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shubham.bagrecha

  • 2 years ago

Prove that: cos(A+B)+sin(A-B)=2sin(45+A)cos(45+B)

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  1. shubham.bagrecha
    • 2 years ago
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    \[ \cos(A+B)+\sin(A-B)=2\sin(45+A)\cos(45+B)\]

  2. Neo92
    • 2 years ago
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    first write cos(A+B) as sin[90-(A+B)] then use sinC + sinD = 2sin[(C+D)/2]cos[(C-D)/2] try and tell me what u get

  3. mathslover
    • 2 years ago
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    ok ... for this first of all tell me what is cos(A+B)

  4. shubham.bagrecha
    • 2 years ago
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    @Neo92 ok i will solve

  5. mathslover
    • 2 years ago
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    cos(A+B)=sin(90-A-B) sin(90-A-B) + sin(A-B) = 2sin(45+A)cos(45+B) we know that sin(x) + sin(y)=2sin([x+y]/2)cos([x-y]/2)

  6. mathslover
    • 2 years ago
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    so we get that sin(90-A-B) + sin(A-B)=2sin(45-B)cos(45-A) but we know that sin(45-B)=cos(90-(45-b))=cos(45+B) and cos(45-A)=sin(90-(45-A))=sin(45+A) hence we get that cos(A+B) + sin(A-B) = 2sin(45+A)cos(45+B)

  7. klimenkov
    • 2 years ago
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    \[\cos(A+B)+\sin(A-B)=\cos A\cos B-\sin A\sin B+\sin A \cos B-\sin B\cos A=\] \[=\cos A(\cos B-\sin B)+\sin A (\cos B-\sin B)=(\cos A+\sin A)(\cos B-\sin B)\] Now use: \(\cos A+\sin A=\sqrt2\cdot(\frac1{\sqrt2}\cos A+\frac1{\sqrt2}\sin A)=\sqrt2\sin(45^\circ +A)\) and \(\cos B-\sin B=\sqrt2\cdot(\frac1{\sqrt2}\cos B-\frac1{\sqrt2}\sin B)=\sqrt2\cos(45^\circ +B)\). Now multiply this: \[(\cos A+\sin A)(\cos B-\sin B)=\sqrt2\sin(45^\circ +A)\cdot \sqrt2\cos(45^\circ +B)=2\sin(45^\circ +A)\cdot\]\[\cdot\cos(45^\circ +B)\]

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