Question

Q] prove that :-

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

Answer 1

Explanation:

Given Problem:-

Prove that

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

Proof:-

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

Now, LHS : Sin ( A + B ).

=>Sin (60° + 30° )

=>Sin 90°

=>1

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

RHS: sin A cos B + cos A sin BB

=>Sin60° Cos30° + Cos60° Sin 30°

=>(√3/2)×(√3/2) + (1/2)(1/2)

=>[(√3×√3)/(2×2)] + [ (1×1)/(2×2)]

=>(3/4)+(1/4)

=>(3+1)/4

=>4/4

=>1

RHS = 1 ----------------(2)

From (1) &(2)

LHS = RHS

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

Hence , Proved.

Used formulae:-

• Sin 30° = 1/2

• Cos 60° = 1/2

• Sin60° = √3/2

• Cos 30°= √3/2

Answer 2

To Prove :

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

Solution :

As by the Given diagram,

$\sf \angle CPN$ and $\sf \angle OPM$ are vertically opposite angles. Then $\sf \angle CPN = \angle OPM$

$\sf \angle CNP = \angle OMP = 90^{circ}$

That means the third angle $\sf \angle PCN = \angle POM$ will also be equal.

(In two triangle two if respective angles are equal then the third angle has to be equal.)

Sum of interior angles of the triangle is 180°

Hence , $\sf \angle CPN = \angle A$$\sf sin(A+B) = \dfrac{CM}{OC}=\dfrac{CR+RM}{OC}$

• The quadrilateral RNQM is rectangle because RN is originally upperpendicular at line CM that means $\sf \angle MRN$ is 90° then, $\sf \angle NRM$ is also 90° and $\sf \angle NQM$ and $\sf \angle RNQ$ are also 90° because sum of four angles is equal to 360° in a quadrilateral. It mean RM = NQ. So we can substitute NQ in place of RM
• $\sf \triangle CRN, \angle CRN$ is a right angle then, $\sf \dfrac{CR}{CN}$ will actually give us cos. Let's divide and multiply first expression by CN. Similarly if we see RM which is equal to NQ, $\sf \dfrac{NQ}{ON}$ we will get SinA

$\sf \dfrac {CR}{CN} \times \dfrac{CN}{OC}= \dfrac {NQ}{ON} \times \dfrac{ON}{OC}$

• In $\sf \triangle CRN ,\angle NCR$ is equal to $\sf \angle A$

$\sf \dfrac{CR}{CN} = cos A$ _______(1)

• In $\sf \triangle OLC ,$ CN is perpendicular and OC is hypotenuse for $\sf \angle B$

$\sf \dfrac{CN}{OC} = sinB$ ________(2)

• NQ and ON both are the part of $\triangle ONQ$ when NQ is perpendicular for $\sf \angle A$ and ON is the hypotenuse

$\sf \dfrac{NQ}{ON} = sinA$ ________(3)

• ON and OC these two lines are the part of $\triangle ONC$ and ON is the base for $\sf \angle B$ and OC is the hypotenuse.

$\sf \dfrac{ON}{OC} = cos B$ ________(4)

• Substitute (1),(2),(3) and (4) in the equation.

$\implies \sf \dfrac {CR}{CN} \times \dfrac{CN}{OC}= \dfrac {NQ}{ON} \times \dfrac{ON}{OC}$

$\implies \sf cosA.sinB + sinA.cosB$

$\implies \sf sinA.cosB+ cosA.sinB$

Hence proved that

$\boxed{\sf sin(A+B)= sinA.cosB+ cosA.sinB}$

Proof In calculations :

Let A = 45° and B = 45°

$\implies \sf sin(A+B)$

$\implies \sf sin(45^{\circ}+45^{\circ})$

$\implies \sf sin(90^{\circ}) \implies 1$ _____________(LHS)

$\sf \implies sinA.cosB+cosA.sinB$

$\sf \implies sin45^{\circ}.cos45^{\circ}+cos45^{\circ}.sin45^{\circ}$

$\implies \[\left(\dfrac{1}{\sqrt{2}}\right)\]\[\left(\dfrac{1}{\sqrt{2}}\right)\]+\[\left(\dfrac{1}{\sqrt{2}}\right)\]\[\left(\dfrac{1}{\sqrt{2}}\right)\]$

$\implies \[\left(\dfrac{1}{(\sqrt{2})(\sqrt{2})}\right)\]+\[\left(\dfrac{1}{(\sqrt{2})(\sqrt{2})}\right)\]$

$\implies \[\left(\dfrac{1}{2}\right)\]+\[\left(\dfrac{1}{2}\right)\]$

$\implies \[\left(\dfrac{\cancel{2}}{\cancel{2}}\right)\]\implies 1$ _______(RHS)

LHS = RHS, hence Proved