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Question:

If $$\cos 25^\circ + \sin 25^\circ = k,$$ then what is $\cos 20^\circ$?


What I did:

I tried to square both sides, and obtained that $\sin 50 = k^2 -1$, however, this didn't get me anywhere. Then I tried splitting 25 into 20 + 5 but that didn't get me anywhere either. Can someone just point me in the right direction?

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3 Answers 3

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Hint

Take into account that $20 = 45 -25$. Develop $\displaystyle\cos(45^\circ-25^\circ)$ and remember that $\displaystyle\cos(45^\circ)=\sin(45^\circ)=\frac{\sqrt 2}2$ and see what happens.

I am sure that you can take from here.

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  • $\begingroup$ You mean 25 = 45-20?? $\endgroup$ Commented Jul 6, 2014 at 14:28
  • $\begingroup$ Sorry for the typo ! It is fixed. $\endgroup$ Commented Jul 6, 2014 at 14:30
  • $\begingroup$ Just checking. Thank you very much!!!! Wait... I did 25 = 45-20 and still got an answer? $\endgroup$ Commented Jul 6, 2014 at 14:31
  • $\begingroup$ Why not ? I think my solution is slightly simpler. Cheers :) $\endgroup$ Commented Jul 6, 2014 at 14:32
  • $\begingroup$ Nice hint $(+1)$ $\endgroup$ Commented Jul 6, 2014 at 16:24
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Then we have $$k^2=(\cos 25^\circ + \sin 25^\circ)^2=1+2\cos 25^\circ \sin 25^\circ=1+\sin 50^\circ.$$ Note that $\cos 40^\circ=\cos(90^\circ-50^\circ)=\sin 50^\circ$. Hence, we have $$\tag{1}\cos 40^\circ=k^2-1.$$ Now $$\tag{2}\cos 40^\circ=2\cos^2 20^\circ-1.$$ Now combining $(1)$ and $(2)$ gives you the answer.

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  • $\begingroup$ Hey that's another way to do it. Guess I missed that when I squared both sides. $\endgroup$ Commented Jul 7, 2014 at 9:18
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$cos 25^\circ+ sin 25^\circ=k$

$\implies cos (45-20) + sin (45-20) =k$

$\implies cos 45\cdot cos 20+ sin 45\cdot sin 20 + sin45\cdot cos 20 -cos 45\cdot sin 20=k$

$2\cdot(\frac{1}{\sqrt2})cos 20^\circ=k$

$cos 20^\circ=k\cdot(\frac{\sqrt{2}}{2})$

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  • $\begingroup$ I asked for a hint, not an answer. $\endgroup$ Commented Jul 7, 2014 at 9:17
  • $\begingroup$ Understand that this is a community, and the answer may or may not be important to you, but it may be for some one whose problem is exactly like yours. $\endgroup$
    – MonK
    Commented Jul 7, 2014 at 9:19
  • $\begingroup$ Even for them, I don't think that an answer should be given. The community is here to help, not do the homework. Please take that into consideration. $\endgroup$ Commented Jul 7, 2014 at 9:52
  • $\begingroup$ I understand what you are trying to say, but that is just how I work. It might help you with just a hint, some one else might not. And with Maths, not everyone is a pro or a quick grabber. No offence to you. $\endgroup$
    – MonK
    Commented Jul 7, 2014 at 10:10

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