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If $\sin\alpha + \cos\alpha = 0.2$, find the numerical value of $\sin2\alpha$. How do I find a value for $\sin\alpha$ or $\cos\alpha$ so I can use a double angle formula?

I know how to solve a problem like

"If $\cos\alpha = \frac{\sqrt{3}}{2}$ , find $\sin2\alpha$"

by using the 'double angle' formula: $\sin2\alpha = 2\sin\alpha\cos\alpha$ like this:

Start by computing $\sin\alpha$ $$\sin^2\alpha = 1 -\cos^2\alpha = 1-(\frac{\sqrt{3}}{2})^2 = \frac{1}{4}$$ so $$\sin\alpha = \pm\frac{1}{2}$$ then it's just a simple matter of plugging $\sin\alpha = \pm\frac{1}{2}$ and $\cos\alpha=\frac{\sqrt{3}}{2}$ into $$\sin2\alpha = 2\sin\alpha\cos\alpha$$ to get $$\sin2\alpha = \pm\frac{\sqrt{3}}{2}$$

Where I can not make progress with the question

"If $\sin\alpha + \cos\alpha = 0.2$, find the numerical value of $\sin2\alpha$".

Is how do I find a value for $\sin\alpha$ or $\cos\alpha$ so I can use a double angle formula?

What I have tried:

If $\sin\alpha+\cos\alpha = 0.2$ then $\sin\alpha=0.2-\cos\alpha$ and $\cos\alpha=0.2-\sin\alpha$. Should I start by by computing $\sin\alpha$ using $\sin^2\alpha = 1 -\cos^2\alpha = 1-(0.2-\cos\alpha)^2$?

mikoyan
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  • $2 \sin\alpha = 0.4 - 2 \cos \alpha$, and $\sin^2 \alpha + \cos^2 \alpha = 1$. Two equations, two variables. –  Jul 24 '13 at 17:15
  • Do you really mean $2 \sin \alpha$? Because you give many indications that you are looking for $\sin 2\alpha$ instead. Are you aware that these are different quantities? – Erick Wong Jul 24 '13 at 21:50
  • You are quite correct. I did mean $\sin2\alpha$. I will edit the question. – mikoyan Jul 25 '13 at 23:20

7 Answers7

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We have $$s^2=(\cos\alpha+\sin\alpha)^2=1+2\cos\alpha\sin\alpha=(0.2)^2$$ so we find $$p=\sin\alpha\cos\alpha=-0.48$$ hence $\sin\alpha$ and $\cos\alpha$ are roots of the quadratic equation: $$x^2-sx+p=x^2-0.2x-0.48=0$$

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You also know that $\sin^2 \alpha + \cos^2 \alpha=1$, so square what you are given, getting $\sin^2 \alpha + 2 \sin \alpha \cos \alpha + \cos^2 \alpha = 0.04, 2 \sin \alpha \cos \alpha=-0.96=\sin (2\alpha)$

Ross Millikan
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  • I understand how you got the value of $\sin2\alpha$ (that's what even I thought of doing). The question however asks for the minimum value of $\sin2\alpha$. Is that minimum redundant? Because, doing this way, we reach only one answer. – Parth Thakkar Jul 28 '13 at 17:47
  • Squaring goes one way but not the other. There can't be any other answers. With this argument, we have not shown that this value works-it might be spuriously introduced by the squaring. But you can observe that $\sin 0 + \cos 0 =1, \sin( -\frac \pi2) + \cos( -\frac \pi2)=-1$ and the mean value theorem to argue there must be a solution. – Ross Millikan Jul 29 '13 at 01:02
  • @RossMillikan. No. Squaring will give extraneous values for $\sin \alpha$ and $\cos \alpha$, but, you have definitely proved that, if $\sin\alpha + \cos\alpha = 0.2$, then $\sin 2\alpha = -0.96$. In the other direction, $\sin 2\alpha = -0.96$ implies $\sin\alpha + \cos\alpha = \pm 0.2$ which has nothing to do with the original problem. – Steven Alexis Gregory Aug 05 '15 at 06:53
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$$\cos \alpha + \sin \alpha = \sqrt{2} (\frac{\cos \alpha}{\sqrt{2}} + \frac{\sin \alpha}{\sqrt{2}}) = \sqrt{2}(\sin \frac{\pi}{4} \cos\alpha + \cos \frac{\pi}{4} \sin \alpha) = \sqrt{2}\sin(\frac{\pi}{4} + \alpha) = .2 $$

Taking the inverse sin of each side yields $$ \alpha = 2.2143 + 2 \pi n_1 \mid n_1 \in \mathbb{Z} \,\,\,\text{or}\,\,\, \alpha = 2\pi n_2 - .643501 \mid n_2 \in \mathbb{Z}$$

Anthony Peter
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    Hint: If you precede your trig functions with backslashes, they come out in the correct font. So \cos gives $\cos$ instead of $cos$ and so on. – Ross Millikan Jul 24 '13 at 17:25
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If $\sin\alpha + \cos\alpha = 0.2$, then squaring both sides and simplifying will produce $2 \sin \alpha \cos \alpha = -0.96$.

It follows that $\sin^2 \alpha - 2 \sin(\alpha) \cos(\alpha) + \cos^2 \alpha = 1.96$, from which we conclude $\sin\alpha - \cos\alpha = \pm 1.4$

From $\left\{ \begin{array}{l} \sin\alpha + \cos\alpha = 0.2\\ \sin\alpha - \cos\alpha = 1.4 \end{array} \right \}$ we find $(\sin \alpha, \cos \alpha) = (0.8, -0.6)$

From $\left\{ \begin{array}{l} \sin\alpha + \cos\alpha = 0.2\\ \sin\alpha - \cos\alpha = -1.4 \end{array} \right \}$ we find $(\sin \alpha, \cos \alpha) = (-0.6, 0.8)$

Note that neither of these solutions is extraneous.

Steven Alexis Gregory
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$$\sin\alpha + \cos\alpha = 0.2$$

$$\sin\alpha + \sqrt {1-\sin^2\alpha}= 0.2$$

$$\sqrt{1-\sin^2\alpha}= 0.2-\sin\alpha$$

$$1-\sin^2\alpha=0.04-0.4\sin\alpha+\sin^2\alpha$$

$$2\sin^2\alpha-0.4\sin\alpha-0.96=0$$

$$\sin^2\alpha-0.2\sin\alpha-0.48=0$$

$$\sin\alpha=\frac{0.2\pm1.392...}{2}$$

$$2\sin\alpha=1,592...$$ $$2\sin\alpha=-1,192...$$

Adi Dani
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Hint: $$\sin { \alpha } +\cos { \alpha } =0.2,\quad \quad (1)\\ \sin { \alpha } -\cos { \alpha } =x,\quad\ \quad (2)$$

Sum up $(1)$ and $(2)$: $$2\sin { \alpha } =0.2+x$$ Multiply $(1)$ and $(2)$:$$2\sin ^{ 2 }{ \alpha } -1=0.2x$$

newzad
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\begin{align} \sin \alpha+\cos \alpha &= 0.2 \\ (\sin \alpha+\cos \alpha)^2 &= 0.04 \\ \sin^2 \alpha + \cos^2 \alpha + 2\sin \alpha \cos\alpha &= 0.04 \\ 1+\sin 2\alpha &= 0.04 \\ \sin 2\alpha &= -0.96 \end{align}

and that would be straight forward to proceed

Ng Chung Tak
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Sivakumar Krishnamoorthi
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