Verify that sin(A+B) + sin(A-B) = 2sinA sinB ?

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Verify that sin(A+B) + sin(A-B) = 2sinA sinB ?

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Answer 1 · 2018-04-14T19:45:42

$see explanation$ $"see explanation"$

Explanation:

$using the addition formulae for sin$ $"using the "color(blue)"addition formulae for sin"$

$∙ x sin (A \pm B) = sin A cos B \pm cos A sin B$ $•color(white)(x)sin(A+-B)=sinAcosB+-cosAsinB$

$\Rightarrow sin (A + B) = sin A cos B + cos A sin B$ $rArrsin(A+B)=sinAcosB+cosAsinB$

$\Rightarrow sin (A - B) = sin A cos B - cos A sin B$ $rArrsin(A-B)=sinAcosB-cosAsinB$

$\Rightarrow sin (A + B) + sin (A - B) = 2 sin A cos B$ $rArrsin(A+B)+sin(A-B)=2sinAcosB$

$\neq 2 sin A sin B \leftarrow check your question$ $!=2sinAsinBlarr"check your question"$

Answer 2 · 2018-04-14T19:50:34

It is not an identity.

Explanation:

It is not an identity.

$A = 90° , B = 0°$
LS: $sin(A+B) + sin(A-B) = sin (90°+0°) + sin ( 90°-0°) = 2$
RS: $2sinA sinB = 2 sin 90° sin 0° = 2 xx1xx0 = 0$
$2!=0$

$= 2sinA sinB$

$sin(A+B) + sin(A-B) = 2sinA sinB$

$LHS: sin(A+B) + sin(A-B)$

$sinAcosB + cosAsinB + sinAcosB - cosAsinB =$

$sinAcosB + sinAcosB = 2sinAcosB$

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Verify that sin(A+B) + sin(A-B) = 2sinA sinB ?

2 Answers

Explanation:

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