Question

Without using the trigonometric tables, prove that $\sin 37{}^\circ \cos 53{}^\circ +\cos 37{}^\circ \sin 53{}^\circ =1$

Answer 1

Hint: We can solve this question easily by using the formula $\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B$ and then we can see that if we substitute the value of A = 37 degree and B = 53 degree then we will get the same expression as given in the question. And hence we can prove the above statement.

Complete step-by-step solution -
Let’s start solving this question.
Let’s first look at the formula that we are going use,
$\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B$
Now we will substitute the value A = 37 degree and B = 53 degree in the above formula.
Therefore, after substituting we get,
$\sin \left( 37{}^\circ+53{}^\circ \right)$ = $\sin 37{}^\circ \cos 53{}^\circ +\cos 37{}^\circ \sin 53{}^\circ$
Now we can see that the RHS of the above equation is the same as given in the question.
Hence, we get
$\sin \left( 90{}^\circ \right)$ = $\sin 37{}^\circ \cos 53{}^\circ +\cos 37{}^\circ \sin 53{}^\circ$
Now we know that $\sin 90{}^\circ = 1$,
So, substituting the value in $\sin \left( 90{}^\circ \right)$ = $\sin 37{}^\circ \cos 53{}^\circ +\cos 37{}^\circ \sin 53{}^\circ$
we get,
$\sin 37{}^\circ \cos 53{}^\circ +\cos 37{}^\circ \sin 53{}^\circ =1$
Hence, we have proved $\sin 37{}^\circ \cos 53{}^\circ +\cos 37{}^\circ \sin 53{}^\circ =1$.

Note: We have just used the sin(A+B) formula and substituted the value, so just by looking at question one must be able to deduce what formula is going to be applicable here and then we just have to use that formula where needed without any mistakes. Formula that we have used must be remembered by the student. One can also convert sin into cos using $\sin \left( 90-x \right)=\cos x$ and then use ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ to prove the statement.