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What is $\tan15\cdot \tan 30 \cdot \tan 45 \cdot \tan 60 \cdot \tan 75$ equal to (in degrees)?

Here is how I tried to solve it:

I assumed $x = \tan 15 (x= 0.27)$, so I rewrote it as:

$x\cdot 2x \cdot 3x \cdot 4x \cdot 5x = 120x^5$

$0.27^5 = 0.001323$

$120 \cdot 0.001323 = 0.16$

But Google Calculator gives me $-1.19576279$

What is the right way to calculate this trig expression?

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Why are you thinking that the tangent values increase linearly? –  Dilip Sarwate May 18 '12 at 19:59
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Tangent is not linear! P.S. Google Calculator computed the result using radians, not degrees. –  Arturo Magidin May 18 '12 at 20:03
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4 Answers

Note that if $0\leq \alpha\leq \frac{\pi}{2}$ ($0^{\circ}$ to $90^{\circ}$), then $$\cos\left(\frac{\pi}{2}-\alpha\right) = \cos\frac{\pi}{2}\cos\alpha + \sin\frac{\pi}{2}\sin\alpha = \sin\alpha.$$ In terms of degrees, $\cos(90-\alpha) = \sin(\alpha)$.

So if you look at $\tan(15^{\circ})$ and $\tan(75^{\circ}) = \tan(90^{\circ}-15^{\circ})$, we have: $$\tan(15)\tan(75) = \frac{\sin(15)}{\cos(15)}\;\frac{\sin(75)}{\cos(75)} = \frac{\sin(15)}{\sin(75)}\;\frac{\sin(75)}{\sin(15)} = 1.$$

Similarly with $\tan 30 \tan 60$. So the entire product is just equal to $\tan(45)$. And $\tan(45^{\circ})$ is equal to...

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First note that $\tan(2x) \neq 2 \tan(x)$ and similarly, $\tan(3x) \neq 3 \tan(x)$ and in general, $\tan(kx) \neq k \tan(x)$ (unless when $k=1$). For this problem, all you need are the following identities: $$\tan(x^{\circ}) = \cot(90^{\circ}-x^{\circ})$$ and $$\cot(y^{\circ}) = \frac1{\tan(y^{\circ})}$$ The above two identities combine to give us $$\tan(x^{\circ}) = \frac1{\tan(90^{\circ}-x^{\circ})}$$ For instance, $\tan(15^{\circ}) = \dfrac1{\tan(75^{\circ})}$.

Can you now complete it?

Also, in calculators, typically, the input for any trigonometric function is in radians and not degrees unless you specify it explicitly. Hence, what google calculator evaluated is $$\tan(15^c) \tan(30^c) \tan(45^c) \tan(60^c) \tan(75^c)$$ where $x^c$ stands for $x$ radians. It is not the same as $$\tan(15^{\circ}) \tan(30^{\circ}) \tan(45^{\circ}) \tan(60^{\circ}) \tan(75^{\circ})$$

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If you know the formulas for $\tan(x+y)$ and $\tan(x-y)$, apply them to the case when $x = 45$ degrees or $\pi/4$ radians, remembering that $\tan(45) = 1$.

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EDIT : Arturo's solution is really cool. Use that instead, I'll let my solution stay just as a reference.

$$\tan(2A) \neq 2\times \tan(A)$$ You might want to read/revise double angle formulas. Assuming you need to do this without a calculator,

$$\begin{align*} \tan 30 &= \frac{1}{\sqrt{3}}\\ \tan 45 &= 1\\ \tan 60 &= \sqrt{3}\\ \tan 60 &= \sqrt{3}\\ \tan(A+B) &= \frac{\tan(A)+\tan(B)}{1-\tan(A)\tan(B)},\\ \tan(15+15) &= \tan(30) \\ &= \frac{1}{\sqrt{3}} \\ &= \frac{\tan(15)+\tan(15)}{1-\tan(15)\tan(15)} \end{align*}$$

Solving $\frac{2x}{1-x^2} = \frac{1}{\sqrt{3}}$ will give you x = 0.2679

Similarly, you can write $\tan(75) = \tan(90-15)$ and you can work out the product.

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