Let $$0$$≤a,b,c,d≤π where b and c are not complementary such that $$2$$ $$cos$$ a $$+$$ $$6cos$$ 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 $$3cos$$⁡(a+d)$$cos$$⁡(b+c) is

Question

Let $$0$$≤a,b,c,d≤π where b and c are not complementary such that $$2$$ $$cos$$ a $$+$$ $$6cos$$ 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 $$3cos$$⁡(a+d)$$cos$$⁡(b+c) is

Infinity Learn · Accepted Answer

From the given equations, we $$have(2cos$$⁡$$a+9cos$$⁡$$d)2=(6cos$$⁡$$b+7cos$$⁡$$c)2and$$  (2sin$$⁡a−$$9sin$$⁡$$d)2=(6sin$$⁡b−$$7sin$$⁡$$c)2Adding$$, we have $$36$$ $$cos$$ $$(a$$ $$+$$ $$d)=84$$ $$cos$$$$(b$$ $$+$$ $$c)or$$  $$cos$$⁡$$(a+d)$$cos$$⁡$$(b+c)=73$$

Compound angles

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Let $0 \leq a, b, c, d \leq π$ where b and c are not complementary such that 2 cos a + 6cos 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 $3 \frac{\cos (a + d)}{\cos (b + c)}$ is

From the given equations, we have
$\begin{aligned} (2 \cos a + 9 \cos d)^{2} = (6 \cos b + 7 \cos c)^{2} \\ and (2 \sin a - 9 \sin d)^{2} = (6 \sin b - 7 \sin c)^{2} \end{aligned}$
Adding, we have 36 cos (a + d)=84 cos(b + c)
or $\frac{\cos (a + d)}{\cos (b + c)} = \frac{7}{3}$

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