Question

Prove that the theorem cos (x + y) = cos x cos y – sin x sin y.​

Answer 1

Solution :-

Let's take a unit circle with centre O

Unit circle :- A unit circle means a circle with radius one unit .

◉ Figure 1 :-

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Choosing point ( $\sf P_1$ ) in quadrant 2 . Assuming the angle made by radius $\sf OP_1$ with positive direction of x - axis as x°

Now , choose another point $\sf P_2$ in 3rd quadrant such that angle made by radius $\sf OP_2$ with radius $\sf OP_1$ as y°.

Let's assume another point $\sf P_3$ again in 3rd quadrant , the angle made by radius $\sf OP_3$ with positive direction of x-axis is y° but angle goes in clock - wise direction from positive direction of x - axis . So, it will be -y°

Assuming the point of intersection of unit circle with positive x - axis as $\sf P_4$

Now writing total information ,

$\sf \angle P_4OP_2=x+y \begin{cases}\sf\angle P_4OP_1=x\\\\ \sf\angle P_1OP_2=y \end{cases}\\\\\sf \angle P_4OP_3 =-y$

◕ Finding co - ordinates of each point :-

Co-ordinates of $\sf P_1$ would be , its x co-ordinate , radius of this circle times cosine of ∠x .

So , co-ordinates of $\sf P_1$ will be $\sf ( cosx,sinx)$

So , similarly for all the points

• $\sf P_2$⇒ [ cos(x+y),sin(x+y) ]
• $\sf P_3$⇒ [ cos(-y),sin(-y) ]
• $\sf P_4$⇒ ( 1 , 0 )

Let consider two triangles $\sf P_1OP_3 \& P_2OP_4$ from fig 1.

Now , drawing fig. 2

◉ Figure 2 :-

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Now by observing the given figure

$\sf \because Both \;have \;same \;radii\begin{cases}\sf OP_1 = OP_2 = 1\\\\\sf OP_3 = OP_4 = 1\end{cases}$

Here , we have an angle $\theta$ , we can see that $\sf\angle P_1 OP_3 = y+\theta = \angle P_2OP_4$

Now here sides & one angle of two triangles are equal . So , by SAS congruency , the both triangles are congruent .

• ∆$\sf P_1OP_3 \cong$∆$\sf P_2OP_4$

So here the corresponding sides are equal Now taking longest side of congruent triangles $\to \sf P_1P_3 = P_2P_4$

Now using distance formula$\displaystyle\to\sf \sqrt{[cosx-cos(-y)]^2+[sinx-sin(-y)]^2}=\sqrt{[cos(x+y)-1]^2+[sin(x+y)-0]^2}$

On squaring on both sides we have ,

The answer will be continued by @HelperToAll

Answer 2

Solution: Continuity of @ItzArchimedes' answer...

$\to$ ( cosx - cosy )² + ( sinx + siny )² = [ cos ( x + y ) - 1 ]² + [ sin( x + y ) ]²

$\to$ cos²x + cos²y - 2cosxcosy + sin²x + sin²y + 2sinxsiny = cos²(x + y) - 2cos(x + y) + 1 + sin²( x + y )

$\to$ 2 + [ - 2cosxcosy ] + 2sinxsiny = 2 - 2cos(x + y)

$\to$ cos ( x + y ) = cosxcosy - sinxsiny

Hence, proved.