4

What is the value of: $$\sin100^\circ+\cos70^\circ\over\cos80^\circ-\cos20^\circ$$

I've done trigonometry in my earlier years of high school but I forgot a lot of rules. This is where I'm stuck on this problem:

$\large{{\sin100^\circ+\cos70^\circ\over\cos80^\circ-\cos20^\circ}=\\{\sin(90^\circ+10^\circ)+\cos(60^\circ+10^\circ)\over\cos(90^\circ-10^\circ)-\cos(30^\circ+10^\circ)}=\\{\sin90^\circ\cos10^\circ+\cos90^\circ\sin10^\circ+\cos60^\circ\cos10^\circ-\sin60^\circ\sin10^\circ\over\cos90^\circ\cos10^\circ+\sin90^\circ\sin10^\circ-\cos30^\circ\cos10^\circ+\sin30^\circ\sin10^\circ}=\\{\cos10^\circ+{1\over2}\cos10^\circ-{\sqrt3\over2}\sin10^\circ\over\sin10^\circ-{\sqrt3\over2}\cos10^\circ+{1\over2}\sin10^\circ}=\\{{3\over2}\cos10^\circ-{\sqrt3\over2}\sin10^\circ\over{3\over2}\sin10^\circ-{\sqrt3\over2}\cos10^\circ}}$

Not sure what I should do further with this.

Aleksa
  • 861

3 Answers3

5

It is: $$\frac{\sin 100^\circ +\cos 70^\circ}{\cos 80^\circ-\cos 20^\circ}=\frac{\sin 80^\circ +\sin 20^\circ}{\cos 80^\circ-\cos 20^\circ}=\frac{2\sin 50^\circ\cos 30^\circ}{-2\sin 50^\circ\sin 30^\circ}=-\sqrt{3}.$$

farruhota
  • 31,482
2

Given $$\dfrac{\sin100+\cos100}{\cos80-\cos20}$$

Now to solve the denominator use the formula $\cos A-\cos B=-2\sin\left(\dfrac{A+B}{2}\right)\sin\left(\dfrac{A-B}{2}\right)$

$\cos80-\cos20=-2\sin\left(\dfrac{80+20}{20}\right)\sin\left(\dfrac{80-20}{2}\right)$ $$=\dfrac{\sin100+\cos70}{-2\sin\left(\dfrac{80+20}{20}\right)\sin\left(\dfrac{80-20}{2}\right)}$$ $$=-\dfrac{\sin100+\cos70}{\sin50}$$

Now to solve the numerator use the identity $\sin A+\sin B=2\cos\left(\dfrac{A-B}{2}\right)\sin\left(\dfrac{A+B}{2}\right)$

$\sin100+\sin20=2\cos\left(\dfrac{100-20}{2}\right)\sin\left(\dfrac{100+20}{2}\right)$ $$=-\dfrac{2\cos\left(\dfrac{100-20}{2}\right)\sin\left(\dfrac{100+20}{2}\right)}{\sin50}$$ $$=-\dfrac{2\cos40\sin60}{\sin50}$$ $$=-\dfrac{2\cos40\cdot\dfrac{\sqrt{3}}{2}}{\sin50}$$ $$-\dfrac{\sqrt{3}\sin(90-40)}{\sin50}=-\sqrt{3}$$

Therefore, $$\dfrac{\sin100+\cos100}{\cos80-\cos20}=-\sqrt{3}$$

Key Flex
  • 9,475
  • 7
  • 17
  • 34
2

Since $\cos x=\sin(90-x)$, $$\frac{\sin100+\cos70}{\cos80-\cos20}=\frac{2\sin\frac{100+20}2\cos\frac{100-20}2}{2\sin\frac{80+20}2\sin\frac{20-80}2}=-\frac{\sin60\cos40}{\sin50\sin30}=-\frac{\frac{\sqrt3}2\sin50}{\frac12\sin50}=-\sqrt3$$ using $$\sin x+\sin y=2\sin\frac{x+y}2\cos\frac{x-y}2\\\cos x-\cos y=-2\sin\frac{x+y}2\sin\frac{x-y}2$$

ə̷̶̸͇̘̜́̍͗̂̄︣͟
  • 27,005