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Trigonometric Ratios of Multiple and Submultiple Angles — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSTrigonometric Ratios of Multiple and Submultiple Angles3 Marks
Question
If $\cos\text{A}=-\frac{24}{25},$ $\cos\text{B}=-\frac{12}{13},$ where A and B both lie in second quadrant,find the value of $\sin\text{(A+B)}$.
  1. $\sin\text{(A+B)}$
  2. $\cos\text{(A+B)}$
✓

Answer

We have, $\cos\text{A}=-\frac{24}{25}$ and $\cos\text{B}=\frac{3}{5}$ $\therefore\sin\text{A}=\sqrt{1-\cos^2\text{A}}$ and $\sin\text{B}=\sqrt{1-\cos^2\text{B}}$ [ in the second quadreant $\cos\theta$ in negetive] $\Rightarrow\sin\text{A}=-\sqrt{1-\Big(-\frac{24}{25}\Big)^2}$ and $\sin\text{B}=\sqrt{1-\Big(\frac{-3}{5}\Big)^2}$ $\Rightarrow\sin\text{A}=-\sqrt{1-\frac{576}{625 }}$ and $\sin\text{B}=\sqrt{1-\frac{9}{25}}$ $\Rightarrow\sin\text{A}=\sqrt{\frac{576}{625}}$ and $\sin\text{B}=\sqrt{\frac{9}{25}}$ $\Rightarrow\sin\text{A}=-\sqrt{\frac{49}{625}}$ and $\sin\text{B}=-\sqrt{\frac{16}{25}}$ $\Rightarrow\sin\text{A}=-\frac{7}{25}$ and $\Rightarrow\sin\text{B}=-\frac{4}{5}$ Now,
  1. $\sin(\text{A+B})=\sin\text{A}\cos{\text{B}+\cos\text{A}}\sin\text{B}$
$=\frac{7}{25}\times\frac{3}{5}-\frac{24}{25}\times\Big(-\frac{4}{5}\Big)$
$=-\frac{21}{25}+\frac{96}{125}$
$=-\frac{75}{125}$
$=\frac{3}{5}$
  1. $\cos(\text{A+B})=\cos\text{A}\cos{\text{B}-\sin\text{A}}\sin\text{B}$
$=-\frac{24}{25}\times\frac{3}{5}-\Big(-\frac{7}{25}\Big)\times\Big(-\frac{4}{5}\Big)$
$=-\frac{72}{125}-\frac{28}{125}$
$=-\frac{72-28}{125}$
$=-\frac{100}{125}$
$=-\frac{4}{5}$

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