Chapter EP, Problem 14.3.14EP

To determine

To prove: the equation is in the form of identity.

Answer to Problem 14.3.14EP

The both L.H.S AND R.H.S are equal or identity to each other.

Explanation of Solution

Given:

  $\cos \left({180}^{\text{0}}\text{-θ}\right)\text{=-cosθ}$ .

Concept used:

The exact value of the trigonometric expression will be defined from the unit circle.

Some unit circle doesn’t provide enough value of angle so the angle should be taken reference to angle in the unit circle by taking difference or sum to make the appropriate angle which has value in the unit circle.

Trigonometric formula:

  $\cos \left(\text{A-B}\right)\text{=cosAcosB-sinAsinB}$

Calculation:

  $\cos \left({180}^{\text{0}}\text{-θ}\right)\text{=-cosθ}$

Taking L.H.S:

  $\Rightarrow \cos \left({180}^{\text{0}}\text{-θ}\right)$ .

According to the Trigonometric formula:

  $\cos \left(\text{A-B}\right)\text{=cosAcosB-sinAsinB}$ .

  $\text{cos}\left({\text{180}}^{\text{0}}\text{-θ}\right)\text{=cos180cosθ-sin180sinθ}$ .

  $\Rightarrow \text{cos180cosθ-sin180sinθ}$ .

From the above unit circle the value of the angle can be deduced.

  $\Rightarrow \text{cos180cosθ-sin180sinθ}$ .

  $\Rightarrow \left(-1\right)\text{cosθ-}\left(0\right)\text{sinθ}$ .

$\text{L}\text{.H}\text{.S=-cosθ}\text{.}$ ………. $\left(\text{i}\right)$

According to the given:

$\text{R}\text{.H}\text{.S=-cosθ}\text{.}$ ……….. $\left(\text{ii}\right)$

By comparing the two equation.

  $\text{L}\text{.H}\text{.S=R}\text{.H}\text{.S}$ .

So, the both L.H.S AND R.H.S. are equal or identity to each other.