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Find sin(a+b) if cos(a)=3/5, sin(b)= -4/5, a is in quadrant 1 and b is in quadrant 3.
cos^-1(cos(a)=3/5)cos^-1 => a = 53.13010235. sin-1(sin(b)= -4/5) => -53.13010235

a + b = 0, and sin(0) = 0. But my answer of 0 is marked wrong. What am I missing?

Respuesta :

LammettHash

[tex]\sin^{-1}(\sin x)=x[/tex] only for [tex]-90^\circ\le x\le90^\circ[/tex]. The angle [tex]b[/tex] you found lies in quadrant 4, not 3. Subtract 90 degrees from the angle you found and you get an angle in quadrant 3 whose sine is still -4/5, so that

[tex]b=\sin^{-1}\left(-\dfrac45\right)-90^\circ\approx-143.13^\circ[/tex]

or equivalently about 216 degrees (which does lie in quadrant 3).

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Without looking for the exact values of [tex]a[/tex] and [tex]b[/tex], we can make two observations:

  • [tex]a[/tex] is in quadrant 1, so we know [tex]\cos a[/tex] and [tex]\sin a[/tex] are both positive, and
  • [tex]b[/tex] is in quadrant 3, so we know [tex]\cos b[/tex] and [tex]\sin b[/tex] are both negative

Next, we find the values of [tex]\sin a[/tex] and [tex]\cos b[/tex] from the given values using the Pythagorean identity:

  • [tex]\sin a=\sqrt{1-\cos^2a}=\dfrac45[/tex]
  • [tex]\cos b=-\sqrt{1-\sin^2b}=-\dfrac35[/tex]

Then using the angle sum identity for sine, we get

[tex]\sin(a+b)=\sin a\cos b+\cos a\sin b=\dfrac45\left(-\dfrac35\right)+\dfrac35\left(-\dfrac45\right)=-\dfrac{24}{25}\neq0[/tex]

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