Question

In \triangle ABC, given \sin(B+\dfrac{C}{2})=\dfrac{4}{5}, find the value of \cos(A-B)

Collected in the board: Trigonometry

Steven Zheng posted 1 month ago


Answer

\sin(B+\dfrac{C}{2})=\dfrac{4}{5}

\cos(B+\dfrac{C}{2})=\sqrt{1-(\dfrac{4}{5})^2} = \dfrac{3}{5}

\sin(2B+C) = 2\sin(B+\dfrac{C}{2})\cos(B+\dfrac{C}{2})= \dfrac{12}{25}

B+C = \pi -A

\sin(2B+C) = \sin (B +\pi-A)= \sin(B-A) = -\sin(A-B)

\therefore \sin(A-B) = \dfrac{12}{25}

\cos(A-B) = \sqrt{1-\sin^2 (A-B)}

=\sqrt{1-(\dfrac{12}{25})^2} = \dfrac{7}{25}


Steven Zheng posted 1 month ago

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