Usethe indentity sin (A + B) = sin (A)cos (B) + cos (A)sin (B) to expand the given expression. Use the above indentities to simplify each term in the above expression. sin (arccos (-1/2)) = √ (1 - (- 1/2) 2) = √3/2 (we have used sin (arccos (x)) = √ (1 - x 2 )) Substitute and calculate.

If \[\sin A = \frac{4}{5}\] and \[\cos B = \frac{5}{13}\], where 0 < A, \[B < \frac{\pi}{2}\], find the value of the following sin A + B Given \[ \sin A = \frac{4}{5}\text{ and }\cos B = \frac{5}{13}\]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,\[ \sin\left A + B \right = \sin A \cos B + \cos A \sin B\]\[ = \frac{4}{5} \times \frac{5}{13} + \frac{3}{5} \times \frac{12}{13}\]\[ = \frac{20}{65} + \frac{36}{65}\]\[ = \frac{56}{65}\]
Teksvideo. Di sini ada soal trigonometri diketahui Sin a 4/5 sin B 5/13 sudut a dan b keduanya merupakan sudut lancip. Carilah nilai cos buka kurung a minus B Pertama kita Gambarkan segitiga siku-siku untuk melukis Sin a 4/5 kita lihat segitiganya Sin A 4 di hadapan sisi miring 4/5 dengan Tripel pythagoras kita dapatkan 3 kemudian segitiga yang kedua segitiga B sudutnya 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 Lesidentités trigonométriques de base. À partir du triangle rectangle, il est possible de définir différents rapports trigonométriques que l'on peut ensuite transposer dans le cercle trigonométrique. Ainsi : À ces trois rapports, on ajoute les trois suivants qui découlent des trois premiers. On peut utiliser cot ou cotan pour la >>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

sin(A ± B) = sin A cos B ± sin B cos A. cos (A ± B) = cos A cos B ∓ sin A sin B. Download Solution PDF. Share on Whatsapp Stay updated with the Mathematics (x2 – 6x + 13) is equal to-Q5. If tanh (x + iy) = p + iq, then \(\frac{p}{q}\) = Q6. If x = nπ, n ∈ I then cot x is-Q7. Let B = 2 sin2x - cos 2x, then. Q8. If A = cos2θ

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.
Inright triangle ABC, mZC = 90°. If cos B = equals? 13 (1) tan A (2) tan B (3) sin A (4) sin B 5 13' which fu > Receive answers to your questions 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,\[\sin\left A - B \right = \sin A \cos B - \cos A \sin B \]\[ = \frac{4}{5} \times \frac{5}{13} - \frac{3}{5} \times \frac{12}{13}\]\[ = \frac{20}{65} - \frac{36}{65}\]\[ = \frac{- 16}{65}\] page08 MARKS 13. sinThe right-angled triangle in the diagram is such that x= 2 11 and x π 0
given, cosA+B=4/5, thus tanA+B=3/4 from triangle sinA-B=5/13,thus tanA-B=5/12. then tan2A=tanA+B+A-B =tanA+B+tanA-B/1-tanA+BtanA-B =3/4+5/12/1-3/45/12 = 56/33.
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