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I tried to solve it but I can’t get the answer. Please help me in proving this trig identity:

If $$x \cos\theta+y\sin\theta=a$$ $$x\sin\theta-y\cos\theta=b$$ then $$\tan\theta=\frac{bx+ay}{ax-by}$$

I've spent many hours trying. Thanks in advance.

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  • $\begingroup$ Please preview your post before submitting it.As I can see no image has been added to the post.Edit it. $\endgroup$
    – user517784
    Feb 16, 2018 at 3:34
  • $\begingroup$ Hint: consider $a/b$. $\endgroup$
    – David
    Feb 16, 2018 at 3:36
  • $\begingroup$ There is no need to link to an image. You can format the question using the advice here: math.stackexchange.com/help/notation $\endgroup$
    – David K
    Feb 16, 2018 at 3:47
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    $\begingroup$ Somehow in the many hours you tried, you had some ideas you can share with us? At least to help us avoid telling you something you already know? (My hint would be: write $bx+ay$ and $ax-by$ in terms of $x,$ $y,$ and $\theta.$) $\endgroup$
    – David K
    Feb 16, 2018 at 3:52

6 Answers 6

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For fun, here's a trigonograph:

enter image description here

$$\tan\theta = \tan(\phi + \psi) = \frac{\tan\phi+\tan\psi}{1-\tan\phi\tan\psi} = \frac{\;\dfrac{y}{x}+\dfrac{b}{a}\;}{\;1-\dfrac{y}{x}\dfrac{b}{a}\;}=\frac{ay+bx}{ax-by}$$

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We first try to find $\sin \theta $ in terms of $a, b, x, y$

Multiply equation first by $y$ and the equation $2^{nd}$ by $x$ and add the resultant equations to get $$\sin \theta= \frac {bx+ay}{x^2+y^2}$$

Now multiply the first equation by $x$ and second equation by $y$ to get following equations

$$x^2\cos \theta+xy\sin \theta=ax$$ And $$xy\sin\theta-y^2\cos\theta=by$$

Hence we get $$\cos\theta=\frac {ax-by}{x^2+y^2}$$ By subtracting resultant $2^{nd}$ equation from the resultant $1^{st}$ equation.

Using these values of $\sin\theta$ and $\cos\theta$ we get $$\tan\theta = \frac {bx+ay}{ax-by}$$

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$x \cos\theta + y\sin\theta = a, \,\, x \sin\theta - y\cos\theta = b$. Therefore:

$$\dfrac{x \cos\theta + y\sin\theta}{x \sin\theta - y\cos\theta} = \dfrac{a}{b}$$ $$\implies \dfrac{x + y\tan\theta}{x \tan\theta - y} = \dfrac{a}{b}$$ $$\implies bx + by\tan\theta = ax \tan\theta - ay$$ $$\implies (ax - by) \tan\theta = bx + ay$$ $$\implies \tan\theta = \dfrac{bx + ay}{ax - by}$$

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Hint:

Solve the two simultaneous linear equation for $\sin\theta,\cos\theta$

See https://brilliant.org/wiki/system-of-linear-equations/ or https://revisionmaths.com/gcse-maths-revision/algebra/simultaneous-equations

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$$x\cos \theta+y\sin\theta=a$$

$$\implies x+y\tan\theta=a\sec\theta$$

$$\implies \sec \theta=\frac{x+y\tan\theta}{a} ...(1)$$

$$x\sin\theta-y\cos\theta=b$$

$$\implies x\tan\theta-y=b\sec\theta$$

$$\implies \sec\theta=\frac{x\tan\theta-y}{b}...(2)$$ From $1$ and $2$,

$$\frac{x+y\tan\theta}{a}=\frac{x\tan\theta-y}{b}$$

$$\implies bx+by\tan\theta=ax\tan\theta-ay$$

$$\implies \tan\theta(ax-by)=bx+ay$$

$$\tan\theta=\frac{bx+ay}{ax-by}$$

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We may write the given equations as: $$\begin{pmatrix}\cos \theta & - \sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \begin{pmatrix}x \\ - y \end{pmatrix} = \begin{pmatrix}a \\ b \end{pmatrix} $$ and recognise that the first matrix represents a counterclockwise rotation by $\theta$ about the origin. We can then use the formula for the tangent of the difference of two angles to see that $$\tan \theta = \frac{\frac{b}{a}- \frac{-y}{x}}{1+\frac{b}{a} \frac{-y}{x}}=\frac{bx+ay}{ax-by}$$

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