Math, asked by riiuwu, 3 months ago

from chapter 3 trignometric functions class 11.

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Answers

Answered by senboni123456
2

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)}

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