Question

cosx × cos(60-x) ×cos(60+x)step by step please ​

Answer 1

Answer:

Hi Good morning....

Step-by-step explanation:

suffering from the same question

Answer 2

Answer:

\textsf{consider}consider

\mathsf{cosx\;cos(60+x)\;cos(60-x)}cosxcos(60+x)cos(60−x)

\mathsf{=\frac{1}{2}cosx[2\;cos(60+x)\;cos(60-x)]}=

2

1

cosx[2cos(60+x)cos(60−x)]

\textsf{Using}Using

\boxed{\mathsf{cos(A+B)+cos(A-B)=2\,cosA\,cosB}}

cos(A+B)+cos(A−B)=2cosAcosB

\mathsf{=\frac{1}{2}cosx[cos(60+x+60-x)+cos(60+x-(60-x))]}=

2

1

cosx[cos(60+x+60−x)+cos(60+x−(60−x))]

\mathsf{=\frac{1}{2}cosx[cos\,120+cos\,2x]}=

2

1

cosx[cos120+cos2x]

\mathsf{=\frac{1}{2}cosx[\frac{-1}{2}+cos\,2x]}=

2

1

cosx[

2

−1

+cos2x]

\mathsf{=\frac{-1}{4}cosx+\frac{1}{2}cosx\,cos\,2x}=

4

−1

cosx+

2

1

cosxcos2x

\mathsf{=\frac{-1}{4}cosx+\frac{1}{4}[2\,cosx\,cos\,2x]}=

4

−1

cosx+

4

1

[2cosxcos2x]

\mathsf{=\frac{-1}{4}cosx+\frac{1}{4}[cos(x+2x)+cos(x-2x)]}=

4

−1

cosx+

4

1

[cos(x+2x)+cos(x−2x)]

\mathsf{=\frac{-1}{4}cos\,x+\frac{1}{4}cos\,3x+\frac{1}{4}cos(-x)}=

4

−1

cosx+

4

1

cos3x+

4

1

cos(−x)

\mathsf{=\frac{-1}{4}cos\,x+\frac{1}{4}cos\,3x+\frac{1}{4}cos\,x}=

4

−1

cosx+

4

1

cos3x+

4

1

cosx

\mathsf{=\frac{1}{4}cos\,3x}=

4

1

cos3x

\implies\boxed{\mathsf{cosx\;cos(60+x)\;cos(60-x)==\frac{1}{4}cos\,3x}}⟹

cosxcos(60+x)cos(60−x)==

4

1

cos3x