13. Problem: If a, b, c are nonzero real numbers and a, ß are solutions of the equation
a cos theta +b sin theta =c then show that
1.sin alpha +sin beta = 2bc/a^2+b^2
2.sin alpha. sin beta=c^2-a^2/a^2+b^2
Answers
Answer:
acosθ+bsinθ=c ...(1)
acos
2
θ+bsin
2
θ+2asinθcosθ=c
a
2
(1+cos2θ)
+b
2
(1−cos2θ)
+asin2θ=c
a+acos2θ+b−bcos2θ+2asin2θ=2c
(a−b)cos2θ+2asin2θ=2c−(a+b) ...(2)
Squaring (1) on both sides
a
2
cos
2
θ+b
2
sin
2
θ+absin2θ=c
2
a
2
2
(1+cos2θ)
+b
2
2
(1−cos2θ)
+absin2θ=c
2
(a
2
−b
2
)cos2θ+2absin2θ=2c
2
−(a
2
+b
2
) ....(3)
From equation (2) sin2θ=
2a
[2c−(a+b)]−[(a−b)cos2θ]
putting in (3)
(a
2
−b
2
)cos2θ+b[(2c−(a+b))−((a−b)cos2θ)]=2c
2
−a
2
−b
2
(a
2
−b
2
)cos2θ+b[2c−a−b−(a−b)cos2θ]=2c
2
−a
2
−b
2
(a
2
−b
2
)cos2θ+2bc−ab−b
2
−
Step-by-step explanation:
acosθ+bsinθ=c ...(1)
acos
2
θ+bsin
2
θ+2asinθcosθ=c
a
2
(1+cos2θ)
+b
2
(1−cos2θ)
+asin2θ=c
a+acos2θ+b−bcos2θ+2asin2θ=2c
(a−b)cos2θ+2asin2θ=2c−(a+b) ...(2)
Squaring (1) on both sides
a
2
cos
2
θ+b
2
sin
2
θ+absin2θ=c
2
a
2
2
(1+cos2θ)
+b
2
2
(1−cos2θ)
+absin2θ=c
2
(a
2
−b
2
)cos2θ+2absin2θ=2c
2
−(a
2
+b
2
) ....(3)
From equation (2) sin2θ=
2a
[2c−(a+b)]−[(a−b)cos2θ]
putting in (3)
(a
2
−b
2
)cos2θ+b[(2c−(a+b))−((a−b)cos2θ)]=2c
2
−a
2
−b
2
(a
2
−b
2
)cos2θ+b[2c−a−b−(a−b)cos2θ]=2c
2
−a
2
−b
2
(a
2
−b
2
)cos2θ+2bc−ab−b
2
−b(a−b)cos2θ=2c
2
−a
2
−b
2
a
2
cos2θ−b
2
cos2θ+2bc−ab−abcos2θ+b
2
cos2θ=2c
2
−a
2
(a
2
−ab)cos2θ=2c
2
−a
2
+ab−2bc
cos2θ=
a
2
−ab
2c
2
−a
2
+ab−2bc
=
a
2
−ab
2c
2
−2bc−(a
2
−ab)
cos2θ=
a(a−b)
2c(c−b)
−1
θ=
2
1
cos
−1
(
a(a−b)
2c(c−b)
−1)