Prove that (a) $sin (x - \frac{3 π}{2}) = cos x$ $sin(x-(3pi)/2)=cosx$ and (b) $sin (θ - \frac{π}{3}) + cos (θ - \frac{π}{6}) = sin θ$ $sin(theta-pi/3)+cos(theta-pi/6)=sintheta$ ?

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Prove that (a) $sin (x - \frac{3 π}{2}) = cos x$ $sin(x-(3pi)/2)=cosx$ and (b) $sin (θ - \frac{π}{3}) + cos (θ - \frac{π}{6}) = sin θ$ $sin(theta-pi/3)+cos(theta-pi/6)=sintheta$ ?

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Answer 1 · 2018-02-07T07:12:45

Verified the above relations below.

Explanation:

(a) LHS
= $sin (x - \frac{3 π}{2})$ $sin(x-(3pi)/2)$

= $sin x cos (\frac{3 π}{2}) - cos x sin (\frac{3 π}{2})$ $sinxcos((3pi)/2)-cosxsin((3pi)/2)$

As $cos (\frac{3 π}{2}) = 0$ $cos((3pi)/2)=0$ and $sin (\frac{3 π}{2}) = - 1$ $sin((3pi)/2)=-1$

Substituting we have

= $sin x \times 0 - cos x \times (- 1)$ $sinx xx0-cosx xx(-1)$

= $0 + cos x$ $0+cosx$

= $cos x$ $cosx$ =RHS

(b) $L H S = sin (θ - \frac{π}{3}) + cos (θ - \frac{π}{6})$ $LHS=sin(theta-pi/3)+cos(theta-pi/6)$

Expanding using addition formula

$sin (A - B) = sin A cos B - cos A sin B$ $sin (A-B)=sinA cosB - cosA sinB$ and
$cos (A - B) = cos A cos B + sin A sin B$ $cos (A-B)=cosA cosB + sinA SinB$

LHS
= $sin θ cos (\frac{π}{3}) - cos θ sin (\frac{π}{3}) + cos θ cos (\frac{π}{6}) + sin θ sin (\frac{π}{6})$ $sinthetacos(pi/3)-costhetasin(pi/3)+costhetacos(pi/6)+sinthetasin(pi/6)$

As $cos (\frac{π}{6}) = sin (\frac{π}{3}) = \frac{\sqrt{3}}{2}$ $cos(pi/6)=sin(pi/3)=sqrt3/2$ and $cos (\frac{π}{3}) = sin (\frac{π}{6}) = \frac{1}{2}$ $cos(pi/3)=sin (pi/6)=1/2$

Therefore LHS= $sin θ \times \frac{1}{2} - cos θ \times \frac{\sqrt{3}}{2} + cos θ \times \frac{\sqrt{3}}{2} + sin θ \times \frac{1}{2}$ $sinthetaxx1/2-costhetaxxsqrt3/2+costhetaxxsqrt3/2+sinthetaxx1/2$

= $sin θ (\frac{1}{2} + \frac{1}{2})$ $sintheta(1/2+1/2)$

= $sin θ$ $sintheta$

= RHS

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Prove that (a) $sin (x - \frac{3 π}{2}) = cos x$ $sin(x-(3pi)/2)=cosx$ and (b) $sin (θ - \frac{π}{3}) + cos (θ - \frac{π}{6}) = sin θ$ $sin(theta-pi/3)+cos(theta-pi/6)=sintheta$ ?

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Explanation:

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Prove that (a) $sin (x - \frac{3 π}{2}) = cos x$ $sin(x-(3pi)/2)=cosx$ and (b) $sin (θ - \frac{π}{3}) + cos (θ - \frac{π}{6}) = sin θ$ $sin(theta-pi/3)+cos(theta-pi/6)=sintheta$ ?