CBSE 12th Standard Maths Subject Inverse Trigonometric Functions HOT Questions Fill Ups Questions With Solution 2021
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CBSE 12th Standard Maths Subject Inverse Trigonometric Functions HOT Questions Fill Ups Questions With Solution 2021
12th Standard CBSE
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Reg.No. :
Maths
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Find the principal values of the following: \({ cos }^{ -1 }\left( \frac { \sqrt { 3 } }{ 2 } \right) \)
(a)\(\text { Let } \cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)=y, \text { Then } \cos y=\frac{\sqrt{3}}{2}=\cos \left(\frac{\pi}{6}\right)\)
We know that the range of the principal value branch of cos−1 is
\([0, \pi] \text { and } \cos \left(\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}\)
Therefore the priciple value of \(\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right) \text { is } \frac{\pi}{6}\) -
Find the value of the following: \({ cos }^{ -1 }\left( cos\frac { 13\pi }{ 6 } \right) \)
(a)\(\pi\)/6
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CBSE 12th Standard Maths Subject Inverse Trigonometric Functions HOT Questions Fill Ups Questions With Solution 2021 Answer Keys
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Let \(\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)=\theta \Rightarrow \cos \theta=\frac{\sqrt{3}}{2}\)
We know that the principal value branch of cos-1 is \( [0, \pi] \)
\(\therefore \quad \cos \theta=\frac{\sqrt{3}}{2}=\cos \frac{\pi}{6} \Rightarrow \theta=\frac{\pi}{6} \)
\(\Rightarrow \cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi}{6} \in[0, \pi] \)
Hence, the principal value of \(\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right) \text { is } \frac{\pi}{6}\)Let \(\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)=\theta \Rightarrow \cos \theta=\frac{\sqrt{3}}{2}\)
We know that the principal value branch of cos-1 is \( [0, \pi] \)
\(\therefore \quad \cos \theta=\frac{\sqrt{3}}{2}=\cos \frac{\pi}{6} \Rightarrow \theta=\frac{\pi}{6} \)
\(\Rightarrow \cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi}{6} \in[0, \pi] \)
Hence, the principal value of \(\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right) \text { is } \frac{\pi}{6}\) -
\( \cos ^{-1}(\cos x)=x \text { if } x \in[0, \pi], \) which is the principal value branch of cos-1x
\(\text {Here, } \frac{13 \pi}{6} \notin[0, \pi]\)
\(\text {Now, }\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right) \)
\(\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)=\cos ^{-1}\left[\cos \left(2 \pi+\frac{\pi}{6}\right)\right]=\cos ^{-1}\left[\cos \left(\frac{\pi}{6}\right)\right], \text { where } \frac{\pi}{6} \in[0, \pi]\)
\(\therefore \cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)=\cos ^{-1}\left[\cos \left(\frac{\pi}{6}\right)\right]=\frac{\pi}{6}\)\( \cos ^{-1}(\cos x)=x \text { if } x \in[0, \pi], \) which is the principal value branch of cos-1x
\(\text {Here, } \frac{13 \pi}{6} \notin[0, \pi]\)
\(\text {Now, }\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right) \)
\(\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)=\cos ^{-1}\left[\cos \left(2 \pi+\frac{\pi}{6}\right)\right]=\cos ^{-1}\left[\cos \left(\frac{\pi}{6}\right)\right], \text { where } \frac{\pi}{6} \in[0, \pi]\)
\(\therefore \cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)=\cos ^{-1}\left[\cos \left(\frac{\pi}{6}\right)\right]=\frac{\pi}{6}\)
