# Question

Evaluate $$\dfrac{2\cos 80^\circ-\sin 70^\circ}{\cos 70^\circ}$$

## My Working

I tried converting $$80^\circ$$ to $$2\cdot 40^\circ$$ and $$70^\circ$$ to $$30^\circ+40^\circ$$. This led to

$$\frac{2\cos (2\cdot 40^\circ)-(\frac{\sqrt3}{2}\sin40^\circ+\frac12\cos40^\circ)}{\frac{\sqrt3}{2}\cos40^\circ-\frac12\sin40^\circ}$$

I do not know which double angle formula to use for $$2\cos 80^\circ$$, and how to simply the expression. Could someone please help? Thank you!

\begin{align}\dfrac{2\cos 80^\circ-\sin 70^\circ}{\cos 70^\circ}&=\dfrac{2\sin 10^\circ-\cos 20^\circ}{\sin 20^\circ}\\ \\ &=\frac{2}{\sin 20^\circ}(\sin10^\circ-\sin30^\circ\cos20^\circ)\\ \\ &=\frac{2}{\sin 20^\circ}\left(\sin10^\circ-\frac{\sin50^\circ+\sin10^\circ}2\right)\tag{1}\\ \\ &=\frac{\sin10^\circ-\sin50^\circ}{\sin 20^\circ}\\ \\ &=\frac{-2\sin(20^\circ)\cos30^\circ}{\sin 20^\circ}\tag{2}\\ \\ &=-\sqrt3 \end{align}

$$(1)$$ use: $$\sin A\cos B=\dfrac12\left[\sin(A+B)+\sin(A-B)\right]$$

$$(2)$$ use: $$\sin A-\sin B=2\sin\left(\dfrac{A-B}2\right)\cos\left(\dfrac{A+B}2\right)$$

• +1 , Answer checks out numerically. My Suggestion : It might be great to include the four formulas you used here !
– Prem
Jun 10 at 11:22
• Thank you, yes, I will add it. Jun 10 at 11:23

Proof with $$4$$ steps instead of $$6$$ steps.

$$\require{cancel}\dfrac{2\cos80\unicode{176}-\sin70\unicode{176}}{\cos70\unicode{176}}=$$

$$=\dfrac{\cos80\unicode{176}+\sin10\unicode{176}-\sin70\unicode{176}}{\sin20\unicode{176}}=$$

$$\dfrac{\cos80\unicode{176}-\color{blue}{\cancel2}\cos40\unicode{176}\color{blue}{\cancel{\sin30\unicode{176}}}}{\sin20\unicode{176}}=\quad\color{brown}{(1)}$$

$$=\dfrac{-2\sin60\unicode{176}\color{blue}{\cancel{\sin20\unicode{176}}}}{\color{blue}{\cancel{\sin20\unicode{176}}}}=\quad\color{green}{(2)}$$

$$=-\sqrt3\,.$$

$$\color{brown}{(1)}\;$$ we have used: $$\;\color{brown}{\sin p\!-\!\sin q=2\cos\left(\dfrac{p\!+\!q}2\right)\sin\left(\dfrac{p\!-\!q}2\right)}$$

$$\color{green}{(2)}\;$$ we have used: $$\;\color{green}{\cos p\!-\!\cos q=\!-2\sin\left(\dfrac{p\!+\!q}2\right)\sin\left(\dfrac{p\!-\!q}2\right)}$$

We have by basic trigonometric relations

• $$\cos 80^\circ=\cos (60^\circ+20^\circ)=\cos 60^\circ\cos 20^\circ-\sin 60^\circ\sin 20^\circ=\frac{1}2\cos 20^\circ-\frac{\sqrt 3}2 \sin 20^\circ$$
• $$\sin 70^\circ =\cos(90^\circ-70^\circ)=\cos 20^\circ$$
• $$\cos 70^\circ =\sin(90^\circ-70^\circ)=\sin 20^\circ$$

then

$$\require{cancel} \dfrac{2\cos 80^\circ-\sin 70^\circ}{\cos 70^\circ}=\dfrac{\color{red}{\cancel{\cos 20^\circ}} \color{blue}{-\sqrt 3 \cancel{\sin 20^\circ}}\color{red}{-\cancel{\cos 20^\circ}}}{\color{blue}{\cancel{\sin 20^\circ}}}=-\sqrt 3$$

Generalization

$$2\cos(150^\circ -t)=2(\cos150^\circ\cos t+\sin150^\circ\sin t)=\sin t-\sqrt3\cos t$$

as

• $$\cos150^\circ=\cos(180^\circ-30^\circ)=-\cos30^\circ=?$$
• $$\sin150^\circ=\sin(180^\circ-30^\circ)=+\sin30^\circ=?$$

Can you recognize $$t$$ here?