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If $\cos\theta = \sin\theta$, then what is $\cos2\theta$?

I am stuck on this problem, please help.

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Where are you stuck? There are two values on the unit circle where the cosine and sine are equal... –  The Chaz 2.0 Mar 26 '12 at 23:25
    
@TheChaz So does it equal $\sin2\theta$? –  David Mar 26 '12 at 23:26
    
sin = opp/hyp, cos = adj/hyp. Does this ringing a bell? –  user2468 Mar 26 '12 at 23:27
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Alternatively, you could divide both sides by $\cos \theta$ to get $\tan \theta = 1$, and maybe this equation would be easier for you to solve by looking at a unit circle. –  The Chaz 2.0 Mar 26 '12 at 23:28
    
No, the answer is zip, zilch, nada. How you arrive at that answer is the most important part of this question. –  The Chaz 2.0 Mar 26 '12 at 23:28
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3 Answers

up vote 11 down vote accepted

Double angle formula: $\cos(2\theta)=\cos^2\theta-\sin^2\theta=0$.

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I would think that if he knew of the existence of the double angle formula, he wouldn't be asking this question. –  hollow7 Mar 26 '12 at 23:32
    
@dragon: I know. It's the slickest way, and if the OP is learning trigonometry he will learn it eventually. –  anon Mar 26 '12 at 23:35
    
@dragoncharmer: if the OP knows the sum formula for $\cos$ they either know or can easily derive the double angle formula, so I'd say this is a good answer, too. –  robjohn Mar 27 '12 at 0:15
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We have $\cos\theta=\sin\theta$ if $\theta=\pi/4+2k\pi$ or $\theta=3\pi/4+2k\pi$. In both of these situations, $\cos 2\theta=0$.

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How did you arrive at the first sentence? [meta-pedagogy] –  The Chaz 2.0 Mar 26 '12 at 23:31
    
@The Chaz: Good point. Let's say "common knowledge" about values of $\sin$, $\cos$ at standard special angles. –  André Nicolas Mar 26 '12 at 23:36
    
In light of the comment by the OP (David): "does it equal $\sin (2 \theta)$", I suspect that this knowledge might not be common to all present for this discussion. Thoughts, David? –  The Chaz 2.0 Mar 26 '12 at 23:38
    
Well it's at least as helpful as the "trig identity" approach (documented twice here, for redundancy's sake)! –  The Chaz 2.0 Mar 26 '12 at 23:42
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@The Chaz: Sorry, had to go away. Gave not best answer to your question. New answer. We know that the picture of $\cos$ is $\sin$ pushed backward, and of course we know the picture of $\sin$. Therefore $\dots$. –  André Nicolas Mar 27 '12 at 2:45
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The tips made above are perfectly reasonable but let me provide an algebraic answer.

Using the addition formula $\cos(x+y) = \cos(x) \cos(y) - \sin(x) \sin(y)$, we know that $\cos(2x) = \cos(x+x) = \cos(x) \cos(x) - \sin(x) \sin(x)$

Does that help you now?

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