# Find the value of: sin(100) + sin(125) + sin(100)•sin(125)

This question has been troubling me since last few days. Can anyone pls solve without a calculator? Thanks in advance. All values are in degrees. Also, the question has the following options: 0,0.5,-1,1,2

Ps: the question asked to solve all the problems which were possible to solve without a calculator and this is a part of it. I don't know whether it is possible to solve this without a calculator. Here the second part of the question: Find the value of: tan(100) + tan(125) + tan(100)•tan(125)

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Can you solve it with a calculator? In all known three standard trig modes (rad, deg, grad) one of the options appears. –  gammatester May 6 at 11:00
I forgot to mention that the values are in degrees. Sorry. –  Harshal Gajjar May 6 at 11:14
And I also forgot to add the correct answer as an option. Sorry (again) –  Harshal Gajjar May 6 at 11:16
Anyways I just wanted to know whether solving this is possible without a calculator or not? –  Harshal Gajjar May 6 at 11:18
I think it's not possible to solve this without a calculator, because the answer coming is 2.-- and not exactly 2. You can try out this question as the answer coming is exactly 1 but I am unable to reach it. :) math.stackexchange.com/questions/783436/… –  Harshal Gajjar May 6 at 11:25

Still incorrect problem. Without a calculator you know that both $\sin$ values are in the interval $(\frac{\sqrt{2}}{2},1)$, so the expression is at least $2\frac{\sqrt{2}}{2} + (\frac{\sqrt{2}}{2})^2 \approx 1.4 + 0.5 = 1.9\dots$. Therefore the best guess option would be 2. But this far away from the true value $2.610667\dots$

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This is an extremely ugly computation without a calculator. The exact value starts out $$2.6106670814127985928729299199359494568369290021781737 \ldots$$

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First all terms are less than $1$, so the sum is less than $3$. And they are all positive.

Also, all angles are relatively close to $90°$, for which sine is $1$, so the sum must be close to $3$, and certainly greater than $2$. Actually, it's approximately $2{,}61$.

To be more accurate, $180°-125°=55°$, almost $60°$, for which sine is $\sqrt{3}/2 \approx 0{,}866$ (you should know $\sqrt{3} \approx 1{,}732$). For $100°$, angle is much closer to $90°$, so its sine is still closer to $1$. Hence the sum of the two first terms is close to $1{,}8$ (actually it's $\approx 1{,}804$). Then the last term is close to $1 \times 0{,}866$, and you could have guessed the $2{,}6$.

And to be extra accurate, you could have done some simple approximations.

First, $\sin 100° = \cos 10°$, and since the angle $10°$ is small, you can use the approximation (in radians)

$$\cos \epsilon \approx 1-\frac{\epsilon^2}2$$

Or in degrees

$$\cos \epsilon ° \approx 1-\frac{(\epsilon ° \times \pi/180)^2}2$$

And you can compute by hand that $180/\pi \approx 180/3{,}14 \approx 57$, and that $57 \times 57 = 3249 \approx 3250$. Thus your cosine is close to $1 - \frac{100}{6500}=1 - \frac{1}{65} \approx 0{,}985$.

That's for $\sin 100°$, now $\sin 125° = \sin 55°$. We know $\sin 60° \approx 0{,866}$ and the derivative is $\cos 60° = 0{,}5$. So, a linear approximation gives

$$\sin 55° \approx \sin 60° - \cos 60° \times (5° \times \pi/180) \approx 0{,}866 - \frac{5}{57 \times 2}\\=0{,}866 - \frac{5}{114} \approx 0{,}866 - 0{,}044 = 0{,}822$$

Then the first two terms add to $0{,}985 + 0{,}822 = 1{,}807$, and the last is $\approx 0{,}985 \times 0{,}822 \approx 0{,}810$, so the whole sum is approximately

$$1{,}807 + 0{,}810 = 2{,}617$$

Given that the true value is $2{,}6106670...$, it's not too bad.

For tangents, it's more interesting. You have $\tan (a+b) =\frac{\tan a + \tan b}{1-\tan a \tan b}$, so if $\tan (a+b)=1$, you have then $\tan a + \tan b = 1- \tan a \tan b$, or

$$\tan a + \tan b + \tan a \tan b = 1$$

Here, $a+b=225°=180°+45°$, and since tangent is $\pi$-periodic (or $180°$-periodic if you prefer), $\tan 225° = \tan 45°=1$, so the previous equality holds and your sum is actually $1$.

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