Compound Angles

**GrigOrig99** · October 30th 2012, 10:53 AM

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

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

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

**skeeter** · October 30th 2012, 02:19 PM

Originally Posted by GrigOrig99

(ii) 1st: $\cos(45^{\circ}-\alpha-\beta) = \color{red}{\cos[45 -(\alpha+\beta)] \rightarrow \cos45\cdot \cos(\alpha+\beta)+\sin45\cdot\sin(\alpha+\beta)}$

correction

...

**GrigOrig99** · October 30th 2012, 03:03 PM

Ok, thank you.

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