That Powered Up Quickly

Geometry Level 1

If θ \theta is an acute angle and tan θ + cot θ = 2 \tan\theta+\cot\theta=2 , find the value of tan 7 θ + cot 7 θ \tan^7\theta+\cot^7\theta .

0 1 2 Not defined

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.

2 solutions

Rishabh Jain
Jun 20, 2016

Relevant wiki: Trigonometric Equations - Problem Solving - Easy

tan θ + 1 tan θ = 2 ( cot θ = 1 tan θ ) \tan\theta+\dfrac1{\tan\theta}=2~~~\left(\small{\because\cot\theta=\frac1{\tan\theta}}\right)

( tan θ 1 ) 2 = 0 tan θ = 1 \implies (\tan\theta-1)^2=0\implies \tan\theta=1

tan 7 θ + cot 7 θ = 1 + 1 = 2 \tan^7\theta+\cot^7\theta=1+1=\large\color{#3D99F6}{\boxed 2}

Akshat Sharda
Jun 20, 2016

tan θ + cot θ = 2 sin 2 θ + cos 2 θ sin θ cos θ = 2 1 = 2 sin θ cos θ 1 = sin 2 θ θ = π 4 tan 7 π 4 + cot 7 π 4 = 2 \tan \theta+\cot \theta=2 \Rightarrow \frac{\sin^2 \theta+\cos^2 \theta}{\sin \theta \cos \theta}=2 \\ 1=2 \sin \theta \cos \theta \Rightarrow 1=\sin 2 \theta \Rightarrow \theta =\frac{\pi}{4} \\ \Rightarrow \tan^7 \frac{\pi}{4}+\cot ^7 \frac{\pi}{4}=\boxed{2}

Small note: θ = π 4 + k π \theta = \dfrac{\pi}{4}+k \pi where k k is an integer. However, since tan \tan and cot \cot have period π \pi , this doesn't raise any issues.

Michael Fuller - 4 years, 11 months ago

Log in to reply

Can you explain in a bit more?

vivek kushal - 4 years, 10 months ago

Log in to reply

Sure, the tan \tan and cot \cot graphs have period π \pi , which means that tan ( θ ) tan ( θ + k π ) \tan(\theta) \equiv \tan(\theta+k \pi) where k k is an integer. In other words, the tan \tan graph repeats itself after every 18 0 180^{\circ} .

Similarly, the sin \sin graph repeats itself every 2 π 2\pi , so 1 = sin 2 θ 2 θ = π 2 + 2 k π θ = π 4 + k π 1=\sin 2 \theta \Rightarrow 2 \theta = \dfrac{\pi}{2} \color{#D61F06}{+ 2k \pi} \Rightarrow \theta = \dfrac{\pi}{4} + k \pi as I mentioned.

I always find these types of questions easier when visualising a sketch. Think about the line y = 1 y=1 and y = sin x y=\sin x . An obvious solution is π 2 \dfrac{\pi}{2} , but there's another solution every ± 2 k π \pm 2k\pi .

Michael Fuller - 4 years, 10 months ago

Log in to reply

@Michael Fuller Got it...thanks Michael :)

vivek kushal - 4 years, 10 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...