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Given $\tan \theta =b/a$, then find $a\cos2\theta+b\sin2\theta$ in terms of $a$ and $b$.

I tried to solve the problem by first converting $\sin2\theta$ and $\cos2\theta$ in the $\tan$ terms (applying formula) and then simplifying it. But I did not get the correct answer, which is $a$.

Then I tried to substituting like this

$$a\cos2\theta + \tan\theta\sin2\theta.$$

I got a quadratic equation which I solved and substituted the values but failed to get the correct answer.

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    $\begingroup$ @TheGreatSeo I think that is $a \cdot \cos{2\theta}$, not $\arccos{2\theta}$. $\endgroup$
    – user124708
    Jul 31, 2014 at 11:45
  • $\begingroup$ you changed the question. -1 $\endgroup$
    – MonK
    Jul 31, 2014 at 12:25
  • $\begingroup$ Looks simple. $\endgroup$
    – hola
    Jul 31, 2014 at 12:34

4 Answers 4

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Hint

You could use the tangent half-angle substitution (Weierstrass substitution). Using $t=\tan(\theta)$, you have $$\sin(2\theta)=\frac{2 t}{1+t^2}$$ $$\cos(2\theta)=\frac{1-t^2}{1+t^2}$$

I am sure that you can take from here.

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  • $\begingroup$ I tried that also please read my whole question. Maybe I made a mistake somewhere but finally I got $\endgroup$
    – geek101
    Jul 31, 2014 at 11:58
  • $\begingroup$ (a((a^2-b^2)+2a(b)^2)/a^2+b^2 $\endgroup$
    – geek101
    Jul 31, 2014 at 12:02
  • $\begingroup$ Please show the working $\endgroup$
    – geek101
    Jul 31, 2014 at 12:02
  • $\begingroup$ Well : you know that $t=b/a$, I gave you $\sin(2\theta)$ and $\cos(2\theta)$ as functions of $t$. Just write your expression and replace. $\endgroup$ Jul 31, 2014 at 12:08
  • $\begingroup$ Well I did that and I am getting the same $\endgroup$
    – geek101
    Jul 31, 2014 at 12:11
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$$\begin{align}a\cos2\theta+b\sin2\theta&= a\ (\cos^2\theta-\sin^2\theta)+b\ (2\sin\theta\cos\theta)\\&= \cos^2\theta\ (a-a\tan^2\theta+2b\tan\theta)\\&= \frac{a-a\tan^2\theta+2b\tan\theta}{1+\tan^2\theta}\\&= a.\end{align}$$

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  • $\begingroup$ Ohh thanks a lot finally understood:) $\endgroup$
    – geek101
    Jul 31, 2014 at 12:30
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Assume, $\cos 2\theta+b\sin 2\theta=k$

$\implies 2b\sin \theta\cos\theta+a(\cos^2 \theta-sin^2\theta)=k$

$\implies 2b\tan \theta +a(1-tan^2\theta)=\large \frac {k}{cos^2\theta}$

$\implies 2b\tan \theta +a(1-tan^2\theta)={k}{sec^2\theta}$

$\implies 2b\tan \theta +a(1-tan^2\theta)={k}{(1+tan^2 \theta)}$

Now,substitute for $tan \theta=\frac{b}a$ and solve for $k$

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  • $\begingroup$ It is much simpler if you use Weierstrass. Cheers :) $\endgroup$ Jul 31, 2014 at 12:17
  • $\begingroup$ Where is the a? $\endgroup$
    – geek101
    Jul 31, 2014 at 12:22
  • $\begingroup$ Here in your question: $tan \theta=b/a$ then find the value of $cos2\theta+bsin2\theta$. :) $\endgroup$
    – MonK
    Jul 31, 2014 at 12:23
  • $\begingroup$ there is an a before cos2theta $\endgroup$
    – geek101
    Jul 31, 2014 at 12:26
  • $\begingroup$ It was not there previously when i took the job of solving it. You added it later. Or may be it was a mistake to Highlight a wrong question in the block. read my edit. $\endgroup$
    – MonK
    Jul 31, 2014 at 12:28
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Observe that $\tan \theta=b/a$ implies $(a,b)=\sqrt{a^2+b^2}(\cos\theta,\sin\theta)$. So $$a\cos 2\theta+b \sin 2\theta=\sqrt{a^2+b^2}\left(\cos\theta \cos 2\theta+\sin\theta \sin2\theta\right)$$ which we can simplify by use of an angle addition identity. The answer then follows directly.

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