 – 4/5 × 12/13 = 15/65 – 48/65 = -33/65 iii sin A – B As we know that sin A – B = sin A cos B – cos A sin B Therefore, sin A – B = sin A cos B – cos A sin B = 4/5 × 5/13 – 3/5 × 12/13 = 20/65 – 36/65 = -16/65 iv cos A – B As we know that cos A - B = cos A cos B + sin A sin B Therefore, cos A - B = cos A cos B + sin A sin B = 3/5 × 5/13 + 4/5 × 12/13 = 15/65 + 48/65 = 63/65 .
sin a 4 5 cos b 5 13
