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anonymous

  • 4 years ago

If sin A+ sin B=c and cosA+cosB =c, where c is not equal to 0, express sinB + cosB in terms of c.

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  1. asnaseer
    • 4 years ago
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    \[\begin{align} \sin(A)+\sin(B)&=c\implies \sin(A)=c-\sin(B)\\ \therefore \sin^2(A)&=(c-\sin(B))^2\\ &=c^2-2c\sin(B)+sin^2(B) \end{align}\]similarly, from the second equation we can get:\[\cos^2(A)=c^2-2c\cos(B)+\cos^2(B)\]using \(\cos^2(A)+\sin^2(A)=1\) we can then get:\[\begin{align} 1&=2c^2-2c\sin(B)-2c\cos(B)+\sin^2(B)+\cos^2(B)\\ &=2c^2-2c(\sin(B)+\cos(B))+1\\ \therefore 2c(\sin(B)+\cos(B))&=2c^2\\ \therefore \sin(B)+\cos(B)&=c \end{align}\]

  2. asnaseer
    • 4 years ago
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    this was notan easy derivation to spot. maybe someone else has found a quicker/better derivation?

  3. mathmate
    • 4 years ago
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    Here's and alternative: Let \[A=\theta+\phi, B=\theta-\phi\] \[\sin(\theta+\phi)+\sin(\theta-\phi)=\cos(\theta)+\cos(\theta-\phi)=c\] Expand by sum/difference formulae Let\[P=\sin(\theta+\phi)+\sin(\theta-\phi)\]\[=\sin(\theta)\cos(\phi)+\cos(\theta)\sin(\phi)+\sin(\theta)\cos(\phi)-\cos(\theta)\sin(\phi)\]\[=2\sin(\theta)\cos(\phi)\] \[Q=\cos(\theta+\phi)+\cos(\theta-\phi)\]\[=\cos(\theta)\cos(\phi)-\sin(\theta)\sin(\phi)+\cos(\theta)\cos(\phi)-\sin(\theta)\sin(\phi)\]\[=2\cos(\theta)\cos(\phi)\] Equate \[P=Q=c\] \[\sin(\theta)\cos(\phi)=\cos(\theta)\cos(\phi)\]\[\sin(\theta)=\cos(\theta)\]\[\theta=\frac{\pi}{4} or \frac{5\pi}{4}\] From which we show that cos(B)=sin(A) and so sin(A)+cos(B)=sin(A)+sin(B)=c

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