trigo problem no. 20

**afeasfaerw23231233** · March 3rd 2008, 01:33 AM

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!

**Soroban** · March 3rd 2008, 09:03 AM

Hello, afeasfaerw23231233!

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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