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Thread: Compound Angles

  1. #1
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    Compound Angles

    My answer for (i) matches that of the text book. In (ii), however, I have not arrived at the correct result. Can anyone help me out?

    Many thanks.

    Q. (i)
    When $\displaystyle \sin\alpha=\frac{5}{13}$ & $\displaystyle \cos\beta=\frac{4}{5}$, $\displaystyle \alpha$ & $\displaystyle \beta<90^{\circ}$, express $\displaystyle \sin(\alpha+\beta)$ in the form $\displaystyle \frac{a}{b}$. (ii) Hence, or otherwise show that $\displaystyle \cos(45^{\circ}-\alpha-\beta)=\frac{89\sqrt{2}}{130}$.

    Attempt: (i) 1st: If $\displaystyle \sin^2\alpha+\cos^2\alpha=1$, then $\displaystyle \cos^2\alpha=1-(\frac{5}{13})^2\rightarrow\cos\alpha=\sqrt{1-\frac{25}{169}}\rightarrow\cos\alpha=\frac{12}{13}$
    2nd: Similarly, $\displaystyle \sin^2\beta+\cos^2\beta=1\rightarrow\sin^2\beta=1-(\frac{4}{5})^2\rightarrow\sin\beta=\frac{3}{5}$
    3rd: $\displaystyle \sin(\alpha+\beta)=\sin\alpha\cdot\cos\beta+\cos \alpha\cdot\sin\beta\rightarrow(\frac{5}{13})( \frac{4}{5} )+(\frac{12}{13})(\frac{36}{65})\rightarrow \frac{56}{65}$

    (ii) 1st: $\displaystyle \cos(45^{\circ}-\alpha-\beta)\rightarrow\cos45\cdot\cos(\alpha-\beta)+\sin45\cdot\sin(\alpha-\beta)$
    2nd: Taking the values from (i), $\displaystyle \cos(\alpha-\beta)=\cos\alpha\cdot\cos\beta+\sin\alpha\sin\bet a=( \frac{12}{13} )(\frac{4}{5})+(\frac{5}{13})(\frac{3}{5})=\frac{6 3}{65}$
    3rd: Similarly, $\displaystyle \sin(\alpha-\beta)=\sin\alpha\cdot\cos\beta-\sin\alpha\cdot\sin\beta=(\frac{5}{13})( \frac{4}{5} )-(\frac{12}{13})(\frac{36}{65})=\frac{-16}{65}$
    4th: Thus, from the 1st step: $\displaystyle (\frac{1}{\sqrt{2}})(\frac{63}{65})+(\frac{1}{ \sqrt{2} })(\frac{-16}{65})=\frac{47}{65\sqrt{2}}=\frac{47\sqrt{2}}{1 30}$
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  2. #2
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    Re: Compound Angles

    Quote Originally Posted by GrigOrig99 View Post
    (ii) 1st: $\displaystyle \cos(45^{\circ}-\alpha-\beta) = \color{red}{\cos[45 -(\alpha+\beta)] \rightarrow \cos45\cdot \cos(\alpha+\beta)+\sin45\cdot\sin(\alpha+\beta)}$

    correction
    ...
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  3. #3
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    Re: Compound Angles

    Ok, thank you.
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