Question

Write the value of ${\cos ^{ - 1}}\left( {\cos 350^\circ } \right) - {\sin ^{ - 1}}\left( {\sin 350^\circ } \right)$ ?

Answer 1

Hint: We will change the given equation according to our ease, so that we can take the help of range and identities of trigonometric functions.We rewrite $350^\circ$ as $360^\circ - 10^\circ = 350^\circ$ and apply the trigonometric identities to solve this question.

Complete step-by-step answer:
It is given that,
${\cos ^{ - 1}}\left( {\cos 350^\circ } \right) - {\sin ^{ - 1}}\left( {\sin 350^\circ } \right)$
We will write the given equation as
$\Rightarrow {\cos ^{ - 1}}\left( {\cos \left( {360^\circ - 10^\circ } \right)} \right) - {\sin ^{ - 1}}\left( {\sin \left( {360^\circ - 10^\circ } \right)} \right)$, where $360^\circ - 10^\circ = 350^\circ$
As we know that, the value of $360^\circ = 2\pi$ in radians ,where $\pi = 180^\circ$.
Since, $2\pi$is also known as one revolution which means the value of $2\pi - \theta$ or $360^\circ - \theta$ will always lie in the 4th quadrant.
Since, in the 4th quadrant the values of $\cos \theta$ and $\sec \theta$ is positive only whereas the values of all other trigonometric functions are negative. Hence, it implies that the
$\cos \left( {360^\circ - \theta } \right) = \cos \theta$ and $\sin \left( {360^\circ - \theta } \right) = - \sin \theta$
Unlikely, $\dfrac{\pi }{2}$ or $\dfrac{{3\pi }}{2}$ the trigonometric function will not change while solving these identities i.e. for $\pi$ and $2\pi$ the sine will remain sine and the cosine will remain cosine.
Now, evaluating the equation by applying all the trigonometric identities mentioned above
$\Rightarrow {\cos ^{ - 1}}\left( {\cos 10^\circ } \right) - {\sin ^{ - 1}}\left( { - \sin 10^\circ } \right)$
Take negative sign out from the bracket we will get,
$\Rightarrow {\cos ^{ - 1}}\left( {\cos 10^\circ } \right) + {\sin ^{ - 1}}\left( {\sin 10^\circ } \right)$
Now, the sine as well as the cosine will cancel out with their inverses and we have left with only
$\Rightarrow 10^\circ + 10^\circ = 20^\circ$
Hence, by evaluating the given equation we are now able to find the value of the equation as $20^\circ$.
Hence, our answer is $20^\circ$.

Note: As you can see, this is the easiest and fastest approach to solve these kinds of questions, that’s why it is highly recommended to learn and retain all the trigonometric identities, their domain and their range. The identities to shift the angles which are used above are called cofunction or periodicity identities in degrees.