Q 131 - Precalculus Enhanced with Graphing Utilities

Question

Establish the identity: 

${\left(\sin \theta \cos \varphi \right)}^2+{\left(\sin \theta \sin \varphi \right)}^2+{\cos }^2\theta =1$

Step-by-Step Solution

Answer

By using the identity ${\sin }^2\theta +{\cos }^2\theta =1$, we proved that :-

${\left(\sin \theta \cos \varphi \right)}^2+{\left(\sin \theta \sin \varphi \right)}^2+{\cos }^2\theta =1$.

Step 1. Given Information

We have to establish the following identity :-

${\left(\sin \theta \cos \varphi \right)}^2+{\left(\sin \theta \sin \varphi \right)}^2+{\cos }^2\theta =1$.

We will use the identity ${\sin }^2\theta +{\cos }^2\theta =1$, to prove the required identity.

Step 2. To prove the identity

We have to establish the identity :-

${\left(\sin \theta \cos \varphi \right)}^2+{\left(\sin \theta \sin \varphi \right)}^2+{\cos }^2\theta =1$.

Take left hand side :-

${\left(\sin \theta \cos \varphi \right)}^2+{\left(\sin \theta \sin \varphi \right)}^2+{\cos }^2\theta$

Open the brackets :-

${\left(\sin \theta \cos \varphi \right)}^2+{\left(\sin \theta \sin \varphi \right)}^2+{\cos }^2\theta ={\sin }^2\theta {\cos }^2\varphi +{\sin }^2\theta {\sin }^2\varphi +{\cos }^2\theta$

Take ${\sin }^2\theta$ as common, then we have :-

${\left(\sin \theta \cos \varphi \right)}^2+{\left(\sin \theta \sin \varphi \right)}^2+{\cos }^2\theta ={\sin }^2\theta ({\cos }^2\varphi +{\sin }^2\varphi )+{\cos }^2\theta$

Put ${\cos }^2\varphi +{\sin }^2\varphi =1$ by using the identity, then we have :-

$$\begin{gathered} {\left(\sin \theta \cos \varphi \right)}^2+{\left(\sin \theta \sin \varphi \right)}^2+{\cos }^2\theta ={\sin }^2\theta \left(1\right)+{\cos }^2\theta \\ \Rightarrow {\left(\sin \theta \cos \varphi \right)}^2+{\left(\sin \theta \sin \varphi \right)}^2+{\cos }^2\theta ={\sin }^2\theta +{\cos }^2\theta \end{gathered}$$

Use the above identity again, then we have :-

${\left(\sin \theta \cos \varphi \right)}^2+{\left(\sin \theta \sin \varphi \right)}^2+{\cos }^2\theta =1$.

Hence proved.