4
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

$\endgroup$
4
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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 $$\sin x+\sin y=2\sin\frac{x+y}2\cos\frac{x-y}2\\\cos x-\cos y=-2\sin\frac{x+y}2\sin\frac{x-y}2$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.