Question

If a cos θ - b sin θ = c, then show that a sin θ + b cos θ = ±$\sqrt{a^{2} + b^{2} - c^{2}}$

Answer 1

Given, $acos\theta-bsin\theta=c$

squaring both sides,

$(acos\theta-bsin\theta)^2=c^2$

$a^2cos^2\theta+b^2sin^2\theta-2absin\theta cos\theta=c^2$..........(1)



now, Let $asin\theta+bcos\theta=x$

squaring both sides,

$a^2sin^2\theta+b^2cos^2\theta+2absin\theta cos\theta=x^2$......(2)

adding equations (1) and (2),

$a^2(cos^2\theta+sin^2\theta)+b^2(cos^2\theta+sin^2\theta)=c^2+x^2$

we know, sin²x + cos²x = 1

so, $a^2+b^2=c^2+x^2$

or, a² + b² - c² = x²

taking square root both sides,

x = $\pm\sqrt{a^2+b^2-c^2}$

hence, asin θ + bcos θ = ±$\sqrt{a^{2} + b^{2} - c^{2}}$

Answer 2

HELLO DEAR,



GIVEN:-
a cos θ - b sin θ = c

on squaring both side,
a²cos²θ + b²sin²θ - 2absinθcosθ = c²----------( 1 )

let
a sin θ + b cos θ = p

on squaring both side,
a²sin²θ + b²cos²θ + 2absinθcosθ = p²---------( 2 )


now, adding---------( 1 ) & ----------( 2 )

a²(sin²θ + cos²θ) + b²(sin²θ + cos²θ) = p² + c²

[sin²x + cos²x = 1]

=> a² + b² - c² = p²

=> p = ±√(a² + b² - c²)


I HOPE IT'S HELP YOU DEAR,
THANKS