Given-sin-A-sin-B-0-7-cos-A-cos-B-0-8-find-the-value-of-A-and-B-

Question Number 134581 by EDWIN88 last updated on 05/Mar/21

Given {\begin{cases} \sin A + \sin B = 0.7 \\ \cos A + \cos B = 0.8 \end{cases}

find the value of A and B .

Answered by liberty last updated on 05/Mar/21

\Leftrightarrow 2 + 2 \cos (A - B) = \frac{49 + 64}{100}

\Leftrightarrow 2 \cos (A - B) = - \frac{87}{100}

\Leftrightarrow \cos (A - B) = - \frac{87}{200}

\Leftrightarrow A - B \approx 115.8 ° + k . 360 °

\Leftrightarrow A \approx B + 115.8 ° + k . 360 °, k ϵ Z

Answered by mr W last updated on 05/Mar/21

\sin A + \sin B = p (0.7)

\cos A + \cos B = q (0.8)

A = \frac{A + B}{2} + \frac{A - B}{2} = u + v

B = \frac{A + B}{2} - \frac{A - B}{2} = u - v

\sin (u + v) + \sin (u - v) = p

\Rightarrow \sin u \cos v = \frac{p}{2} \dots (I)

\cos (u + v) + \cos (u - v) = q

\Rightarrow \cos u \cos v = \frac{q}{2} \dots (I I)

(I) \boldsymbol \div (I I) :

\tan u = \frac{p}{q}

\Rightarrow u = m π + \tan^{- 1} \frac{p}{q}

{(I)}^{2} + {(I I)}^{2} :

\cos^{2} v = \frac{p^{2} + q^{2}}{4}

\cos v = \pm \frac{\sqrt{p^{2} + q^{2}}}{2}

\Rightarrow v = n π \pm \cos^{- 1} \frac{\sqrt{p^{2} + q^{2}}}{2}

\Rightarrow A = u + v = (m + n) π + \tan^{- 1} \frac{p}{q} \pm \cos^{- 1} \frac{\sqrt{p^{2} + q^{2}}}{2}

\Rightarrow B = u - v = (m - n) π + \tan^{- 1} \frac{p}{q} \mp \cos^{- 1} \frac{\sqrt{p^{2} + q^{2}}}{2}

Answered by bobhans last updated on 05/Mar/21

\Rightarrow 2 \sin (\frac{A + B}{2}) \cos (\frac{A - B}{2}) = \frac{7}{10}

\Rightarrow 2 \cos (\frac{A + B}{2}) \cos (\frac{A - B}{2}) = \frac{8}{10}

(i) \tan (\frac{A + B}{2}) = \frac{7}{8}; A + B = 82.24 ° + 2 k π

(ii) 2 + 2 \cos (A - B) = \frac{113}{100}

\Rightarrow \cos (A - B) = - \frac{87}{200}; A - B = 115.78 ° + 2 k π

Thus {\begin{cases} A = 99.01 ° + 2 k π \\ B = - 16.77 ° + 2 k π \end{cases}

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