Thesum formula of sin is sin(A + B) = sin A cos B + sin B cos A. Let us see the derivation of sin2x step by step: Substitute A = B = x in the formula sin(A + B) = sin A cos B + sin B cos A, sin(x + x) = sin x cos x + sin x cos x. ⇒ sin2x = 2 sin x cos x. Hence, we have derived the formula of sin2x.
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? 00JEEMain 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)
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 = 63/654 Soal UN Fisika SMA 2012/2013 SA 67 No.5. Sebuah benda bermassa 5,0 kg ditarik dengan tali ke atas bidang miring yang kasar oleh sebuah gaya 71 N (g = 10 m.s-2, sin 37 o = 0,6, cos 37 o = 0,8). Jika koefisien gesekan antara benda dan bidang adalah 0,4, percepatan yang dialami benda adalah A. 0,5 ms-2. B. 2 ms-2. C. 2,5 ms-2
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=120/1692 If `sin α = 4/5` (in Quadrant I) and `cos β = -12/13` (in Quadrant II) evaluate `cos(β − α).` [This is not the same as Example 2 above. This time we need toWe 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
mo m What Did Mrs. Margarine Think About Her Sister' Husband? For each exercise, select the correct ratio from the four choices given. Write the letter of the correct choice
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.
sina b 4 5 and sin a b 5 13 prove that tan 2a 63 16 - Maths - a9j2osuu. If acos θ = bcos (θ+120°) = c cos (θ+240°) , then find the value of 1/a+1/b+1/c If 13θ= π , find the value of 3θ+cos5θ + 2cosθcos9θ Prove that, If find the value of
a) [latex] text { v }(text { t })=10 cos left(4 t-60^{circ}right)[/latex] [latex]mathrm{i}(mathrm{t})=4 sin left(4
Ifsin A= 4/5 and cos B =- 12/13, where A and B lie in first and third quadrant respectively, then cos (A+B)=.. A) (56)/(65) B) -(1
sinA4/5 A=53.13˚ cosB=5/13 B=67.38˚ A+B=53.13+67.38=120.51˚ tan (A+B)=tan (120.51˚)=-1.6969 as calculated: tan (A+B)=-56/33=-1.6969Ptolemy's theorem states that the sum of the products of the lengths of opposite sides is equal to the product of the lengths of the diagonals. When those side-lengths are expressed in terms of the sin and cos values shown in the figure above, this yields the angle sum trigonometric identity for sine: sin(α + β) = sin α cos β + cos α sin β. 4gJq2.
sin a 4 5 cos b 5 13
