🥃 Sin A 4 5 Cos B 5 13

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.

Question13. Evaluate: sin θ.sec(9O – θ) Solution: Question 14. Find the value of (cosec² θ – l).tan²θ Solution: Short Answer Type Question I [2 Marks] Question 15. Prove the following identity:sin³ θ+cos³ θ/sin θ+cos θ= 1 – sin Solution: Short Answer Type Questions II [3 Marks] Question 16.

Here, colorgreenI^st Quadrant=> 0 all+ve sina=5/13=>cosa=sqrt1-sin^2a=sqrt1-25/169=12/13 cosb=4/5=>sinb=sqrt1-cos^2b=sqrt1-16/25=3/5 colorredisina+b=sinacosb+cosasinb colorwhiteisina+b=5/13xx4/5+12/13xx3/5=20/65+36/65=56/65 colorblueiicosa-b=cosacosb+sinasinb colorwhiteiicosa-b=12/13xx4/5+5/13xx3/5=48/65+15/65=63/65 colorvioletiiicosb/2=sqrt1+cosb/2=sqrt1+4/5/2=sqrt9/10=3/sqrt10 colororangeivsin2a=2sinacosa=2xx5/13xx12/13=120/169 ^{f= (50 × 10 −3)(10π) 2 (5 × 10 −2) = 2,5 n Baca juga : Contoh soal dan pembahasan fisika kelas 11 lengkap dengan latihan soalnya Sekian dulu ya Contoh soal dan pembahasan gerak harmonik sederhana . periode ayunan pada bandul dan getaran pegas adalah materi gerak harmonik sederhana yang wajib kita pelajari. untuk itu tetap semangat ya} 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 Ifcos(A + B) = 4/5 and sin(A - B) = 5/13 and A,B lies between 0 and pi/4. asked Jun 24, 2014 in TRIGONOMETRY by anonymous. trigonometric-functions; if sinx=7/5 and angle x is in quadrant 2 and cos y=12/13 and angle y is in quadrant 1 find sin (x+y) asked Nov 26, 2013 in TRIGONOMETRY by harvy0496 Apprentice. 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 Jawaban Diketahui : cos a = 3/5 dan sin b = 5/13. jika a sudut lancip dan b sudut tumpul. Ditanya : tentukan nilai dari \sin (\alpha-\beta) ! sin(α−β)! Pembahasan : \cos a=\frac {3} {5}=\frac {samp} {mir} cosa = 53 = mirsamp maka \sin a=\frac {dep} {mir}=\frac {4} {5} sina = mirdep = 54. depan=\sqrt {5^2-3^2}=\sqrt {16}=4 depan = 52 −32 Byju's AnswerStandard XIIMathematicsComposition of Trigonometric Functions and Inverse Trigonometric FunctionsIf cos a+b=4 ...QuestionOpen in AppSolutiongiven, cosA+B = 4/5, thus tanA+B=3/4. 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/ Corrections20Similar questionsQ. If sinA=45 and cosB=513, where 0
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
$\begingroup$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 . AgqI0h. 25 81 173 127 26 486 97 448 125
sin a 4 5 cos b 5 13