Using all trigonometric functions

Geometry Level 5

sin ( x ) + cos ( x ) + tan ( x ) + csc ( x ) + sec ( x ) + cot ( x ) = 7 \large \sin(x) + \cos(x) + \tan(x) + \csc(x) + \sec(x) + \cot(x) = 7

Suppose a real number x x satisfy the equation above.

And for positive constants a , b , c a,b,c , we have sin ( 2 x ) = a b c \sin(2x) = a - b \sqrt c with c c square-free. Find the value of a b + 2 c a-b+2c .


The answer is 28.

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2 solutions

Tanishq Varshney
May 30, 2015

Moderator note:

Yes, we just need to group them such that we can apply the substitution sin ( 2 A ) = 2 sin ( A ) cos ( A ) \sin(2A) = 2\sin(A) \cos(A) .

Bonus question: With the same given equation, can you write a least degree polynomial such that it has a root of tan ( x 2 ) \tan\left(\frac x2 \right) ?

To make your work neater, you can factor it nicely as:

7 sin ( 2 x ) 2 = ( sin x + cos x ) ( 2 + sin ( 2 x ) ) 7\sin(2x)-2=(\sin x+\cos x)(2+\sin(2x))

Now, if we consider A = sin ( 2 x ) A=\sin(2x) and square both sides of that equation we have, we get, after a bit of simplification,

A 3 44 A 2 + 36 A = 0 A = 0 , 22 ± 8 7 A^3-44A^2+36A=0\implies A=0,22\pm 8\sqrt7

We can easily rule out the two extraneous solutions A = 0 A=0 and A = 22 + 8 7 A=22+8\sqrt 7 using the reason you provided. We get the answer as A = 22 8 7 A=22-8\sqrt 7 .

Prasun Biswas - 6 years ago

I used the substitution sin ( x ) + cos ( x ) = A \sin(x) + \cos(x) = A

So, sin ( x ) cos ( x ) = A 2 1 2 \sin(x)\cdot\cos(x) = \frac{A^2 - 1}{2}

-> Find A

-> Evaluate A 2 1 A^2 - 1

Renjith Joshua - 6 years ago

Didn't think of solvable to sin 2 x which I suppose to realize as asked. Nevertheless, a general way is still workable. I think you made your own a solvable question. Nice to know this equation.

Lu Chee Ket - 5 years, 7 months ago
Lu Chee Ket
Nov 2, 2015

x = 1.07764686747605+

Sin 2 x = 0.83398951148327+ = 22 - 8 Sqrt (7)

22 - 8 + 2 (7) = 28

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