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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!

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4 Answers 4

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


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

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

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    $\begingroup$ +1 , Answer checks out numerically. My Suggestion : It might be great to include the four formulas you used here ! $\endgroup$
    – Prem
    Jun 10, 2023 at 11:22
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    $\begingroup$ Thank you, yes, I will add it. $\endgroup$
    – MathFail
    Jun 10, 2023 at 11:23
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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: $\;\color{green}{\cos p\!-\!\cos q=\!-2\sin\left(\dfrac{p\!+\!q}2\right)\sin\left(\dfrac{p\!-\!q}2\right)}$

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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

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

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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?

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