Question

prove geometrically cos(x+y)=cosx×cosy-sinx×siny and hence show that cos 2x=cos²x-sin²x​

Answer 1

SOLUTION

TO PROVE

1. Prove geometrically

$\displaystyle \sf{ \cos \:(x + y) =\cos x \: \cos y - \sin x \: \sin y }$

2. Hence prove that

$\sf{ \cos \:2x={\cos}^{2} x - { \sin}^{2} x }$

EVALUATION

PROVE TO QUESTION : 1

Figure : Figure is referred to the attachment

In the figure

∠AOB = x and ∠ BOC = y

So that ∠ AOC = x + y

Clearly

∠ TPR + ∠ PRT = 90°

∠ ORT + ∠ PRT = 90°

Thus we have ∠ TPR = ∠ ORT = x

Now

$\displaystyle \sf{ \cos \:(x + y) \: = \frac{OQ}{OP} }$

$\displaystyle \sf{ \implies \: \cos \:(x + y) \: = \frac{OS - QS}{OP} }$

$\displaystyle \sf{ \implies \: \cos \:(x + y) \: = \frac{OS - TR}{OP} }$

$\displaystyle \sf{ \implies \: \cos \:(x + y) \: = \frac{OS}{OP} - \frac{ TR}{OP} }$

$\displaystyle \sf{ \implies \: \cos \:(x + y) \: = \frac{OS}{OR} . \frac{OR}{OP} - \frac{ TR}{PR}. \frac{PR}{OP} }$

$\displaystyle \sf{ \implies \:\cos \:(x + y) =\cos x \: \cos y - \sin x \: \sin y }$

Hence proved

PROOF TO QUESTION : 2

We have proved that

$\sf{\cos \:(x + y) =\cos x \: \cos y - \sin x \: \sin y }$

Putting y = x we get

$\sf{\cos \:(x + x) =\cos x \: .\cos x - \sin x .\: \sin x }$

$\sf{ \implies \: \cos 2x={\cos }^{2} x - {\sin}^{2} x }$

Hence proved

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