If $\frac{\sin(x)}{a}=\frac{\cos(x)}{b}$ then $a\sin(2x)+b\cos(2x)=?$

Question

$b\sin(x)=a\cos(x)$;

$\tan(x)=\frac{a}{b}$.

I couldn't simplify after that.

I'm sure that there is an identity somewhere to solve it that I missed, so let me know.

Answer 1

Though there are lots of answers here, I'd like to put an answer with almost no computations. \begin{align} a\sin(2x)+b\cos(2x)&=\frac{a}{\sin(x)}(\sin(x)\sin(2x)+\cos(x)\cos(2x))\\ &=\frac a{\sin(x)}\cos(x)=b \end{align}

Answer 2

Perhaps the identities used Weierstrass's substitution will help: $$\sin 2x=\frac{2\tan x}{1+\tan^2 x}$$ $$\cos 2x=\frac{1-\tan^2 x}{1+\tan^2 x}$$ Leading to: $$2a\frac{a/b}{1+a^2/b^2}+b\frac{1-a^2/b^2}{1+a^2/b^2}$$ $$\frac{2a^2b}{a^2+b^2}+\frac{b^3-a^2b}{a^2+b^2}=\frac{b^3+a^2b}{a^2+b^2}=b$$

Answer 3

$\tan{x}=\frac{a}{b}$ and from here $$a\sin2x+b\cos2x=\frac{\frac{2a^2}{b}}{1+\frac{a^2}{b^2}}+\frac{b-\frac{a^2}{b}}{1+\frac{a^2}{b^2}}=b$$

Answer 4

Notice, that by right triangle trigonometry, if $\displaystyle \tan(x)=\frac{a}{b}$, we have $\sin(x)=\frac{a}{\sqrt{a^2+b^2}}$ and $\cos(x)=\frac{b}{\sqrt{a^2+b^2}}$.

We have $\sin(2x)=2\sin(x)\cos(x)=\frac{2ab}{a^2+b^2}$.

We also have that $\cos(2x)=\cos^2(x)-\sin^2(x) =\frac{b^2-a^2}{a^2+b^2}$

We finally have $a\sin(2x)+b\cos(2x)=\frac{2a^2b+b^3-a^2b}{a^2+b^2}=\frac{a^2b+b^3}{a^2+b^2}=\boxed{b}$

Answer 5

We have $b \sin (x) = a \cos (x).$ Thus $$a \sin(2x) + b \cos(2x) = 2\dfrac{a^2}{b} \cos^2(x) + 2b \cos^2(x) - b = 2\dfrac{a^2+b^2}{b} \cos^2(x) - b.$$ Now, we have $\tan(x) = \dfrac{a}{b}$. Hence $\cos^2(x) = \dfrac{1}{\tan^2(x) +1} = \dfrac{b^2}{a^2 +b^2}$.

Then we obtain $$a \sin(2x) + b \cos(2x) = 2\dfrac{a^2+b^2}{b}\dfrac{b^2}{a^2 +b^2} - b = b. $$

Answer 6

$$\frac{\sin(x)}{a} = \frac{\cos(x)}{b} = k$$ $$\sin(x) = ka$$ $$cos(x) = kb$$

$$\sin(2x) = 2\sin(x) \cdot \cos(x) = 2k^2ab$$

$$\cos(2x) = \cos^2(x) - \sin^2(x) = k^2b^2 - k^2a^2$$ $$a\sin(2x) = 2a\sin(x) \cdot \cos(x) = 2k^2a^2b$$

$$b\cos(2x) = b(\cos^2(x) - \sin^2(x)) = k^2b^3 - k^2a^2b$$ $$a\sin(2x) + b\cos(2x) = k^2a^2b + k^2b^3 = k^2b(a^2 + b^2) = b(k^2a^2 + k^2b^2) = b(1) = b$$

Answer 7

Alternatively: $$a=b\cdot \frac{\sin{x}}{\cos{x}}.$$ $$b\cdot \frac{\sin{x}}{\cos{x}}\cdot 2\sin{x}\cos{x}+b(1-2\sin^2{x})=b.$$