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

Prove that sin⁡θsin⁡(90∘−θ)−cos⁡θcos⁡(90∘−θ)=0\sin \theta \sin (90^\circ - \theta ) - \cos \theta \cos (90^\circ - \theta ) = 0.

Explanation

Solution

In this problem we are to prove sin⁡θsin⁡(90∘−θ)−cos⁡θcos⁡(90∘−θ)=0\sin \theta \sin (90^\circ - \theta ) - \cos \theta \cos (90^\circ - \theta ) = 0. In our left hand side we will use sin⁡(90∘−θ)=cos⁡θ\sin (90^\circ - \theta ) = \cos \theta and cos⁡(90∘−θ)=sin⁡θ\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⁡θsin⁡(90∘−θ)−cos⁡θcos⁡(90∘−θ)=0\sin \theta \sin (90^\circ - \theta ) - \cos \theta \cos (90^\circ - \theta ) = 0
We are given as L.H.S,
sin⁡θsin⁡(90∘−θ)−cos⁡θcos⁡(90∘−θ)\sin \theta \sin (90^\circ - \theta ) - \cos \theta \cos (90^\circ - \theta )
Now, as, sin⁡(90∘−θ)=cos⁡θ\sin (90^\circ - \theta ) = \cos \theta and cos⁡(90∘−θ)=sin⁡θ\cos (90^\circ - \theta ) = \sin \theta ,
So we get,
=sin⁡θcos⁡θ−cos⁡θsin⁡θ= \sin \theta \cos \theta - \cos \theta \sin \theta
Taking, sin⁡θ\sin \theta common from the term, we get,
=sin⁡θ(cos⁡θ−cos⁡θ)= \sin \theta (\cos \theta - \cos \theta )
On simplifying we get,
=sin⁡θ.0= \sin \theta .0
=0= 0
So, we have our answer as, sin⁡θsin⁡(90∘−θ)−cos⁡θcos⁡(90∘−θ)=0\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∘−θ)=g(θ)f(90^\circ - \theta ) = g(\theta )and g(90∘−θ)=f(θ)g(90^\circ - \theta ) = f(\theta ).