First slide

Compound angles

Question

Let $0 \leq a, b, c, d \leq π$ and a,b,c,d are not complimentary such that $2 \cos a + 6 \cos b + 7 \cos c + 9 \cos d = 0$ and $2 \sin a - 6 \sin b + 7 \sin c - 9 \sin d = 0$ . Then the value of $\frac{\cos (a + d)}{\cos (b + c)} =$

Difficult

A

1
B

7
C

$\frac{7}{3}$
D

$\frac{3}{7}$

Solution

We have
$\begin{array}{l} (2 \cos a + 9 \cos d)^{2} = (6 \cos b + 7 \cos c)^{2} & - - - - - (1) \\ (2 \sin a - 9 \sin d)^{2} = (6 \sin b - 7 \sin c)^{2} & - - - - - - - - (2) \end{array}$
Adding we get
$\frac{\cos (a + d)}{\cos (b + c)} = \frac{84}{36} = \frac{7}{3}$

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