(3 puntos) Por lo tanto, N = √ 3 / 2√ 6 √ 2 2(2 √ 3) = √ 32 √ 6 2 √ 2 8 4 √ 3 (6 puntos) 5 UNIVERSIDAD DEL BÍOBÍO FACULTAD DE CIENCIAS DEPARTAMENTO DE MATEMÁTICA JESCEHMHFMNMGT/mh c ) Demostrar la identidad sec( θ ) 1 csc( θ ) = tan( θ ) 1 sen( θ ) Solución Amplificamos la fracción sec( θ= 2√3 √6 – √2/4 Register at BYJU'S to learn all about trigonometry and its formulas and download BYJU'SThe Learning App to study with the helpExample 3 Check that points A(√3, 1), B(0, 0), and C(2, 0) are the vertices of an equilateral triangle Solution Three vertices A, B, and C are vertices of an equilateral triangle if and only if
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3 1/2 x 2/3-Cos 45° = 1/√2 By applying the above values in the first equation, we get sin (60 45) = (√3/2) (1/√2) (1/√2) (1/2) = (√3/2√2) (1/2√2) = (√3 1)/2√2 Hence the value of sin 105 ° = (√3 1)/2√2 cos 105 ° = cos (60 45) cos (A B) = cos A cos B sin A sin BWe know that, cos 45° = 1/√2 = sin 45° cos 30° = √3/2 and sin 30° = 1/2 Now, cos 45° cos 30° sin 45° sin 30° = (1/√2)×(√3/2) (1/√2)(1/2)
After Rationalizing The Denominator Of 2 3 5 2 2 3 3 And Simplifying We Get A 3 15 B 10 4 6 19 Then The Value Of A B Is
写像の合成と行列の積 行列を A= a11 a12 a21 a22, B= b11 b12 b21 b22 とおき，これらの行列によって定義される線形写像を fA(x)=Ax= a11 a12 a21 a22 x y!, fB(x)=Bx= b11 b12 b21 b22 x y! Show that cot(7(1/2)°) = √2 √3 √4 √6 trigonometry; (√31)^2 (√2)^2 2^2 / 2 * (√31) * √2= 2(√31) / 2√2(√31) = √2 / 2 この問題がなぜそうなるのかわかりません。分かる方、途中の式を教えてください。分子(√31)^2(√2)^22^2=32√3124=2(√31)分母2×(√31)×√2=2√2(√31)
√31/2√2√31 simply by rationalizing meathod babu009china is waiting for your help Add your answer and earn points ｜3－√3｜＋｜1－2√2｜＝2＋√2になるらしいです。よくわからないので教えてくれませんか？後同じような感じですが・・｜1－√2｜＝√2－1になるらしいです。どうか教えて下さい。 数学 解決済 教えて！goo x = 3/2, y = 1/2 ⇒ √ 2 √3 = √ 3/2 √ 1/2 Algebra teachers usually want us to rationalize denominators, so you can multiply each radical on the righthand side by √ 2/2 to get √ 2 √3 = (1/2)(√ 6 √ 2) Example 3 √ 2 − √3 Solution This is exactly the same deal as Example 2, so I'll hurry through it
= √3/2√2 1/2√2 = (√3 1) / 2√2 Fun Facts Other trigonometric ratios of 15 0 can be determined using the relation of other trigonometric ratios with sine of angle 15 0 Using the abovementioned derivation steps with a little modification, trigonometric ratios of 75 0 can be calculated FAQ (Frequently Asked Questions)√ 3− 1 1 √ 3×1 = √ 3− 1 √ 31 It would be more usual to tidy this result up to avoid leaving a root in the denominator This can be done by multiplying top and bottom by the same quantity, as follows √ 3−1 √ 31 = (√ 3− 1) √ 31 × (√ 3−1) √ 3−1 = 3− √ 3− √ 31 3−1 = 4− 2 √ 3 2 = 2− √ 3Share It On Facebook Twitter Email 1 Answer 1 vote answered by RamanKumar (499k points) selected by Anjali01 Best answer We have to prove that ← Prev
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Math\text{Let}\,\, \sin^{1} \left(\dfrac{\sqrt{3} 1}{2\sqrt{2}}\right) = \theta/math math\implies \sin \theta = \dfrac{\sqrt{3} 1}{2\sqrt{2}}/math mathRS Aggarwal Solutions for Class 10 Maths Chapter 11 Trigonometric Ratios of Some Particular Angles Question 5 Solution We know that, cos 30° = √3/2 ⇨ cos2 30° = 3/4 cos 60° = 1/2 ⇨ cos2 60° = 1/4 sec 30° =(2/√3) ⇨ sec2 30° = 4/3 tan 45° = 1『3行3列行列の対角化相異なる3個の固有値を持つとき』 15/12 3行3列正方行列 A TrA=A112A33 detA
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c √1/2 * √2 * √3 = pierw z 6/2 d √2/3 * √5/6 * √30 = pierw z 300/pierw z 18 = 2 pierw z 75 /3pierw z 2 e √12/√3 = pierw z 3*4/pierw z 3 = 2pierw z 3 = 2
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