JEE Determinants 3

Geometry Level 5

∣ 3 x 2 x 2 + x cos ⁡ θ + c o s 2 θ x 2 + x sin ⁡ θ + sin ⁡ 2 θ x 2 + x cos ⁡ θ + cos ⁡ 2 θ 3 cos ⁡ 2 θ 1 + sin ⁡ 2 θ 2 x 2 + x sin ⁡ θ + sin ⁡ 2 θ 1 + sin ⁡ 2 θ 2 3 sin ⁡ 2 θ ∣ = 0 \left| \begin{matrix} 3{ x }^{ 2 } & x^{ 2 }+x \cos\theta +{ cos }^{ 2 }\theta & x^{ 2 }+x \sin\theta +{ \sin }^{ 2 }\theta \\ x^{ 2 }+x \cos\theta +{ \cos }^{ 2 }\theta & 3 \cos^{ 2 }\theta & 1+\frac { \sin2\theta }{ 2 } \\ x^{ 2 }+x \sin\theta+{ \sin }^{ 2 }\theta & 1+\frac { \sin2\theta }{ 2 } & 3{ \sin }^{ 2 }\theta \end{matrix} \right| =0 are

Given the roots of the above determinant are f ( θ ) f(\theta) and g ( θ ) g(\theta) . Evaluate ( f ( π 29 ) ) 2 + ( g ( π 29 ) ) 2 (f(\frac{\pi}{29}))^{2}+(g(\frac{\pi}{29}))^{2}

You may use calculator for the calculations.

Do like and share

Also see Refer to dictionary


The answer is 1.00.

Tanishq Varshney by Tanishq Varshney , 23, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Ariel Gershon
Mar 16, 2015

If you let x = sin ⁡ θ x = \sin{\theta} , the determinant becomes: ∣ 3 sin ⁡ 2 θ 1 + sin ⁡ θ cos ⁡ θ 3 sin ⁡ 2 θ 1 + sin ⁡ θ cos ⁡ θ 3 cos ⁡ 2 θ 1 + sin ⁡ 2 θ 2 3 sin ⁡ 2 θ 1 + sin ⁡ 2 θ 2 3 sin ⁡ 2 θ ∣ \begin{vmatrix} 3\sin^2\theta & 1 + \sin\theta \cos\theta & 3\sin^2\theta \\ 1 + \sin\theta \cos\theta & 3\cos^2\theta & 1+\frac{\sin{2\theta}}{2} \\ 3\sin^2\theta & 1+\frac{\sin{2\theta}}{2} & 3\sin^2\theta \end{vmatrix}

Now since 1 + sin ⁡ θ cos ⁡ θ = 1 + sin ⁡ 2 θ 2 1 + \sin\theta \cos\theta = 1 + \frac{\sin{2\theta}}{2} , the top and bottom rows are identical. Hence the determinant is 0 0 when x = sin ⁡ θ x = \sin{\theta} .

Similarly, if you let x = cos ⁡ θ x = \cos{\theta} , the determinant becomes: ∣ 3 cos ⁡ 2 θ 3 cos ⁡ 2 θ 1 + sin ⁡ θ cos ⁡ θ 3 cos ⁡ 2 θ 3 cos ⁡ 2 θ 1 + sin ⁡ 2 θ 2 1 + sin ⁡ θ cos ⁡ θ 1 + sin ⁡ 2 θ 2 3 sin ⁡ 2 θ ∣ \begin{vmatrix} 3\cos^2\theta & 3\cos^2\theta & 1 + \sin\theta \cos\theta \\ 3\cos^2\theta & 3\cos^2\theta & 1+\frac{\sin{2\theta}}{2} \\ 1 + \sin\theta \cos\theta & 1+\frac{\sin{2\theta}}{2} & 3\sin^2\theta \end{vmatrix} In this case, the first 2 2 rows are identical, so once again the determinant is 0 0 .

Therefore, f f and g g must be sin ⁡ θ \sin\theta and cos ⁡ θ \cos\theta . Hence f 2 ( θ ) + g 2 ( θ ) = 1 f^2(\theta)+g^2(\theta) = \boxed{1} for all θ \theta .

13 Helpful 0 Interesting 0 Brilliant 0 Confused

What encouraged you to put x = sin ⁡ θ x= \sin \theta or x = cos ⁡ θ x= \cos \theta and not anything else?

Md Zuhair - 2 years, 11 months ago

can u generalise ur approach instead of directly putting values of x

Tanishq Varshney - 6 years, 2 months ago

Log in to reply

I just happened to notice that in the top and bottom rows we have 3 x 2 3x^2 , x 2 + x sin ⁡ θ + sin ⁡ 2 θ x^2 + x \sin \theta + \sin^2\theta and 3 sin ⁡ 2 θ 3\sin^2\theta which are all equal when x = sin ⁡ θ x = \sin \theta . So I decided to try it, and it worked. Similarly in the first 2 rows, we have 3 x 2 = x 2 + x cos ⁡ θ + cos ⁡ 2 θ = 3 cos ⁡ 2 θ 3x^2 = x^2 + x \cos \theta + \cos^2\theta = 3\cos^2\theta when x = cos ⁡ θ x = \cos \theta . I admit this is a guess-and-check approach, but sometimes that helps!

Ariel Gershon - 6 years, 2 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...