Question

If \[\sin 15^\circ = \cos (n \times 15^\circ )\], then find $n$.
A) $1$
B) $2$
C) $5$
D) $0$

Answer 1

Hint:
We can solve this question using the trigonometric relation between sine and cosine. We can convert sine of an angle to cosine and vice versa by replacing the angle by its complementary angle. And thus we can find the value of $n$.

Useful formula:
In trigonometry we have these results. For any value $\theta $,
$(i)\sin ({90^ \circ } - \theta ) = \cos \theta $
$(ii)\cos ({90^ \circ } - \theta ) = \sin \theta $
Here the angles $\theta $ and $90 - \theta $ are called complementary angles.

Complete step by step solution:
Given that \[\sin 15^\circ = \cos (n \times 15^\circ )\].
We are asked to find the value of $n$.
Here we can use this result.
$(ii)\cos ({90^ \circ } - \theta ) = \sin \theta $
Let $\theta = 15^\circ $
Substituting in the above result we have,
$\cos ({90^ \circ } - 15^\circ ) = \sin 15^\circ $
Simplifying the above equation we get,
$\cos 75^\circ = \sin 15^\circ $
In the question we have,
\[\sin 15^\circ = \cos (n \times 15^\circ )\]
Comparing these two equations we have,
$\cos (n \times 15^\circ ) = \cos 75^\circ $
This gives $n \times 15^\circ = 75^\circ $
Dividing both sides of the equation by $15$ we get,
$ \Rightarrow n = \dfrac{{75^\circ }}{{15^\circ }} = 5$

$\therefore $ We have $n = 5$ and the answer is option C.

Additional information:
Like the above used equations we have other trigonometric equations relating $\tan $ and $\cot $ as well as $\sec $ and $\cos ec$. They are,
$\tan ({90^ \circ } - \theta ) = \cot \theta $
$\sec ({90^ \circ } - \theta ) = \cos ec\theta $
Here $\cos ec,\sec $ and $\cot $ are the reciprocals of $\sin ,\cos $ and $\tan $ respectively.
And also $\tan $ is the ratio of sine and cosine functions.
These trigonometric functions can be related to angles and sides of a right angled triangle.

Note:
Here we have used the second equation, $(ii)\cos ({90^ \circ } - \theta ) = \sin \theta $.
Instead we can use the first equation as well.
$(i)\sin ({90^ \circ } - \theta ) = \cos \theta $
Here we have to let the value of $\theta $ be $75^\circ $.
Substituting in the above equation we get,
$ \Rightarrow \sin ({90^ \circ } - 75^\circ ) = \cos 75^\circ $
Simplifying the above equation we get,
$ \Rightarrow \sin 15^\circ = \cos 75^\circ $
So in either way we can solve this question. The only difference to be taken care of is the correct choice of $\theta $.