Question

Prove that \[\sin \theta \sin (90^\circ - \theta ) - \cos \theta \cos (90^\circ - \theta ) = 0\].

Answer 1

Hint: In this problem we are to prove \[\sin \theta \sin (90^\circ - \theta ) - \cos \theta \cos (90^\circ - \theta ) = 0\]. In our left hand side we will use \[\sin (90^\circ - \theta ) = \cos \theta \] and \[\cos (90^\circ - \theta ) = \sin \theta \] formulas to get ahead with the problem. Then we get both terms are equal and subtracting them we will give the required result.

Complete step-by-step answer:
We have to prove, \[\sin \theta \sin (90^\circ - \theta ) - \cos \theta \cos (90^\circ - \theta ) = 0\]
We are given as L.H.S,
\[\sin \theta \sin (90^\circ - \theta ) - \cos \theta \cos (90^\circ - \theta )\]
Now, as, \[\sin (90^\circ - \theta ) = \cos \theta \] and \[\cos (90^\circ - \theta ) = \sin \theta \],
So we get,
\[ = \sin \theta \cos \theta - \cos \theta \sin \theta \]
Taking, \[\sin \theta \] common from the term, we get,
\[ = \sin \theta (\cos \theta - \cos \theta )\]
On simplifying we get,
\[ = \sin \theta .0\]
\[ = 0\]
So, we have our answer as, \[\sin \theta \sin (90^\circ - \theta ) - \cos \theta \cos (90^\circ - \theta ) = 0\].
Hence, proved.

Note: You may have noticed that the words sine and cosine sound similar. That's because they're cofunctions! The way cofunctions work is exactly what you saw above. In general, if f and g are cofunctions, then,
\[f(90^\circ - \theta ) = g(\theta )\]and \[g(90^\circ - \theta ) = f(\theta )\].