Trigonometry

Geometry Level 3

$\large \begin{cases} a\sin^3 x +b\cos^3 x =\sin x \cos x \\ a \sin x= b\cos x \end{cases}$

$a$ and $b$ are non-zero constants such that for some $x$ , the above equation is satisfied. Find $a^2+b^2$ .

2 solutions

Rishabh Tiwari
Jun 19, 2016

$\large \begin{cases} a\sin^3 x +b\cos^3 x =\sin x \cos x \\ a \sin x= b\cos x \end{cases}$

We can write the first equation as :

$a\sin x \cdot sin^{2} x \ +\ b\cos^{3} x = \sin x \cos x$

$\Rightarrow$ $b\cos x \cdot \sin^2 x + b\cos^3 x = \sin x \cos x$

$\Rightarrow$ $b\cos x (\sin^2 x + \cos^2 x) = \sin x \cos x$

$\Rightarrow$ $b = \sin x$

Putting this value of $b$ in second equation gives us :

$a = \cos x$

Hence ,

$a^2 \ + \ b^2 \ = \color{#20A900}{\sin^2 x + \cos^2 x} = \color{#69047E}{\boxed{1}}$

Exactly same solution

Prince Loomba - 4 years, 11 months ago

Great ! :)

Rishabh Tiwari - 4 years, 11 months ago

@Akshat Sharda From where did you find this question?

Swapnil Das - 4 years, 9 months ago

It came in our coaching exam.

Akshat Sharda - 4 years, 9 months ago

Oh, it was given in our mathematics class work, and our teacher claimed that no one could solve it :P

Swapnil Das - 4 years, 9 months ago

@Swapnil Das – and its so easy haha

Prince Loomba - 4 years, 9 months ago

Margaret Yu
Jun 19, 2016

$\large \begin{cases} a\sin^3 x +b\cos^3 x =\sin x \cos x \\ a \sin x= b\cos x \end{cases}$

$b\cos x \times a\sin^2 x +b\cos x \times b\cos^2 x =\sin x \cos x$

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

$b\cos x =\sin x \cos x$

$\boxed{b =\sin x}$

$a\sin x \times a\sin^2 x +b\cos x \times b\cos^2 x =\sin x \cos x$

$a\sin x x \times a\sin^2 x +a\sin x \times b\cos^2 x =\sin x \cos x$

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

$a\sin x =\sin x \cos x$

$\boxed{a =\cos x}$

$a^{2}+b^{2}= \boxed{1}$