Question

If sin A=-5/13,π<A<3π/2 and cos B=3\5 3π/2< B<2π find first one 1}sin[A+B]​

Answer 1

Given :   sin A= -5/13   π<A<3π/2

cos B=3/5    3π/2< B<2π

To Find : Sin ( A + B)

Solution:

Sin ( A + B) = SinA CosB  + cosA SinB

sin A= -5/13   π<A<3π/2

sin²A + cos²A = 1

=> cosA = - 12/13    as  π<A<3π/2 hence  cos is -ve

cos B=3/5     3π/2< B<2π  

=> sin B = - 4/5  as 3π/2< B<2π    hence sin is -ve

Sin ( A + B) = SinA CosB  + cosA SinB

=>  Sin ( A + B) = (-5/13) (3/5)  + (-12/13) ( -4/5)

=>  Sin ( A + B) = -3/13  + 48/65

=> Sin ( A + B) = 33/65

=> Sin ( A + B) = 0.5077

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Answer 2

$\underline{\textbf{Given:}}$

$\mathsf{sinA=\dfrac{-5}{13},\;\;\pi<A<\dfrac{3\pi}{2}}$

$\mathsf{cosB=\dfrac{3}{5},\;\;\dfrac{2\pi}{2}<B<2}$

$\underline{\textbf{To find:}}$

$\textsf{The value of sin(A+B)}$

$\underline{\textbf{Solution:}}$

$\mathsf{Consider,}$

$\mathsf{sinA=\dfrac{-5}{13}}$

$\mathsf{cos^2A=1-sin^2A}$

$\mathsf{cos^2A=1-\dfrac{25}{169}}$

$\mathsf{cos^2A=\dfrac{144}{169}}$

$\mathsf{cosA=\pm\dfrac{12}{13}}$

$\textsf{Since A lies in III quadrant,}$

$\mathsf{cosA=\dfrac{-12}{13}}$

$\mathsf{cosB=\dfrac{3}{5}}$

$\mathsf{sin^2B=1-cos^2B}$

$\mathsf{sin^2B=1-\dfrac{9}{25}}$

$\mathsf{sin^2B=\dfrac{16}{25}}$

$\mathsf{sinB=\pm\dfrac{4}{5}}$

$\textsf{Since B lies in IV quadrant,}$

$\mathsf{sinB=\dfrac{-4}{5}}$

$\mathsf{Now,}$

$\mathsf{sin(A+B)=sinA\;cosB+cosA\;sinB}$

$\mathsf{sin(A+B)=\dfrac{-5}{13}{\times}\dfrac{3}{5}+\dfrac{(-12)}{13}{\times}\dfrac{(-4)}{5}}$

$\mathsf{sin(A+B)=\dfra{-15}{65}+\dfrac{48}{65}}$

$\implies\boxed{\mathsf{sin(A+B)=\dfrac{33}{65}}}$