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

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?

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.

• You mean 25 = 45-20?? Commented Jul 6, 2014 at 14:28
• Sorry for the typo ! It is fixed. Commented Jul 6, 2014 at 14:30
• Just checking. Thank you very much!!!! Wait... I did 25 = 45-20 and still got an answer? Commented Jul 6, 2014 at 14:31
• Why not ? I think my solution is slightly simpler. Cheers :) Commented Jul 6, 2014 at 14:32
• Nice hint $(+1)$ Commented Jul 6, 2014 at 16:24

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.

• Hey that's another way to do it. Guess I missed that when I squared both sides. Commented Jul 7, 2014 at 9:18

$cos 25^\circ+ sin 25^\circ=k$

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

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

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

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

• I asked for a hint, not an answer. Commented Jul 7, 2014 at 9:17
• 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.
– MonK
Commented Jul 7, 2014 at 9:19
• 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. Commented Jul 7, 2014 at 9:52
• 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.
– MonK
Commented Jul 7, 2014 at 10:10