Is Pythagoras Identities sufficient?

Geometry Level 1

4 sin ⁡ 2 ( x ) + tan ⁡ 2 ( x ) + cot ⁡ 2 ( x ) + csc ⁡ 2 ( x ) = 6 \large 4\sin^2 (x) + \tan^2(x) + \cot^2(x) + \csc^2(x) = 6

Find the number of solutions of x x in the interval [ 0 , 2 π ] [0,2\pi] that satisfy the equation above.


The answer is 4.

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Mehul Chaturvedi by Mehul Chaturvedi , 22, India

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1 solution

Pi Han Goh
Jun 16, 2015

4 sin ⁡ 2 ( x ) + csc ⁡ 2 ( x ) + tan ⁡ 2 ( x ) + cot ⁡ 2 ( x ) = 6 ( 2 sin ⁡ ( x ) − 1 sin ⁡ ( x ) ) 2 + 4 + ( tan ⁡ ( x ) − 1 tan ⁡ ( x ) ) 2 + 2 = 6 ( 2 sin ⁡ ( x ) − 1 sin ⁡ ( x ) ) 2 + ( tan ⁡ ( x ) − 1 tan ⁡ ( x ) ) 2 = 0 \begin{aligned} 4\sin^2(x) + \csc^2(x) + \tan^2 (x) + \cot^2(x) &=& 6 \\ \left(2\sin(x) - \frac1{\sin(x)} \right)^2 + 4 + \left(\tan(x) - \frac1{\tan(x)} \right)^2 + 2&=& 6 \\ \left(2\sin(x) - \frac1{\sin(x)} \right)^2 + \left(\tan(x) - \frac1{\tan(x)} \right)^2 &=& 0 \\ \end{aligned}

Sum of squares of two real numbers equals to 0 implies that both these two numbers equals 0.

2 sin ⁡ ( x ) − 1 sin ⁡ ( x ) = 0 , tan ⁡ ( x ) − 1 tan ⁡ ( x ) = 0 ⇒ x = ± π 4 + π n 2\sin(x) - \frac1{\sin(x)} = 0, \tan(x) - \frac1{\tan(x)} = 0 \Rightarrow x = \pm \frac\pi4 + \pi n

for integer n n . But x x is restricted to only [ 0 , 2 π ] [0,2\pi] , so there are 4 \boxed4 solutions, namely x = π 4 , 3 π 4 , 4 π 4 , 7 π 4 x= \frac\pi4, \frac{3\pi}4, \frac{4\pi}4, \frac{7\pi}4 .

16 Helpful 1 Interesting 2 Brilliant 1 Confused

Moderator note:

This also shows that

4 sin ⁡ 2 ( x ) + csc ⁡ 2 ( x ) + tan ⁡ 2 ( x ) + cot ⁡ 2 ( x ) ≥ 6 4\sin^2(x) + \csc^2(x) + \tan^2 (x) + \cot^2(x) \geq 6

Finding the minimum of the more general case n sin ⁡ 2 ( x ) + csc ⁡ 2 ( x ) + tan ⁡ 2 ( x ) + cot ⁡ 2 ( x ) n\sin^2(x) + \csc^2(x) + \tan^2 (x) + \cot^2(x)

seems to be harder.

Awesome solution! Are you an IMO winner or a mathematician.

Siddharth Singh - 5 years, 12 months ago

Challenge Master: apply double angle formula, cos ⁡ ( 2 A ) = 2 cos ⁡ 2 ( A ) − 1 = 1 − 2 sin ⁡ 2 ( A ) \cos(2A) = 2\cos^2(A) - 1 = 1- 2\sin^2(A) and csc ⁡ 2 ( A ) = cot ⁡ 2 ( A ) − 1 \csc^2(A) = \cot^2(A) - 1 .

Let f n ( x ) f_n(x) denote the function above. Then, we have f n ( x ) = n 2 ( 1 − cos ⁡ ( 2 x ) ) + 4 1 − cos ⁡ ( 2 x ) + 2 ( 1 − cos ⁡ ( 2 x ) ) − 4 f_n(x) = \frac n2(1-\cos(2x)) + \frac4{1-\cos(2x)} + \frac2{(1-\cos(2x)) - 4} . Let z = 1 − cos ⁡ ( 2 x ) z = 1-\cos(2x) .

Apply chain rule, d y d x = d y d z × d z d x \frac {dy}{dx} = \frac{dy}{dz} \times \frac{dz}{dx} . And at extremal point(s), d y d x = 0 \frac{dy}{dx} = 0 .

We are left to solve the root(s) of z z in the interval [ 0 , 2 ] [0,2] for z 4 ( n ) + z 3 ( − 8 n ) + z 2 ( 16 n − 12 ) + z ( 64 ) − 128 = 0 z^4(n) + z^3(-8n) + z^2(16n-12) + z(64) - 128 = 0 . Noting that sin ⁡ ( 2 x ) ≠ 1 \sin(2x) \ne 1 else the denominator of one of the fractions of f n ( x ) f_n(x) is undefined.

With specific n n , we just need to determine whether each of these roots are minimum or maximum by the construction of the second derivative test table.

Pi Han Goh - 5 years, 12 months ago

Expressing everything in sin gives:

4 sin ⁡ 4 x − 12 sin ⁡ 3 x + 9 sin ⁡ 2 x − 2 = 0 4\sin^4 x -12\sin^3 x + 9\sin^2 x - 2 = 0

Let 4 k 3 − 12 k 2 + 9 k − 2 = 0 4k^3 -12k^2 + 9k - 2 = 0

( k − 2 ) 2 ( k − 1 2 ) = 0 (k -2)^2(k -\frac12) =0

As sin ⁡ x ≤ 1 \sin x ≤ 1 , sin ⁡ x = 1 2 . \sin x = \dfrac{1}{\sqrt2}.

Shubhrajit Sadhukhan - 4 months, 2 weeks ago

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