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Given that:

$$ 2\cos(x + 50) = \sin(x + 40) $$

Show, without using a calculator, that:

$$ \tan x = \frac{1}{3}\tan 40 $$

I've got the majority of it:

$$ 2\cos x\cos50-2\sin x\sin50=\sin x\cos40+\cos x\sin40\\ $$ $$ \frac{2\cos50 - \sin40}{2\sin50 + \cos40}=\tan x $$

But then, checking the notes, it says to use $\cos50 = \sin40$ and $\cos40 =\sin50$; which I don't understand. Could somebody explain this final step?

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    $\begingroup$ $\cos(90-x)=\sin(x)$ for any $x$ because $\cos(90)=0$ and $\sin(90)=1$ (in degrees not radians, of course) $\endgroup$ – Ewan Delanoy Jul 14 '15 at 15:27
  • $\begingroup$ @EwanDelanoy Got it, thanks! $\endgroup$ – hohner Jul 14 '15 at 15:30
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Where you have left of using $\cos(90^\circ-x)=\sin x,\sin(90^\circ-x)=\cos x$

$$\frac{2\cos50^\circ - \sin40^\circ}{2\sin50^\circ + \cos40^\circ}=\frac{2\sin40^\circ- \sin40^\circ}{2\cos40^\circ + \cos40^\circ}=?$$

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Using the well-known identity $$\cos (90^{\circ} - x) = \sin x$$ since the cosine function is just a $90^{\circ}$ horizontal translation of the sine function.

Taking $x = 40^{\circ}$ gives us $\cos 50^{\circ} = \sin 40^{\circ}$ and letting $x=50^{\circ}$ establishes the second result. Then $$\begin{align}2\cos x\cos50-2\sin x\sin50=\sin x\cos40+\cos x\sin40 \\ \iff 2 \cos x \sin 40 - 2\sin x\cos 40 = \sin x \cos 40 +\cos x \sin 40 \\ \iff 2 \cos x\sin 40 - \cos x\sin 40 = \sin x\cos 40 + 2\sin x\cos 40 \\ \iff \tan 40 \cos x = 3 \sin x \end{align}$$

so that $$\tan x = \frac{\tan 40^{\circ}}{3}$$

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I guess your angle $x$ is between $0^{\circ}$ and $90^{\circ}$. If that is so, recall the following:

(a) $Cos(x)=Sin(90-x)$; (b) $Sin(x)=Cos(90-x)$; (c) $Tan(x)=\frac{Sin(x)}{Cos(x)}$.

(a), (b) and (c) are standard trig formula, from where you can get that :$$Cos(50)^{\circ}=Sin(90-50)^{\circ}=Sin40^{\circ},$$ $$Cos(40)^{\circ}=Sin(90-40)^{\circ}=Sin50^{\circ},$$ $$Sin(50)^{\circ}=Cos(90-50)^{\circ}=Cos40^{\circ}$$ and $$Sin(40)^{\circ}=Cos(90-40)^{\circ}=Cos50^{\circ}.$$

You know that $2Cos(x+50)=Sin(x+40)$. So apply either (a) or (b). For instance, using (b) shows that $Sin(x+40)=Cos(90-(x+40))=Cos(50-x)$. Hence, $$2Cos(x+50)=Cos(50-x).$$ So, $$2Cos(x+50)-Cos(50-x)=0------(+)$$ You can now use the following: $$Cos(A+B)=CosACosB-SinASinB-----(1)$$ $$Cos(A-B)=CosACosB+SinASinB------(2).$$ For instance, $$2Cos(x+50)=2CosxCos50-2SinxSin50$$ and $$Cos(50-x)=Cos50Cosx+Sin50Sinx$$.

Now, eqn (+) gives: $$CosxCos50-SinxSin50=0-----(++)$$

Recall that $Cos(A+B)=CosACosB-SinASinB$. So (++) becomes: $$Cos(x+50)^{\circ}=0.$$ As $Cos 90^{\circ}=0$, we know that $$x=40^{\circ}.$$

You can finish up from there!

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    $\begingroup$ Note that in reality, what you are actually proving doesn't hold since $x=40^{\circ}$, and there is no way that $$tan40^{\circ}=\frac{1}{3}tan40^{\circ}.$$ $\endgroup$ – Chuks Jul 14 '15 at 16:38

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