Question

Eliminate θ from the following. a=sinθ+cosθ,b=sin 3 θ+cos 3 θ

Answer 1

Answer:

a=SIN θ+ COS θ -----------------— eq (1)

b=SIN³ θ+ COS³ θ ------------------- eq( 2)

b =(SIN θ+ COS θ) (SIN²θ – SIN θ COS θ + COS² θ)

=> (a) ( 1– SIN θ COS θ)

a²=(SIN θ=COS θ)²

a²=1+2SIN θ COS θ

(OR) SIN θ COS θ =1-a²/2

PUTTING THE VALUE OF SIN θ COS θ IN EQUATION ii

b=(a) (1–(1– a²/ 2))

=a [2–1+a²/2]

b=( a) [ 1+ a²/2]

Answer 2

$\huge\pink{ANS} \huge\blue{WER}$

here is ur answer mate:------------------------------------------------------

a=sinθ+cosθ........eq (1)

b=sin³θ+cos³θ........eq (2)

b =>(sinθ+cosθ)(sin²θ-sinθcosθ+cos²θ)

=> A(1-sinθcosθ)

a²=(sinθ=cosθ)²

a²=1+2sinθcosθ

(or)

sinθcosθ=1-a²/2

putting the value of sinθcosθ in equation.....(2)

b=a[1-(1-a²/2)]

=a[2-1+a²/2]

=a[1+a²/2]

I hope this helps u

have a nice day✌☘