 Monique Slaughter

2021-12-31

Find the value of the expression
when $\alpha ={60}^{\circ }$
I don't know if I am supposed to simplify the given expression, but I just put α=60∘ to get
$\mathrm{sin}\left({135}^{\circ }-{120}^{\circ }\right)\cdot \mathrm{sin}\left({60}^{\circ }+{45}^{\circ }\right)+{\mathrm{cos}}^{2}\left({75}^{\circ }-{60}^{\circ }\right)-2\mathrm{sin}\left({60}^{\circ }-{30}^{\circ }\right)=$

$=\mathrm{sin}{15}^{\circ }\mathrm{sin}{105}^{\circ }+\mathrm{cos}{215}^{\circ }-2\mathrm{sin}{30}^{\circ }$

What can I do next? Mary Goodson

This seems to be a simpler question than what is thought to be.
Replace $\alpha$ with ${60}^{\circ }$, just remember

Then,
${\mathrm{sin}15}^{\circ }{\mathrm{sin}15}^{\circ }+{\mathrm{cos}}^{2}{15}^{\circ }-2{\mathrm{sin}30}^{\circ }$
$=1-2\cdot \frac{12}{=}0$ HINT
You could first use the following formula, known as one of the product to sum formulae among other names,
$\mathrm{sin}x\mathrm{sin}y\equiv \frac{12}{\mathrm{cos}\left(x-y\right)-\mathrm{cos}\left(x+y\right)}$
to evaluate ${\mathrm{sin}15}^{\circ }{\mathrm{sin}105}^{\circ }$ without having to explicitly evaluate
You could then use the double angle formula for $\mathrm{cos}$ to write ${\mathrm{cos}215}^{\circ }$ in terms of ${\mathrm{cos}30}^{\circ }$ Vasquez

Hint: Notice that $\mathrm{sin}{105}^{\circ }=\mathrm{cos}{15}^{\circ }$. Then you have to use the formulae:
$\frac{\mathrm{cos}2\theta +1}{2}={\mathrm{cos}}^{2}\theta$
And,
$\mathrm{sin}2\theta =2\mathrm{sin}\theta \mathrm{cos}\theta$
If you do it this way, you won't have to calculate trigonometric ratios for ${15}^{\circ }$,which can be evaluated, but are slightly complicated.

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