Tricky Trigo

Geometry Level 3

The general solution of $\cos^{50} x - \sin^{50} x = 1$ is:

Note : In the answer choices, $n$ is any integer.

n\pi

None of these

2n\pi + \dfrac{\pi}{2}

Doesn't exist

n\pi + \dfrac{\pi}{2}

2n\pi

1 solution

Akhil Bansal
Dec 3, 2015

$\cos^{50} x = 1 + \sin^{50} x$ Observe that L.H.S $\leq$ 1 and R.H.S $\geq$ 1
Hence, we must have sign of equality(for solution to exist) $1 + \sin^{50} x = 1 \Rightarrow \sin x = 0$ $\Rightarrow \boxed{ x = n\pi }$

Moderator note:

As pointed out by Sudeep, the last line is incorrect. In order for there to be a solution, we must have $\cos x = \pm 1$ and $\sin x = 0$ .

Also one must ensure that $\cos ^{50} x = 1$ for these values. Here it is does not create any problem. However had the power been 49 instead of 50, the answer would have been $2n \pi$ because $\cos ^{49} x = -1$ for $x = (2n+1)\pi$ . Thus it is important to create a solution for both the cases and take their intersection.

Sudeep Salgia - 5 years, 6 months ago

Right,and in this case if $cos^{50}x \neq 1$ when $\sin^{50} x = 0$ , then there will be no solution.
Hence,there is only one possible case.

Akhil Bansal - 5 years, 6 months ago

Nice observation, Sudeep. Can you post your version as a problem? Thanks!

Calvin Lin Staff - 5 years, 6 months ago

Tricky Trigo

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