cosx × cos(60-x) ×cos(60+step by step please
Answers
Step-by-step explanation:
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