Chapter 7.5, Problem 5AYU

To determine

Verify that $\cos \left(\alpha +\beta \right)=\cos \alpha \cos \beta \_\_\_\_\sin \alpha \sin \beta$ .

Answer to Problem 5AYU

  $\boxed{\cos \left(\alpha -\beta \right)=\cos \alpha \cos \beta -\sin \alpha \sin \beta }$

Explanation of Solution

Given information:

Verify that $\cos \left(\alpha +\beta \right)=\cos \alpha \cos \beta \_\_\_\_\sin \alpha \sin \beta$

Calculation:

Consider the given expression,

  $\cos \left(\alpha +\beta \right)=\cos \alpha \cos \beta \_\_\_\_\sin \alpha \sin \beta$

Rewrite the expression,

Let us consider point A on the circle X axis. Therefore the coordinates of point A will be $\left(1,0\right)$ .

Let us assume another point B on the circle. Point B lies in the first quadrant of circle,

Let $\angle BOA\text{}=\beta$

Let line segment BC $\perp$ OA intersects OA at point C

In right $△BOA\text{}$

  $\frac{OC}{OB\text{}}=\cos \beta$

  $OC\text{}=1\times \cos \beta \left[OB=radiusofcircle=1\left(Assumed\right)\right]$

  = $\cos \beta$

  $\begin{array}{l}\frac{BC}{OB}=\sin \beta \\ BC=1\times \sin \beta \left[OB=radiusofcircle=1\left(Assumed\right)\right]\\ =\sin \beta \end{array}$

Hence the coordinates of point B are $\left(\cos \beta ,\sin \beta \right)$ .

Let us assume another point D on the circle.Point D lies in the first quadrant of circle,

Let $\angle DOA=\alpha$

Let line segment DF $\perp OA$ intersects OA at point D

In right $△DOF$

  $\begin{array}{l}\frac{OF}{OD}=\cos \alpha \\ OF=1\times \cos \alpha \left[OD=radiusofcircle=1\left(Assumed\right)\right]\\ =\cos \alpha \end{array}$

  $\begin{array}{l}\frac{DF}{OD}=\sin \alpha \\ DF=1\times \sin \alpha \left[OD=radiusofcircle=1\left(Assumed\right)\right]\\ =\sin \alpha \end{array}$

Hence coordinates of point D are $\left(\cos \alpha ,\sin \alpha \right)$ .

We know that the length of segment with coordinates $\left(x_{1,}{y}_1\right)and\left(x_{2,}{y}_2\right)is$

  $\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2}$

Therefore length of line segment

BD $\begin{array}{l}=\sqrt{{\left(\cos \alpha -\cos \beta \right)}^2+{\left(\sin \alpha -\sin \beta \right)}^2}\\ =\sqrt{{\cos }^2\alpha +{\cos }^2\beta -2\cos \alpha \cos \beta +{\sin }^2\alpha +{\sin }^2\beta +2\sin \alpha \sin \beta }\\ =\sqrt{1+1-2\cos \alpha \cos \beta +2\sin \alpha \sin \beta }\\ =\sqrt{2-2\cos \alpha \cos \beta +2\sin \alpha \sin \beta }\end{array}$

Let line segment $BE\perp OD$ intersects OD at point E

In right $\Delta OEB$

  $\begin{array}{l}\frac{OE}{OB}=\cos \left(\alpha -\beta \right)\\ OE=1\times \cos \left(\alpha -\beta \right)\\ =\cos \left(\alpha -\beta \right)\\ \frac{BE}{OB}=\sin \left(\alpha -\beta \right)\\ BE=1\times \sin \left(\alpha -\beta \right)\\ =\sin \left(\alpha -\beta \right)\end{array}$

In right $\Delta BED$

ED = OD-OE

  $1-\cos \left(\alpha -\beta \right)$

Using Pythagoras theorem

  $B{D}^2=B{E}^2+E{D}^2$

  $\begin{array}{l}{\left(\sqrt{2-2\cos \alpha \cos \beta +2\sin \alpha }\sin \beta \right)}^2={\left(\sin \left(\alpha -\beta \right)\right)}^2+{\left(1-\cos \left(\alpha -\beta \right)\right)}^2\\ 2-2\cos \alpha \cos \beta +2\sin \alpha \sin \beta ={\sin }^2\left(\alpha -\beta \right)+1+{\cos }^2\left(\alpha -\beta \right)-2\cos \left(\alpha -\beta \right)\\ 2-2\cos \alpha \cos \beta +2\sin \alpha \sin \beta =1+1-2\cos \left(\alpha -\beta \right)\end{array}$

  $\begin{array}{l}-2\cos \left(\alpha -\beta \right)=2-2\cos \alpha \cos \beta +2\sin \alpha \sin \beta -2\\ -2\cos \left(\alpha -\beta \right)=-2\left(\cos \alpha \cos \beta -\sin \alpha \sin \beta \right)\end{array}$

  $\boxed{\cos \left(\alpha -\beta \right)=\cos \alpha \cos \beta -\sin \alpha \sin \beta }$ .

Hence, $\boxed{\cos \left(\alpha -\beta \right)=\cos \alpha \cos \beta -\sin \alpha \sin \beta }$ .