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Question

# What are the formulae of $\left(1\right)1+\mathrm{cos}2x\left(2\right)1-\mathrm{cos}2x$.

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Solution

## Step 1. Find the formula for $1+\mathrm{cos}2x$.As we know that,$\mathrm{cos}\left(a+b\right)=\mathrm{cos}a\mathrm{cos}b-\mathrm{sin}a\mathrm{sin}b$Substitute $a=b=x$ in the above equation.$\mathrm{cos}\left(x+x\right)=\mathrm{cos}x\mathrm{cos}x-\mathrm{sin}x\mathrm{sin}x\phantom{\rule{0ex}{0ex}}â‡’\mathrm{cos}2x={\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x$$âˆ´1+\mathrm{cos}2x=1+{\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x\phantom{\rule{0ex}{0ex}}â‡’={\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x+{\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x\left[{\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x=1\right]\phantom{\rule{0ex}{0ex}}â‡’=2{\mathrm{cos}}^{2}x$Thus, $1+\mathrm{cos}2x=2{\mathrm{cos}}^{2}x$Step 2. Find the formula for $1-\mathrm{cos}2x$.$âˆ´1-\mathrm{cos}2x=1-\left({\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x\right)\phantom{\rule{0ex}{0ex}}â‡’=1-{\mathrm{cos}}^{2}x+{\mathrm{sin}}^{2}x\phantom{\rule{0ex}{0ex}}â‡’={\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x-{\mathrm{cos}}^{2}x+{\mathrm{sin}}^{2}x\left[{\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x=1\right]\phantom{\rule{0ex}{0ex}}â‡’=2{\mathrm{sin}}^{2}x$Thus, $1-\mathrm{cos}2x=2{\mathrm{sin}}^{2}x$Hence,$\mathbf{1}\mathbf{+}\mathbf{cos}\mathbf{}\mathbf{2}\mathbf{x}\mathbf{=}\mathbf{2}{\mathbf{cos}}^{\mathbf{2}}\mathbf{}\mathbf{x}\phantom{\rule{0ex}{0ex}}\mathbf{1}\mathbf{-}\mathbf{cos}\mathbf{}\mathbf{2}\mathbf{x}\mathbf{=}\mathbf{2}{\mathbf{sin}}^{\mathbf{2}}\mathbf{}\mathbf{x}$

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