sincos 5 θθ−= ，試求下列各式之值： (1) sin cosθθ⋅ (2)sin cosθθ+ (3)sin cos. 33. 13 ； 12 13 ； 5 12 2、(1)36；(2) 12 5 − 3、36 4、(1) 9 4 ；(2)1 5、 51+ 6、8 7、2 8、
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 Vềchúng tôi. Chủ quản: Hệ thống giáo dục HOCMAI Tòa nhà 25T2 Trung Hòa - Cầu Giấy - Hà Nội Giấy phép MXH số 597/GP-BTTTT Bộ TT&TT cấp 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,\[\sin\left A - B \right = \sin A \cos B - \cos A \sin B \]\[ = \frac{4}{5} \times \frac{5}{13} - \frac{3}{5} \times \frac{12}{13}\]\[ = \frac{20}{65} - \frac{36}{65}\]\[ = \frac{- 16}{65}\]
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The 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.
Sinus Cosinus dan Tangent digunakan untuk menghitung sudut dengan perbandingan trigonometri sisi di segitiga. Dengan gambar segitiga diatas, nilai Sinus, Cosinus dan Tangent diperoleh dengan cara sebagai berikut: Soal No. 13. Seorang anak berdiri 20 meter dari sebuah menara seperti gambar berikut. Perkirakan ketinggian menara dihitung dari
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+ϕ=tanθ+tanϕ1-tanθtanϕAlso,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.

cosA < 0. ∴ cos A = `- 4/5` sin B = `4/5` We know that, cos 2 B = 1 – sin 2 B = `1 - (4/5)^2` = `1 - 16/25 = 9/25` ∴ cos B = `± 3/5` Since B lies in the second quadrant. cos B < 0. ∴ cos B = `-3/5` ∴ 4cos A + 3cos B = `4(-4/5) + 3(-3/5)` = `-16/5 - 9/5` = `-25/5` = – 5.

$\begingroup$I've used the angle sum identity to end up with $\cos A \cos B -\sin A \sin B = \frac{5}{13} = \frac{3}{5}\cos B -\frac{4}{5} \sin B$, but don't know how to proceed from here. Any tips? asked Aug 10, 2017 at 1020 $\endgroup$ 1 $\begingroup$Hint $\cos B=\cosA+B-A$ use the compound angle formula answered Aug 10, 2017 at 1029 David QuinnDavid gold badges19 silver badges48 bronze badges $\endgroup$ $\begingroup$ $$A=\arcsin\dfrac45=\arccos\dfrac35$$ $$A+B=\arccos\dfrac5{13}=\arcsin\dfrac{12}{13}$$ $$B=\arccos\dfrac5{13}-\arccos\dfrac35$$ answered Aug 10, 2017 at 1056 $\endgroup$ You must log in to answer this question. Not the answer you're looking for? Browse other questions tagged .
JEEMain 2019: If cos (α+β) =(3/5), sin(α-β) = (5/13) and 0 < α, β < (π/4) then tan (2α ) is equal to: (A) (21/16) (B) (63/52) (C) (33/52) (D)

>>Class 11>>Maths>>Trigonometric Functions>>Trigonometric Functions of Sum and Difference of Two angles>>If cos A = 4/5 , cos B = 12/13 , 3pi/Open in AppUpdated on 2022-09-05SolutionVerified by TopprA and B both lie in the IV quadrant.=> are negativei iiSolve any question of Trigonometric Functions with-Was this answer helpful? 00More From ChapterLearn with Videos Practice more questions

Pada soal diketahui sin A = 5 4 dan cos B = 13 5 untuk A sudut tumpul dan B sudut lancip. Sehingga sudut A berada pada kuadran II sedangkan sudut B berada pada kuadran I. Sehingga untuk cos A bernilai negatif: sin A sisi miring sisi depan = = → → 5 4 5 4 sisi depan = 4 sisi miring = 5

Open in Appwe have the value of and but we don't have the value of and so, first we find the value of and let side opposite to angle hypotenuse where is any positive integer So, by Pythagoras theorem we can find the third side of a triangle taking positive square root as side cannot be negative So, Base we know that side adjacent to angle hypotenuse so, now we have to find the we know that let side adjacent to angle hypotenuse where is any positive integer so, by Pythagoras theorem, we can find the third side of a triangle taking positive square root since, side cannot be negative so, perpendicular we know that Now putting the values, we get Was this answer helpful? 00

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    • sin a 4 5 cos b 5 13