Question

If $\sin \left( {7\phi + {9^ \circ }} \right) = \cos 2\phi$, find a value of $\phi$.

Answer 1

We will find the value of $\phi$ by using the relation between sine and cosine. Here, we will convert the right hand side of the given equation into the sine function by using the suitable trigonometric identity. We will then equate the angles and solve it further to get the required value of $\phi$.

Complete step-by-step answer:
The relation that is given to us to find the value of $\phi$ is
$\sin \left( {7\phi + {9^ \circ }} \right) = \cos 2\phi$
First of all we will convert the cosine on the right hand side into sine by using the relation $\sin ({90^ \circ } - x) = \cos x$ where $x = 2\phi$. Therefore, we get
$\Rightarrow \sin \left( {7\phi + {9^ \circ }} \right) = \sin \left( {{{90}^ \circ } - 2\phi } \right)$
We will take the $\sin$ term from left to right and get the following equation
$\Rightarrow \left( {7\phi + {9^ \circ }} \right) = {\sin ^{ - 1}}\left( {\sin \left( {{{90}^ \circ } - 2\phi } \right)} \right)$
We know that ${\sin ^{ - 1}}\left( {\sin \left( x \right)} \right) = x$. Therefore, using the relation, we get
\Rightarrow 7\phi + {9^ \circ } = {90^ \circ } - 2\phi $$$$$$ Taking all the\phi $value on one side, we get$ \Rightarrow 7\phi + 2\phi = {90^ \circ } - {9^ \circ }$Adding and subtracting the like terms, we get$ \Rightarrow 9\phi = {81^ \circ }$Dividing both side by 9, we get$\begin{array}{l} \Rightarrow \phi = \dfrac{{{{81}^ \circ }}}{9}\\ \Rightarrow \phi = {9^ \circ }\end{array}$$

Note: We know that the “cosine” is the “sine of complement” which means that $\sin ({90^ \circ } - x) = \cos x$ or we can say “sine” is the “cosine of complement” which means $\sin x = \cos ({90^ \circ } - x)$. Here, we have also used the condition of inverse where ${\sin ^{ - 1}}\sin (x) = x$ to get our answer.
We can use another way to solve the question that is by changing sine value on the left side to cosine.
$\sin \left( {7\phi + {9^ \circ }} \right) = \cos 2\phi$
Using the relation $\sin \theta = \cos ({90^ \circ } - \theta )$, we get
$\Rightarrow \cos \left( {{{90}^ \circ } - \left( {7\phi + {9^ \circ }} \right)} \right) = \cos 2\phi$
Now taking cos on right hand side, we get
$\Rightarrow \left( {{{90}^ \circ } - 7\phi - {9^ \circ }} \right) = {\cos ^{ - 1}}\left( {\cos 2\phi } \right)$
We know that ${\cos ^{ - 1}}\left( {\cos (x)} \right) = x$. Therefore, using this relation in the above equation, we get
$\Rightarrow {81^ \circ } - 7\phi = 2\phi$
Taking all $\phi$ value on one side we get
$\Rightarrow 9\phi = {81^ \circ }$
Dividing both side by 9, we get
$\begin{array}{l} \Rightarrow \phi = \dfrac{{{{81}^ \circ }}}{9}\\\ \Rightarrow \phi = {9^ \circ }\end{array}$