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 $\sin^2\alpha = 1 -\cos^2\alpha = 1-(0.2-\cos\alpha)^2$?
$\endgroup$ 37 Answers
$\begingroup$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: $$x^2-sx+p=x^2-0.2x-0.48=0$$
$\endgroup$ $\begingroup$You also know that $\sin^2 \alpha + \cos^2 \alpha=1$, so square what you are given, getting $\sin^2 \alpha + 2 \sin \alpha \cos \alpha + \cos^2 \alpha = 0.04, 2 \sin \alpha \cos \alpha=-0.96=\sin (2\alpha)$
$\endgroup$ 3 $\begingroup$$$\cos \alpha + \sin \alpha = \sqrt{2} (\frac{\cos \alpha}{\sqrt{2}} + \frac{\sin \alpha}{\sqrt{2}}) = \sqrt{2}(\sin \frac{\pi}{4} \cos\alpha + \cos \frac{\pi}{4} \sin \alpha) = \sqrt{2}\sin(\frac{\pi}{4} + \alpha) = .2 $$
Taking the inverse sin of each side yields $$ \alpha = 2.2143 + 2 \pi n_1 \mid n_1 \in \mathbb{Z} \,\,\,\text{or}\,\,\, \alpha = 2\pi n_2 - .643501 \mid n_2 \in \mathbb{Z}$$
$\endgroup$ 1 $\begingroup$$$\sin\alpha + \cos\alpha = 0.2$$
$$\sin\alpha + \sqrt {1-\sin^2\alpha}= 0.2$$
$$\sqrt{1-\sin^2\alpha}= 0.2-\sin\alpha$$
$$1-\sin^2\alpha=0.04-0.4\sin\alpha+\sin^2\alpha$$
$$2\sin^2\alpha-0.4\sin\alpha-0.96=0$$
$$\sin^2\alpha-0.2\sin\alpha-0.48=0$$
$$\sin\alpha=\frac{0.2\pm1.392...}{2}$$
$$2\sin\alpha=1,592...$$ $$2\sin\alpha=-1,192...$$
$\endgroup$ $\begingroup$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 $(2)$:$$2\sin ^{ 2 }{ \alpha } -1=0.2x$$
$\endgroup$ $\begingroup$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.
$\endgroup$ $\begingroup$\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
$\endgroup$
