Compound Angles

Jul 2011
140
0
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), \(\displaystyle \cos(\alpha-\beta)=\cos\alpha\cdot\cos\beta+\sin\alpha\sin\beta=( \frac{12}{13} )(\frac{4}{5})+(\frac{5}{13})(\frac{3}{5})=\frac{63}{65}\)
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: \(\displaystyle (\frac{1}{\sqrt{2}})(\frac{63}{65})+(\frac{1}{ \sqrt{2} })(\frac{-16}{65})=\frac{47}{65\sqrt{2}}=\frac{47\sqrt{2}}{130}\)
 

skeeter

MHF Helper
Jun 2008
16,217
6,765
North Texas
(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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