If cos alpha+4 cos beta+9 cos gamma=0 and sin | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given: $\cos \alpha+4 \cos \beta+9 \cos \gamma=0$ and $\sin \alpha+4 \sin \beta+9 \sin \gamma=0$. Let $z_1 = e^{i\alpha}$,$z_2 = e^{i\beta}$,$z_3

Vedclass · Accepted Answer

$1+8 \cos (\beta-\alpha)$

Vedclass · Answer

(A) Given: $\cos \alpha+4 \cos \beta+9 \cos \gamma=0$ and $\sin \alpha+4 \sin \beta+9 \sin \gamma=0$. Let $z_1 = e^{i\alpha}$,$z_2 = e^{i\beta}$,$z_3 = e^{i\gamma}$. The equations can be written as $z_1 + 4z_2 + 9z_3 = 0$. Thus,$4z_2 = -(z_1 + 9z_3)$. Squaring both sides: $16|z_2|^2 = |z_1 + 9z_3|^2 = |z_1|^2 + 81|z_3|^2 + 18 	ext{Re}(z_1 \bar{z_3})$. Since $|z_1|=|z_2|=|z_3|=1$,we have $16 = 1 + 81 + 18 \cos(\alpha - \gamma)$ $\Rightarrow 18 \cos(\alpha - \gamma) = -66$ $\Rightarrow \cos(\alpha - \gamma) = -\frac{11}{3}$. Similarly,$9z_3 = -(z_1 + 4z_2)$ $\Rightarrow 81 = 1 + 16 + 8 \cos(\alpha - \beta)$ $\Rightarrow 8 \cos(\alpha - \beta) = 64$ $\Rightarrow \cos(\alpha - \beta) = 8$. Now,$81 \cos(2\gamma - 2\alpha) = 81(2 \cos^2(\gamma - \alpha) - 1) = 81(2(-\frac{11}{3})^2 - 1) = 81(2 \cdot \frac{121}{9} - 1) = 81(\frac{242-9}{9}) = 9(233) = 2097$. And $16 \cos(2\beta - 2\alpha) = 16(2 \cos^2(\beta - \alpha) - 1) = 16(2(8)^2 - 1) = 16(128 - 1) = 16(127) = 2032$. Thus,$2097 - 2032 = 65$. Checking the options: $1 + 8 \cos(\beta - \alpha) = 1 + 8(8) = 1 + 64 = 65$. Therefore,the correct option is $A$.

If $\cos \alpha+4 \cos \beta+9 \cos \gamma=0$ and $\sin \alpha+4 \sin \beta+9 \sin \gamma=0$,then $81 \cos (2 \gamma-2 \alpha)-16 \cos (2 \beta-2 \alpha)=$

Similar Questions

If $|z - 3 - 4i| = 4$,where $i = \sqrt{-1}$,then the maximum possible value of $|z|$ is:

If for the complex numbers $z$ satisfying $|z-2-2 i| \leq 1$,the maximum value of $|3 i z+6|$ is attained at $a+i b$,then $a+b$ is equal to .... .

The number of values of $z \in \mathbb{C}$, satisfying the equations $|z - (4 + 8i)| = \sqrt{10}$ and $|z - (3 + 5i)| + |z - (5 + 11i)| = 4\sqrt{5}$, is:

The equation $\text{Im}\left( \frac{iz - 2}{z - i} \right) + 1 = 0$,where $z \in \mathbb{C}$ and $z \neq i$,represents a part of a circle having radius equal to

$|{z_1} + {z_2}| = |{z_1}| + |{z_2}|$ is possible if

Vedclass Test Series

Exam Paper Generator

Online Exam Module