Question

If \[\alpha = 3{\sin ^{ - 1}}\dfrac{6}{{11}}\] and \[\beta = 3{\cos ^{ - 1}}\dfrac{4}{9}\] where the inverse trigonometric functions takes only principle values, then the correct option(s) is/are
A) \[\cos \beta > 0\]
B) \[\sin \beta < 0\]
C) \[\cos (\alpha + \beta ) > 0\]
D) \[\cos \alpha < 0\]

Answer 1

Hint:
Firstly calculate for \[\alpha \] and then for \[\beta \] compare the given values with the approximate around known values and from that, we will have a rough idea about the values of \[\alpha \] and \[\beta \]. From there we can mark the options which are correct using the various approximations.

Complete step by step solution:
As the given values are \[\alpha = 3{\sin ^{ - 1}}\dfrac{6}{{11}}\]and \[\beta = 3{\cos ^{ - 1}}\dfrac{4}{9}\],
First checking for \[{\text{}\alpha }\], compare it with the approximate similar value and then get an idea for ranging of \[{\text{}\alpha }\]as,
As we know that,
\[\dfrac{6}{{11}} > \dfrac{6}{{12}}\]
Now taking inverse on both side of the above term
So, as
 \[x > y\] while taking inverse or taking sin both side we get
 \[{\sin ^{ - 1}}x > {\sin ^{ - 1}}y\] or \[\sin x > \sin y\]
Similarly,
\[ \Rightarrow \] \[{\sin ^{ - 1}}\dfrac{6}{{11}} > {\sin ^{ - 1}}\dfrac{6}{{12}}\]
On multiplication of \[3\]on both side so we get
\[ \Rightarrow \] \[3{\sin ^{ - 1}}\dfrac{6}{{11}} > 3{\sin ^{ - 1}}\dfrac{6}{{12}}\]
Hence, from the above given information we can say that
\[ \Rightarrow \] \[\alpha = 3{\sin ^{ - 1}}\dfrac{6}{{11}} > 3{\sin ^{ - 1}}\dfrac{6}{{12}}\]
Now as \[3{\sin ^{ - 1}}\dfrac{6}{{12}} = 3{\sin ^{ - 1}}\dfrac{1}{2}\],
Now substitute the inverse trigonometric value \[{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{2}}} = \dfrac{\pi }{6}\] in the above equation so,
 \[\because 3{\sin ^{ - 1}}\dfrac{1}{2} = 3(\dfrac{\pi }{6}) = \dfrac{\pi }{2}\]
So, we get,
 \[\alpha {\text{ = 3si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{6}}}{{{\text{11}}}}{\text{ > }}\dfrac{\pi }{2}\]
 \[ \Rightarrow \alpha > \dfrac{\pi }{2}\]
Now, doing the same above comparison for \[\beta \] as,
Similarly, for cos functions,
 \[x > y\]
Taking cos on both sides we get
 \[\cos x < \cos y\]
So, from the given we know that,
 \[\dfrac{4}{8} > \dfrac{4}{9}\]
On taking cosecant inverse on both side we get
\[ \Rightarrow \] \[{\cos ^{ - 1}}\dfrac{4}{9} > {\cos ^{ - 1}}\dfrac{4}{8}\]
And On multiplication of \[3\]on both side so we get
\[ \Rightarrow \] \[3{\cos ^{ - 1}}\dfrac{4}{9} > 3{\cos ^{ - 1}}\dfrac{4}{8}\]
Now, proceeding further
\[ \Rightarrow \] \[\beta {\text{ = 3co}}{{\text{s}}^{{\text{ - 1}}}}\dfrac{{\text{4}}}{{\text{9}}}{\text{ > 3co}}{{\text{s}}^{{\text{ - 1}}}}\dfrac{{\text{4}}}{8}\]
Now as, \[{\text{3co}}{{\text{s}}^{{\text{ - 1}}}}\dfrac{{\text{4}}}{8}{\text{ > 3co}}{{\text{s}}^{{\text{ - 1}}}}\dfrac{1}{2}\]
Use the inverse trigonometric ratios, \[{\text{co}}{{\text{s}}^{{\text{ - 1}}}}\dfrac{1}{2} = \dfrac{\pi }{3}\] , we get,
 \[\because {\text{3co}}{{\text{s}}^{{\text{ - 1}}}}\dfrac{{\text{1}}}{{\text{2}}}{\text{ = 3}}(\dfrac{\pi }{3}) = \pi \]
So, we get,
 \[
  \beta {\text{ = 3co}}{{\text{s}}^{{\text{ - 1}}}}\dfrac{{\text{4}}}{{\text{9}}}{\text{} > \pi } \\
   \Rightarrow {\text{}\beta > \pi } \\
 \]
Hence now we have an idea about the range of both \[\alpha \] and \[\beta \].
So now we just need to check options as,
\[\cos \beta > 0\] as we know that the value of \[\beta \] ranges as, \[{\text{}\beta > \pi }\]
Hence, \[\cos \beta > 0\] is false. So, option (A) is incorrect.
For \[\beta > \pi \] the value of \[\sin \beta < 0\], hence option (B) is correct.
As \[\alpha + \beta > \pi + \dfrac{\pi }{2} > \dfrac{{3\pi }}{2}\], so \[\cos (\alpha + \beta ) > 0\] is also correct. So, option (C) is also correct.
As \[\alpha > \dfrac{\pi }{2}\], the value of \[\cos \alpha < 0\] , So, option (D) is also correct.

Hence, option (B, C, D) are correct.

Note:
Remember the general domain and range of the following trigonometric and inverse trigonometric functions.
Drawing the graph of the inverse trigonometric function will be more well and good in order to determine the range of various values.
Inverse trigonometric functions are simply defined as the inverse functions of the basic trigonometric functions which are sine, cosine, tangent, cotangent, secant, and cosecant functions.