3

Find the value using a calculator: $\cos 75°$

At first I thought all I need is to separate the simpler known values like this:

$\cos 75^\circ = \cos 30°+\cos45° = {\sqrt3}/{2} + {\sqrt2}/{2} $

$= {(\sqrt3+\sqrt2)}/{2} $ This is my answer which translates to= $1.5731$ by calculator

However, when I used the calculator directly on $\cos 75°$, I get $0.2588$.

Where I am going wrong?

vadim123
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Sylvester
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    $\cos (x+y) \ne \cos x + \cos y$ in general. – njguliyev Oct 24 '13 at 19:08
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    Whenever you "conjecture" such formula, try it for some values. For example, $\cos \pi=-1$, but $\cos 2\pi=1$. Clearly, $1\neq -2$, so the formula cannot be right! – Pedro Oct 24 '13 at 19:25
  • @tylerc0816 That's not true. You need the extra hypothesis that $f$ is continuous. – Pedro Oct 24 '13 at 19:26
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    It's good to do a 'sanity check' for this sort of thing. If it were true that $\cos(x+y)=\cos x + \cos y$ then we'd have $$1 = \cos 0 = \cos(0+0) = \cos 0 + \cos 0 = 1+1 = 2$$ but as I'm sure you're aware, $1 \ne 2$! – Clive Newstead Oct 24 '13 at 19:29
  • Assuming $\cos(x+y)=\cos(x)+\cos(y)$, continuity of $\cos$ and boundedness of $\cos$ it is easy to show that $\cos(x)=0$ for all $x$. Since $0.2588\ne0$, something must be wrong. – Carsten S Oct 24 '13 at 19:38
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    I want to look past the immediate question. I'm concerned about why you thought that $\cos(x + y) = \cos(x) + \cos(y)$. You made a serious mistake and it wasn't about trigonometry: you assumed that $\cos$ has a property ("distributivity over addition") and you didn't check. Most functions don't have "nice" properties. You should never assume anything about the properties of any mathematical entity. Research them, or prove them yourself, or have them proved to you. If you ever see a technique being used you haven't seen before, challenge it. That will open all the doors in math for you. – Euro Micelli Oct 25 '13 at 02:46

7 Answers7

13

You used a formula $\cos(x+y) = \cos(x) + \cos(y)$ which is false. The correct formula is: $$\cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y)$$

Adam
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9

The answer is simply: No and to see this take $y=0$ we find $$\cos(x)=\cos(x)+1$$ which's obviously false.

5

You can simply plot $cos(x+y)-(cos(x)+cos(y))$ to have your answer:

enter image description here

Franck Dernoncourt
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2

Alternatively: cos(A+B) = cosAcosB-sinAsinB with A = 45 and B is 30

imranfat
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1

As mentioned in other answers, $\cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y)$.

We also have $\cos(x)+\cos(y)=2\cos\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)$.

Since $\cos(0)=1$, we get $\cos(0+0)=1\ne2=\cos(0)+\cos(0)$.

robjohn
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0

The maximum value of cos(x+y)=1.

The maximum values of cos(x) and cos(y) are 1.

Since cos(x)+cos(y)=2cos(x+y) for some values of x and y, it cannot be the case that cos(x)+cos(y)=cos(x+y) except for special values of x and y.

Marc
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-1

If you consider all $x,y$ in $\mathbb R$ such that $\cos(x+y)=\cos(x)+\cos(y)$ holds true, you get this.

your solution

user_194421
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Carlos Cornejo
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    As it’s currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. – Community Dec 21 '21 at 22:26