Find the values of \(\cos^{-1}(\cos\frac{7\pi}{6})\) is equal to
\(\frac{7\pi}{6}\)
\(\frac{5\pi}{6}\)
\(\frac{\pi}{3}\)
\(\frac{\pi}{6}\)
We know that \(\cos^{-1}(cos \,x)=x\) if x∈[0,π] , which is the principal value branch of cos−1x.
Here,7π/6∉x∈[0,π].
Now,cos−1 (cos7π/6)can be written as:
cos−1 (cos7π/6)=cos−1 (cos-7π/6)=cos−1[(cos(2π-7π/6)] [cos(2π+x)=cosx]
=cos−1(cos5π/6) where 5π/6∈[0,π]
therefore cos−1(cos7π/6)=cos−1(cos5π/6)=5π/6.
The correct answer is B.
The elementary properties of inverse trigonometric functions will help to solve problems. Here are a few important properties related to inverse trigonometric functions:
Tan−1x + Tan−1y = π + tan−1 (x+y/ 1-xy), if xy > 1
Tan−1x + Tan−1y = tan−1 (x+y/ 1-xy), if xy < 1
Tan−1x + Tan−1y = tan−1 (x+y/ 1-xy), if xy < 1
Tan−1x + Tan−1y = -π + tan−1 (x+y/ 1-xy), if xy > 1
= x, if x∈[−π/2, π/2]
= π−x, if x∈[π/2, 3π/2]
=−2π+x, if x∈[3π/2, 5π/2] And so on.
= −x, ∈[−π,0]
= x, ∈[0,π]
= 2π−x, ∈[π,2π]
=−2π+x, ∈[2π,3π]
= x, (−π/2, π/2)
= x−π, (π/2, 3π/2)
= x−2π, (3π/2, 5π/2)