Question

Value of cos inverse (cos 13pie/6)?

Answer 1

it would be -π/6 instead of π/6

Answer 2

Answer

$\sf{\cos^{-1} \left(\cos \left(\dfrac{13\pi }{6}\right)\right)}$

• We know cos⁻¹ can be written as arccos (Since Arccosine of y is defined as the inverse cosine of y. In simple Words inverse cosine function is defined as Arccosine)

$\to\sf{\displaystyle\arccos \left(\cos \left(\frac{13\pi }{6}\right)\right)}$

• Rewrite the angles for cos(13π/6)

$\to\sf{\displaystyle\arccos \left(\cos \left(\frac{12+1}{6}\pi\right)\right)}$

$\to\sf{\displaystyle\arccos \left(\cos \left( \left(\frac{12}{6}+\frac{1}{6}\right)\pi\right)\right)}$

$\to\sf{\displaystyle\arccos \left(\cos \left( \left(2+\frac{1}{6}\right)\pi\right)\right)}$

$\to\sf{\displaystyle\arccos \left(\cos \left(2\pi+\frac{1}{6}\pi\right)\right)}$

• Apply the Periodicity of cos : $\sf{\cos (x+2 \pi \cdot k)=\cos (x)}$

$\to\sf{\displaystyle\arccos \left(\cos \left(\frac{1}{6}\pi\right)\right)}$

$\to\sf{\displaystyle\arccos \left(\cos \left(\frac{\pi}{6}\right)\right)}$

• We know :

$\bigstar\:\:\:\sf{\mathsf{If}\:a=\arccos \left(b\right)\quad \Rightarrow \quad \cos \left(a\right)=b,\:\mathsf{for}\:0\le \:a\le \:\pi }$

• From this We get

$\bigstar\:\:\:\sf{\displaystyle x=\arccos \left(\cos \left(\frac{\pi }{6}\right)\right)\quad \Rightarrow \quad \cos \left(x\right)=\cos \left(\frac{\pi }{6}\right),\:0\le \:x\le \:\pi }$

Now solve cos(x) = cos(π/6)

• Using the following Trivial Identity : cos(π/6) = √3/2

$\to\sf{\displaystyle \cos \left(x\right)=\frac{\sqrt{3}}{2}}$

• General solutions for cos(x) = √3/2

(from cos(x) periodicity table with 2πn cycle) :

$\to\sf{\displaystyle x=\frac{\pi }{6}+2\pi n,\:x=\frac{11\pi }{6}+2\pi n}$

• We need solutions for the range 0 ≤ π ≤ 6

$\boxed{\boxed{\to\sf{x=\frac{\pi }{6}}}}$

Therefore,

$\large\boxed{\boxed{\sf{\cos^{-1} \left(\cos \left(\dfrac{13\pi }{6}\right)\right)=\dfrac{\pi }{6}\quad }}}$