How do you verify $sin(x + pi/6) - cos(x + pi/3) = sqrt3 sin x$ ?

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How do you verify $sin(x + pi/6) - cos(x + pi/3) = sqrt3 sin x$ ?

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Answer 1 · 2018-04-08T10:44:09

It is given that the LHS is $sin(x + pi/6) - cos(x + pi/3)$

Applying, $color(magenta)(sin(A+B) = sinAcosB + cosAsinB$ and $color(red)(cos(A+B) = cosAcosB - sinAsinB$

$color(white)(dd$

$=> color(magenta)(sinxcos (pi/6) +cosxsin(pi/6))- (color(red)(cosxcos(pi/3) -sinxsin(pi/3)))$

$color(white)(dd$

$=> sinx(sqrt3/2) +cosx(1/2)- cosx(1/2) +sinx(sqrt3/2)$

$color(white)(dd$

$=>cancel 2xxsinx(sqrt3/cancel2)$

$=>sqrt3 sin x$

Answer 2 · 2018-04-08T10:55:19

See below.

Explanation:

Identities:

$color(red)bb(sin(A+B)=sinAcosB+cosAsinB)$

$color(red)bb(cos(A+B)=cosAcosB-sinAsinB)$

$LHS:$

$sin(x+pi/6)=sin(x)cos(pi/6)+cos(x)sin(pi/6)$

$cos(x+pi/3)=cos(x)cos(pi/3)-sin(x)sin(pi/3)$

$sin(x+pi/6)-cos(x+pi/3)$

$sin(x)cos(pi/6)+cos(x)sin(pi/6)-[cos(x)cos(pi/3)-sin(x)sin(pi/3)]$

$sin(x)cos(pi/6)+cos(x)sin(pi/6)-cos(x)cos(pi/3)+sin(x)sin(pi/3)$

$cos(pi/6)=sqrt(3)/2$
$cos(pi/3)=1/2$
$sin(pi/6)=1/2$
$sin(pi/3)=sqrt(3)/2$

$sin(x)sqrt(3)/2+cos(x)(1/2)-cos(x)(1/2)+sin(x)(sqrt3)/2$

$2sin(x)sqrt(3)/2$

$sqrt(3)sin(x)$

$LHS-=RHS$

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How do you verify $sin(x + pi/6) - cos(x + pi/3) = sqrt3 sin x$ ?

2 Answers

Explanation:

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