Fora, b and R positive and α acute, our equivalent expression is given by: a sin θ + b cos θ ≡ R cos (θ − α) This time there is a difference in the way we obtain α, compared to before. Expanding R cos (θ − α) using our result for the expansion of cos (A − B) gives us: R cos (θ − α) = R cos θ cos α + R sin θ sin α. จงหาค่าของ1.1 cos 75 sin 15 1.2 sin 75 sin 15 1.3 cos 15 cos 75 cos 105 cos 195 1.4 sin 40 sin 80 sin 160 1.5 sin 75 – sin 15 1.6 cos 75 + cos 15 5 1.7 sin sin 1.8 cos 9 2 12 8 1.9 sin 22.5 1.10 tan 22.5 1.11 sin 165 + sin 45 – sin 75 1.12 cos 20 + cos 100 + cos 220 1.13 sin 20 sin 40 sin 60 sin 80 1.14 cos 40 + cos 80 105 460 10. S 10. S What are the sine, cosine, and tangent ratios for LG? 15 cos G h IS Ex: Find the following: 5 sin A = S cos A = tan A = sin C = cos C = tan C = Key Concept Trigonometric Ratios E length sine of LA = c, len length cosine of ZA — tangent of LA = 8-3 Trigonometry Understand what the Sin., and tan. are cos., Use these Sebagaicontoh akan ditentukan nilai minimum dan maksimum fungsi trigonometri f(x) = 3 cos x + 4 sin x + 1. f’(x) = 0 –3 sin x + 4 cos x = 0 –3 sin x = –4 cos x sin x / cos x = –4 / –3 = 4 / 3 tan x = 4 / 3. Proses perhitungan di atas menunjukan hasil bahwa nilai tan x = 3 / 4, sehingga dapat diperoleh dua kondisi. Dua kondisi cosAcosB. (10), (11), and (12) are special cases of (4), (6), and (8) obtained by putting A= B= . Sum and product formulae cosA+ cosB= 2cos A+ B 2 cos A B 2 (13) cosA cosB= 2sin A+ B 2 sin A B 2 (14) sinA+ sinB= 2sin A+ B 2 cos A B 2 (15) sinA sinB= 2cos A+ B 2 sin A B 2 (16) Note that (13) and (14) come from (4) and (5) (to get (13), use (4 Dịch Vụ Hỗ Trợ Vay Tiền Nhanh 1s. >>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 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 =6365 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 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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