If α and β are two different values of θ ( between 0 and 2π ) which satisfy the equation .................?

Question

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6 cos θ + 8 sin θ = 9, Find the value of ..... sin ( α + β )




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Answer 1

6 cos θ + 8 sin θ = 9

=> (3 / 5) cos θ + (4 / 5) sin θ = 9 / 10

=> sin (θ + φ) = 9/10



Clearly if

sin x = sin a

=> x = a or π - a

for given range



=> θ + φ = α + φ or π - (α + φ)

Clearly one of them is equal to ß + φ

=> π - (α + φ) = ß + φ

=> α + ß = π - 2φ

=> sin (α + ß) = sin (π - 2φ)

=> sin (α + ß) = sin 2φ

=> sin (α + ß) = 2 sin φ cos φ = 24 / 25

Answer 2

α and ß are the roots of the equation : 6 cosΘ + 8 sinΘ = 9

=> 6 cosα + 8 sinα = 9 ------------- (1) and

.....6 cos ß + 8 sin ß = 9 ------------- (2)

Subtracting (2) from (1), we have :

6(cosα - cos ß) + 8(sin α - sin ß) = 0

=> 4(sinα - sin ß) = 3(cos ß - cos α)

=> (sin α - sin ß)/(cos ß - cos α) = 3/4

=> [2 cos{(α + ß)/2} sin {(α - ß)/2}] / [2 sin {(α + ß)/2} sin {(α - ß)/2}] = 3/4

=> cot {(α + ß)/2} = 3/4

=> cot²{(α + ß)/2} = 9/16

=> 1 + cot²{(α + ß)/2} = 1 + 9/16 = 25/16

=> cosec²{(α + ß)/2} = 25/16

=> sin² {(α + ß)/2} = 16/25

=> sin {(α + ß)/2} = ± 4/5 --------- (3)

=> 1 - sin²{(α + ß)/2} = 1 - 16/25 = 9/25

=> cos²{(α + ß)/2} = 9/25

=> cos {(α + ß)/2} = ± 3/5 ---------- (4)

=> sin (α + ß) = 2 sin {(α + ß)/2} cos {(α + ß)/2} = ± 2(4/5)*(3/5) = ± 24/25

Answer 3

Pramod - for this we'll have to fall back on the Wave Function i.e. express the LHS in the form kcos(theta - phi) = kcostheta.cosphi + ksintheta.sinphi where k > 0, and comparing that line with the original we see that the coefficient of cos theta is kcos phi and this = 6; similarly the coefficient of sin theta is ksin phi ans this is 8. Squaring these and adding we get k^2(cos^2phi + sin^2phi) = 6^2 + 8^2 = 100 so k^2 = 100 (the bracket = 1) therefore k = 10, and the function is now 10cos(theta - phi); for phi, put 10sin phi over 10cos phi and tan phi = 8/6 = 4/3 so since both sin phi and cos phi are +ve, phi is in Quad.1 so phi = tan^-1(4/3) = 53.1 degrees (from here I'll omit the word degrees) So we now have 10cos(theta - 53.1) = 9, so cos(theta - 53.1) = 0.9, giving theta - 53.1 = 25.8 or 360 - 25.8 i.e.

theta = 53.1 + 25.8 or 53.1 + 334.2 = 78.9 or 387.3, but since both alpha & beta are < 360 this is 387.3 - 360 = 27.3. It's actually very late here just now 0330 in fact) so the brain is just ticking over ! I can't see any way to find sin(alpha + beta) without recourse to a calculator.

Answer 4

Dhambarudhar's approach is correct till the end as



6 cosα + 8 sinα = 9



so cosα and sinα both > 0 ( as if one is -ve then sum < 8) and so α is in 1st quadrant and similarly β



so ( α + β ) is either in 1st or 2nd quadrant so sin ( α + β )is positive and hence -24/25 need to be ruled out and hence it is 24/25.

Answer 5

i) Applying half angle identities,

6{1 - tan²(Ө/2)}/{1 + tan²(Ө/2)} + 8*2tan(Ө/2)/{1 + tan²(Ө/2)} = 9



This simplifies to: 15tan²(Ө/2) - 16tan(Ө/2) + 3 = 0



ii) This is a quadratic in tan(Ө/2), which will have two roots;

let them be tan(α/2) and tan(β/2)



Applying properties of roots of quadratic equation,



Sum of roots: tan(α/2) + tan(β/2) = -b/a = 16/15 and

Product of roots: tan(α/2) * tan(β/2) = c/a = 3/15 = 1/5



==> tan{( α + β)/2} = {tan(α/2) + tan(β/2)}/{1 - tan(α/2) * tan(β/2)} =

= (16/15)/{1 - 1/5) = 4/3



iii) Again by half angle identities, sin(α + β) = 2*tan{(α + β)/2}/[1 + tan²{(α + β)/2}]

Substituting from the above tan{(α + β)/2} = 4/3 and simplifying,



sin(α + β) = 24/25