Question

from chapter 3 trignometric functions class 11.

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{cos(\alpha)+cos(\beta)+cos(\gamma)+cos(\alpha+\beta+\gamma)}$

$\sf{=2cos\bigg(\dfrac{\alpha+\beta}{2}\bigg)cos\bigg(\dfrac{\alpha-\beta}{2}\bigg)+2cos\bigg(\dfrac{\gamma+\alpha+\beta+\gamma}{2}\bigg)cos\bigg(\dfrac{\gamma-\alpha-\beta-\gamma}{2}\bigg)}$

$\sf{=2cos\bigg(\dfrac{\alpha+\beta}{2}\bigg)cos\bigg(\dfrac{\alpha-\beta}{2}\bigg)+2cos\bigg(\dfrac{2\gamma+\alpha+\beta}{2}\bigg)cos\bigg(\dfrac{-\alpha-\beta}{2}\bigg)}$

$\sf{=2cos\bigg(\dfrac{\alpha+\beta}{2}\bigg)cos\bigg(\dfrac{\alpha-\beta}{2}\bigg)+2cos\bigg(\dfrac{2\gamma+\alpha+\beta}{2}\bigg)cos\bigg(\dfrac{\alpha+\beta}{2}\bigg)}$

$\sf{=2cos\bigg(\dfrac{\alpha+\beta}{2}\bigg)\bigg[cos\bigg(\dfrac{\alpha-\beta}{2}\bigg)+cos\bigg(\dfrac{2\gamma+\alpha+\beta}{2}\bigg)\bigg]}$

$\sf{=2cos\bigg(\dfrac{\alpha+\beta}{2}\bigg)\bigg[2cos\bigg\{\dfrac{1}{2}\bigg(\dfrac{\alpha-\beta}{2}+\dfrac{2\gamma+\alpha+\beta}{2}\bigg)\bigg\}\,cos\bigg\{\dfrac{1}{2}\bigg(\dfrac{\alpha-\beta}{2}-\dfrac{2\gamma+\alpha+\beta}{2}\bigg)\bigg\}\bigg]}$

$\sf{=2cos\bigg(\dfrac{\alpha+\beta}{2}\bigg)\bigg[2cos\bigg\{\dfrac{1}{2}\bigg(\dfrac{\alpha-\beta+2\gamma+\alpha+\beta}{2}\bigg)\bigg\}\,cos\bigg\{\dfrac{1}{2}\bigg(\dfrac{\alpha-\beta-2\gamma-\alpha-\beta}{2}\bigg)\bigg\}\bigg]}$

$\sf{=2cos\bigg(\dfrac{\alpha+\beta}{2}\bigg)\bigg[2cos\bigg\{\dfrac{1}{2}\bigg(\dfrac{\alpha+2\gamma+\alpha}{2}\bigg)\bigg\}\,cos\bigg\{\dfrac{1}{2}\bigg(\dfrac{-\beta-2\gamma-\beta}{2}\bigg)\bigg\}\bigg]}$

$\sf{=2cos\bigg(\dfrac{\alpha+\beta}{2}\bigg)\bigg[2cos\bigg\{\dfrac{1}{2}\bigg(\dfrac{2\gamma+2\alpha}{2}\bigg)\bigg\}\,cos\bigg\{\dfrac{1}{2}\bigg(\dfrac{-2\beta-2\gamma}{2}\bigg)\bigg\}\bigg]}$

$\sf{=2cos\bigg(\dfrac{\alpha+\beta}{2}\bigg)\bigg[2cos\bigg\{\dfrac{1}{2}(\gamma+\alpha)\bigg\}\,cos\bigg\{\dfrac{1}{2}(-\beta-\gamma)\bigg\}\bigg]}$

$\sf{=2cos\bigg(\dfrac{\alpha+\beta}{2}\bigg)\bigg[2cos\bigg(\dfrac{\gamma+\alpha}{2}\bigg)\,cos\bigg\{\dfrac{1}{2}(\beta+\gamma)\bigg\}\bigg]}$

$\sf{=2cos\bigg(\dfrac{\alpha+\beta}{2}\bigg)\bigg[2cos\bigg(\dfrac{\gamma+\alpha}{2}\bigg)\,cos\bigg(\dfrac{\beta+\gamma}{2}\bigg)\bigg]}$

$\sf{=2cos\bigg(\dfrac{\alpha+\beta}{2}\bigg)\cdot2cos\bigg(\dfrac{\gamma+\alpha}{2}\bigg)\,cos\bigg(\dfrac{\beta+\gamma}{2}\bigg)}$

$\sf{=4cos\bigg(\dfrac{\alpha+\beta}{2}\bigg)\,cos\bigg(\dfrac{\beta+\gamma}{2}\bigg)\,cos\bigg(\dfrac{\gamma+\alpha}{2}\bigg)}$