Question

How do you solve $\csc \left( {\dfrac{{19\pi }}{4}} \right) = \sqrt 2$ ?

Answer 1

$\;\cos 2\theta = 1 - 2{\sin ^2}\theta$ This is one of the basic trigonometric identities. Also $\csc x = \dfrac{1}{{\sin x}}$. In order to solve the given expression we have to use the above identities and express our given expression in that form and thereby solve it.

Complete step by step solution:
Given
$\csc \left( {\dfrac{{19\pi }}{4}} \right) = \sqrt 2 .........................\left( i \right)$
So here we have to prove LHS=RHS for the given expression.
Now we know
$\csc x = \dfrac{1}{{\sin x}}$
So we can write (i) as:

\right)}}......................\left( {ii} \right)$$ So to find the value of $$\csc \left( {\dfrac{{19\pi }}{4}} \right)$$ let’s first find the value of $$\sin \left( {\dfrac{{19\pi }}{4}} \right)$$ and take its reciprocal. So calculating the value of$$\sin \left( {\dfrac{{19\pi }}{4}} \right)$$. Now let’s assume $\sin \left( {\dfrac{{19\pi }}{4}} \right) = \sin a......................\left( {iii} \right)$ So similarly we can write $\;\cos a = \cos \left( {\dfrac{{19\pi }}{4}} \right)$ $ \Rightarrow \cos 2a = \cos \left( {\dfrac{{38\pi }}{4}} \right)$ We have to find the value of $\cos \left( {\dfrac{{38\pi }}{4}} \right)$such that by using the identity we can then solve the question. So finding the value of$\cos \left( {\dfrac{{38\pi }}{4}} \right)$: We know that $\cos \left( {\dfrac{{38\pi }}{4}} \right)$can be written as $ \cos \left( {\dfrac{{12\left( {3\pi } \right)}}{4} + \dfrac{{2\pi }}{4}} \right) = \cos \left( {3\left( {3\pi } \right) + \dfrac{{2\pi }}{4}} \right).................(iv) \\\ \\\ $ So from (iii) we know that $\cos \left( {3\left( {3\pi } \right) + \dfrac{{2\pi }}{4}} \right)$ would be in the IV Quadrant where cosine is negative. Such that: $$\cos \left( {3\left( {3\pi } \right) + \dfrac{{2\pi }}{4}} \right) = - \cos \left( {\dfrac{{2\pi }}{4}} \right)..................(v)$$ Also we know $$ - \cos \left( {\dfrac{{2\pi }}{4}} \right) = - \cos \left( {\dfrac{\pi }{2}} \right) = 0....................(vi)$$ Now by using the identity $\;\cos 2\theta = 1 - 2{\sin ^2}\theta $ we get $ \Rightarrow \cos 2a = 1 - 2{\sin ^2}a = 0 \\\ \Rightarrow 2{\sin ^2}a = 1 - 0 = 1 \\\ \Rightarrow {\sin ^2}a = \dfrac{1}{2} \\\ \Rightarrow \sin a = \dfrac{1}{{\sqrt 2 }} \\\ $ Now from (ii) $$\sin \left( {\dfrac{{19\pi }}{4}} \right) = \sin a$$ $ \Rightarrow \sin \left( {\dfrac{{19\pi }}{4}} \right) = \dfrac{1}{{\sqrt 2 }}$ Now we have from (ii) $$\csc \left( {\dfrac{{19\pi }}{4}} \right) = \dfrac{1}{{\sin \left( {\dfrac{{19\pi }}{4}} \right)}}$$ $$ \Rightarrow \csc \left( {\dfrac{{19\pi }}{4}} \right) = \dfrac{1}{{\sin \left( {\dfrac{{19\pi }}{4}} \right)}} = \dfrac{1}{{\left( {\dfrac{1}{{\sqrt 2 }}} \right)}} = \sqrt 2 ....................\left( {vii} \right)$$ So hence from (vii) we can say that LHS=RHS. **Therefore the above given steps are used to solve$$\csc \left( {\dfrac{{19\pi }}{4}} \right) = \sqrt 2 $$.** **Note:** Some other equations needed for solving these types of problem are:

1 + {\tan ^2}x = {\sec ^2}x \\
\begin{array}{*{20}{l}}
{\sin \left( {2x} \right) = 2\sin \left( x \right)\cos \left( x \right)} \\
{\cos \left( {2x} \right) = {{\cos }^2}\left( x \right)-{{\sin }^2}\left( x \right) = 1-2{\text{ }}{{\sin
}^2}\left( x \right) = 2{\text{ }}{{\cos }^2}\left( x \right)-1}
\end{array} \\

$$Also while approaching a trigonometric problem one should keep in mind that one should work with one side at a time and manipulate it to the other side. The most straightforward way to do this is to simplify one side to the other directly, but we can also transform both sides to a common expression if we see no direct way to connect the two.$$