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Question

# (cos 0° + cos 45° + sin 30°) (sin 90° – cos 45° + cos 60°)

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Solution

## As we know that, ​$\mathrm{cos}0°=1\phantom{\rule{0ex}{0ex}}\mathrm{cos}45°=\frac{1}{\sqrt{2}}\phantom{\rule{0ex}{0ex}}\mathrm{sin}30°=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\mathrm{sin}90°=1\phantom{\rule{0ex}{0ex}}\mathrm{cos}60°=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{By}\mathrm{substituting}\mathrm{these}\mathrm{values},\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}\left(\mathrm{cos}0°+\mathrm{cos}45°+\mathrm{sin}30°\right)\left(\mathrm{sin}90°-\mathrm{cos}45°+\mathrm{cos}60°\right)=\left(1+\frac{1}{\sqrt{2}}+\frac{1}{2}\right)\left(1-\frac{1}{\sqrt{2}}+\frac{1}{2}\right)\phantom{\rule{0ex}{0ex}}=\left(\left(1+\frac{1}{2}\right)+\left(\frac{1}{\sqrt{2}}\right)\right)\left(\left(1+\frac{1}{2}\right)-\left(\frac{1}{\sqrt{2}}\right)\right)\phantom{\rule{0ex}{0ex}}={\left(1+\frac{1}{2}\right)}^{2}-{\left(\frac{1}{\sqrt{2}}\right)}^{2}\phantom{\rule{0ex}{0ex}}=1+\frac{1}{4}+2\left(1\right)\left(\frac{1}{2}\right)-\frac{1}{2}\phantom{\rule{0ex}{0ex}}=1+\frac{1}{4}+1-\frac{1}{2}\phantom{\rule{0ex}{0ex}}=\frac{4+1+4-2}{4}\phantom{\rule{0ex}{0ex}}=\frac{7}{4}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Hence},\left(\mathrm{cos}0°+\mathrm{cos}45°+\mathrm{sin}30°\right)\left(\mathrm{sin}90°-\mathrm{cos}45°+\mathrm{cos}60°\right)=\frac{7}{4}.$

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