Question

xty
32. If a and ß are the solutions of the equation a cos 0 + b sin 0 = c,
then show that
a² - 6²
(i) cos (a +B) =
a² +62
2c² - (a² +6²)
(ii) cos (a - ) =
a² +6²​

Answer 1

Step-by-step explanation:

We have acosθ=c−bsinθ

Squaring both sides

⇒a 2 (1−sin 2 θ)=c 2 −2bcsinθ+b 2sin 2θ

⇒(a 2 +b 2)sin 2 θ−2bcsinθ+(c 2 −a 2)=0

Since α and β are the values of θ as given,

Since α and β are the values of θ as given,∴ roots of the above equation are sinα and sinβ.

∴sinα+sinβ= Sum of roots = 2bc/a^2+b^2

and sinα.sinβ= Product of roots =

c^2 -a^2/a^2+b^2