Question

givenA=B=45 show that sin(A-B)=sin A .cos B-cos A.sinB​

Answer 1

Step-by-step explanation:

L.H.S=

$\sin(A - B) = \sin(45 - 45) = \sin(0 )$

$\sin(0) = 0$

R.H.S=

$\sin(A) . \cos(B) - \sin(B) . \cos(A)$

$\sin(45) . \cos(45) - \sin(45) . \cos(45)$

$\frac{1}{ \sqrt{2} } . \frac{1}{ \sqrt{2} } - \frac{1}{ \sqrt{2} } . \frac{1}{ \sqrt{2} } = 0$

therefore L.S.H=R.H.S

Answer 2

Answer:

Given,

A = B = 45°

Now,

sin(A + B) = sinA cosB + cosA sinB

Substituting A = 45 , B = 45 °

sin( 45 + 45 ) = sin45 cos45 + cos45 sin45

sin90 = sin45 cos45 + cos45 sin45

We know, Trigonometry ratios of particular angles : sin90 = 1 , cos45 = 1/√2 , sin45 = 1/√2

1 = 1/√2 ( 1/√2 ) + 1/√2 ( 1/√2)

1 = 1/2 + 1/2

1 = 1 .

Both Sides of the equation are equal.

Hence, We proved and verified that sin (A+B)= sin A cos B + cos A sin B holds good for A = B = 45°