
Find the value of \[{\sin ^{ - 1}}\left[ {\cos \left( {{{4095}^ \circ }} \right)} \right]\]
A.\[ - \dfrac{\pi }{3}\]
B.\[\dfrac{\pi }{6}\]
C.\[ - \dfrac{\pi }{4}\]
D.\[\dfrac{\pi }{4}\]
E.\[\dfrac{\pi }{2}\]
Answer
164.1k+ views
Hint: First express 4095 as \[11 \times 360 + 135\], then \[\left[ {\cos \left( {{{4095}^ \circ }} \right)} \right]\]changed to \[\cos \left( {{{135}^ \circ }} \right)\], then write \[\cos \left( {{{135}^ \circ }} \right)\]as \[\cos \left( {\dfrac{\pi }{2} + \dfrac{\pi }{4}} \right)\] and calculate to obtain the required result.
Formula Used:
\[{\sin ^{ - 1}}(\sin x) = x\]
\[\cos \left( {\dfrac{\pi }{2} + x} \right) = - \sin x\]
Complete step by step solution:
The given expression is \[{\sin ^{ - 1}}\left[ {\cos \left( {{{4095}^ \circ }} \right)} \right]\]
Now,
\[{\sin ^{ - 1}}\left[ {\cos \left( {11 \times {{360}^ \circ } + {{135}^ \circ }} \right)} \right]\]
=\[{\sin ^{ - 1}}\left[ {\cos {{135}^ \circ }} \right]\]
=\[{\sin ^{ - 1}}\left[ {\cos \left( {\dfrac{\pi }{2} + \dfrac{\pi }{4}} \right)} \right]\]
=\[{\sin ^{ - 1}}\left[ { - \sin \dfrac{\pi }{4}} \right]\]
=\[ - {\sin ^{ - 1}}\left[ {\sin \dfrac{\pi }{4}} \right]\]
=\[ - \dfrac{\pi }{4}\]
Hence the correct option is C.
Additional Information:
A function of an arc or angle, such as the sine, cosine, tangent, cotangent, secant, or cosecant, can be represented most simply in terms of the ratios of pairs of sides of a right-angled triangle.
The domain and range of the functions determine the characteristics of inverse trigonometric functions. A greater comprehension of this idea and the ability to solve difficulties both depend on certain aspects of inverse trigonometric functions. Recall that "Arc Functions" is another name for inverse trigonometric functions.
To solve any problem in algebra one must have to know the identities and properties of sine, cosine, and tangent functions mainly. As particularly in this problem we are converting cosine function to sine function with the help of the formula \[\cos \left( {\dfrac{\pi }{2} + x} \right) = - \sin x\]
, without using this formula we can not proceed further.
Note: Sometimes students get confused about the signs of the quadrants, for that we need to remember that in the first quadrant sign of all trigonometric functions are positive, in the second quadrant only sine and cosine functions are positive, in the third quadrant only tan and cot functions are positive and in the fourth quadrant the cosine and sec functions are positive.
Formula Used:
\[{\sin ^{ - 1}}(\sin x) = x\]
\[\cos \left( {\dfrac{\pi }{2} + x} \right) = - \sin x\]
Complete step by step solution:
The given expression is \[{\sin ^{ - 1}}\left[ {\cos \left( {{{4095}^ \circ }} \right)} \right]\]
Now,
\[{\sin ^{ - 1}}\left[ {\cos \left( {11 \times {{360}^ \circ } + {{135}^ \circ }} \right)} \right]\]
=\[{\sin ^{ - 1}}\left[ {\cos {{135}^ \circ }} \right]\]
=\[{\sin ^{ - 1}}\left[ {\cos \left( {\dfrac{\pi }{2} + \dfrac{\pi }{4}} \right)} \right]\]
=\[{\sin ^{ - 1}}\left[ { - \sin \dfrac{\pi }{4}} \right]\]
=\[ - {\sin ^{ - 1}}\left[ {\sin \dfrac{\pi }{4}} \right]\]
=\[ - \dfrac{\pi }{4}\]
Hence the correct option is C.
Additional Information:
A function of an arc or angle, such as the sine, cosine, tangent, cotangent, secant, or cosecant, can be represented most simply in terms of the ratios of pairs of sides of a right-angled triangle.
The domain and range of the functions determine the characteristics of inverse trigonometric functions. A greater comprehension of this idea and the ability to solve difficulties both depend on certain aspects of inverse trigonometric functions. Recall that "Arc Functions" is another name for inverse trigonometric functions.
To solve any problem in algebra one must have to know the identities and properties of sine, cosine, and tangent functions mainly. As particularly in this problem we are converting cosine function to sine function with the help of the formula \[\cos \left( {\dfrac{\pi }{2} + x} \right) = - \sin x\]
, without using this formula we can not proceed further.
Note: Sometimes students get confused about the signs of the quadrants, for that we need to remember that in the first quadrant sign of all trigonometric functions are positive, in the second quadrant only sine and cosine functions are positive, in the third quadrant only tan and cot functions are positive and in the fourth quadrant the cosine and sec functions are positive.
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