Question

Q} Prove that :-

cos (A + B) = cos A cos B - sin A sin B​

Answer 1

Explanation:

Given Problem :-

Prove that :

cos (A + B) = cos A cos B - sin A sin B

Solution:-

To prove this , we take A = 60° and B = 30°

then

LHS: Cos (A + B)

=>Cos ( 60° + 30°)

=>Cos 90°

=>0

LHS = 0 ---------------(1)

RHS:cos A cos B - sin A sin B

=> Cos 60° Cos 30° - Sin 60° Sin 30°

=>[(1/2)×(√3/2)] - [(√3/2)×(1/2)]

=>[(1×√3)/(2×2)] - [(√3×1)/(2×2)]

=>(√3/4) - (√3/4)

=>(√3-√3)/4

=>0/4

=>0

RHS = 0 --------------------(2)

From (1)&(2)

LHS = RHS

cos (A + B) = cos A cos B - sin A sin B

Hence, Proved

Used formulae:-

• Sin 30° = 1/2

• Sin 60° = √3/2

• Cos 30° = √3/2

• Cos 60° = 1/2

Answer 2

${\boxed{\underline{\tt{ \orange{Required \: answer:-}}}}}$

${\boxed{\underline{\bf{\red{ L.H.S}}}}}$

LET α = A and β = B

In ∆ OPG:-

$: \implies \sf{Cos (\alpha + \beta) = \frac{OG}{OP} = \frac{OF - HE }{ OP} }$

FG = HE

$: \implies \sf{Cos (\alpha + \beta) = \frac{OF }{ OP} - \frac{HE }{ OP} }$

In ∆ OFE and ∆ OEP:-

$:\implies\sf{ \frac{OF}{OE} × \frac{OE }{ OP} - \frac{HE}{PE} × \frac{PE}{OP} }$

Cos (A) . Cos (B) - Sin (A) . Sin(B)

${\boxed{\underline{\bf{\red{R .H.S}}}}}$

✿USED:-

• Cos (a) = base/hypotenuse