petrusrexcs

2021-12-30

Find the value of ${\mathrm{cos}105}^{\circ }+{\mathrm{sin}75}^{\circ }$
We can write the given trig expression as
$\mathrm{cos}\left({180}^{\circ }-{75}^{\circ }\right)+{\mathrm{sin}75}^{\circ }=-{\mathrm{cos}75}^{\circ }+{\mathrm{sin}75}^{\circ }$
$={\mathrm{sin}75}^{\circ }-{\mathrm{cos}75}^{\circ }$

MoxboasteBots5h

${\mathrm{sin}75}^{\circ }$ can be written as $\mathrm{sin}\left(30+45\right)$
$\mathrm{sin}\left(30+45\right)=\mathrm{sin}30\cdot \mathrm{cos}45+\mathrm{cos}30\cdot \mathrm{cos}45$
Similarly $\mathrm{cos}105=\mathrm{cos}\left(60+45\right)⇒\mathrm{cos}60\cdot \mathrm{cos}45-\mathrm{sin}60\mathrm{sin}45$
Now just add both the above equation and try to use value from the trignometry chart below.

Navreaiw

Following on from where you left off,
${\mathrm{sin}75}^{\circ }-{\mathrm{cos}75}^{\circ }=-\mathrm{sin}\left(-{75}^{\circ }\right)-\mathrm{cos}\left(-{75}^{\circ }\right)$
$=-{\mathrm{sin}75}^{\circ }+\mathrm{cos}\left(-{75}^{\circ }\right)$
Using the formula $\mathrm{sin}\theta +\mathrm{cos}\theta =\sqrt{2}\mathrm{sin}\left(\theta +{45}^{\circ }\right)$ (which can be proven using the addition formula for $\mathrm{sin}$), this becomes
$-\sqrt{2}\mathrm{sin}\left(-{30}^{\circ }\right)=\sqrt{2}{\mathrm{sin}30}^{\circ }=\frac{\sqrt{2}}{2}$

karton

There's some transformation formula I'm listing below, $\mathrm{sin}\left(a+b\right)+\mathrm{sin}\left(a-b\right)=2\mathrm{sin}a\mathrm{cos}b$
$\mathrm{sin}\left(a+b\right)-\mathrm{sin}\left(a-b\right)=2\mathrm{cos}a\mathrm{sin}b$
$\mathrm{cos}\left(a+b\right)+\mathrm{cos}\left(a-b\right)=2\mathrm{cos}a\mathrm{cos}b$
$\mathrm{cos}\left(a-b\right)-\mathrm{cos}\left(a+b\right)=2\mathrm{sin}a\mathrm{sin}b$
Now your problem reduces to $\mathrm{sin}{75}^{\circ }-\mathrm{cos}{75}^{\circ }$, to make use the above formula, we convert $\mathrm{cos}{75}^{\circ }=\mathrm{sin}{15}^{\circ }$ and we write 75 = 45+30 and 15 = 45-30.
So we have $\mathrm{sin}\left({45}^{\circ }+{30}^{\circ }\right)-\mathrm{sin}\left({45}^{\circ }-{30}^{\circ }\right)=2\mathrm{cos}{45}^{\circ }\cdot \mathrm{sin}{30}^{\circ }=2\cdot \frac{1}{\sqrt{2}}\cdot \frac{1}{2}=\frac{1}{\sqrt{2}}$
Alter: You need not break the given form even, as follows
$\mathrm{cos}{105}^{\circ }+\mathrm{sin}{75}^{\circ }=\mathrm{cos}{105}^{\circ }+\mathrm{cos}{15}^{\circ }=\mathrm{cos}\left(60+45\right)+\mathrm{cos}\left(60-45\right)$

Do you have a similar question?