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Thread: trigo problem no. 20

  1. #1
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    trigo problem no. 20

    p165 q14
    i did this question correctly a month ago. but know i try 15 min. and still don't figure out how to do! i write notes on textbook and don't keep a notebook. thanks in advance!
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  2. #2
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    Hello, afeasfaerw23231233!

    Given: .$\displaystyle \begin{array}{cccc}\cos(\theta+\alpha) & = & p & {\color{blue}[1]}\\ \sin(\theta+\beta) &=& q & {\color{blue}[2]}\end{array}$

    (a) Express $\displaystyle \cos\theta\text{ and }\sin\theta$ in terms of $\displaystyle
    \alpha,\:\beta,\:p\text{, and }q.$

    (b) Hence, show that: .$\displaystyle p^2+q^2+ 2pq\sin(\alpha-\beta) \;=\;\cos^2\!(\alpha-\beta) $

    $\displaystyle \begin{array}{ccccc}\text{From }{\color{blue}[1]}: & \cos\alpha\cos\theta - \sin\alpha\sin\theta &=& p & {\color{blue}[3]}\\
    \text{From }{\color{blue}[2]}: & \sin\beta\cos\theta + \cos\beta\sin\theta &=& q & {\color{blue}[4]}\end{array}$

    $\displaystyle \begin{array}{cccc}\text{Multiply }{\color{blue}[3]}\text{ by }\cos\beta: & \cos\alpha\cos\beta\cos\theta - \sin\alpha\cos\beta\sin\theta & = & p\cos\beta \\
    \text{Multiply }{\color{blue}[4]}\text{ by }\sin\alpha: & \sin\alpha\sin\beta\cos\theta + \sin\alpha\cos\beta\sin\theta &=& q\sin\alpha \end{array} $

    Add: .$\displaystyle (\cos\alpha\cos\beta + \sin\alpha\sin\beta)\cos\theta \:=\:p\cos\beta + q\sin\alpha $

    Therefore: .$\displaystyle \cos\theta \;=\;\frac{p\cos\beta + q\sin\alpha}{\cos\alpha\cos\beta + \sin\alpha\sin\beta} \quad\Rightarrow\quad\boxed{\;\cos\theta \;=\;\frac{p\cos\beta + q\sin\alpha}{\cos(\alpha - \beta)}\;}\;\;{\color{red}[1]}$

    In a similar fashion, we find that: .$\displaystyle \boxed{\;\sin\theta \;=\;\frac{q \cos\alpha- p\sin\beta} {\cos(\alpha-\beta)}\;} \;\;{\color{red}[2]}$


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    Square [2]: .$\displaystyle \sin^2\!\theta \;=\;\frac{q^2\cos^2\!\alpha - 2pq\sin\beta\cos\alpha + p^2\sin^2\!\beta}{\cos^2\!(\alpha-\beta)} $

    Square [1]: .$\displaystyle \cos^2\!\theta \;=\;\frac{p^2\cos^2\!\beta + 2pq\sin\alpha\cos\beta + q^2\sin^2\!\alpha}{\cos^2\!(\alpha-\beta)} $

    Add: .$\displaystyle \sin^2\!\theta \:+ \:\cos^2\!\theta \;=\;\frac{p^2(\sin^2\!\beta +\cos^2\!\beta) + q^2(\sin^2\!\alpha +\cos^2\!\alpha) + 2pq(\sin\alpha\cos\beta - \sin\beta\cos\alpha)}{\cos^2\!(\alpha-\beta)}$

    And we have: .$\displaystyle 1 \;=\;\frac{p^2 + q^2 + 2pq\sin(\alpha-\beta)}{\cos^2\!(\alpha-\beta)} $

    Therefore: . $\displaystyle \boxed{p^2+q^2+2pq\sin(\alpha-\beta) \;=\;\cos^2\!(\alpha-\beta)} $

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