## anonymous 5 years ago First I proved that cos 2Q=1-tan^2 Q/1+tan^2 Q But the problem is that the second part says, Hence show without calculator that; tan 67.5=1+square root of 2.

1. anonymous

Okay...bear with me...

2. anonymous

It hinges on you recognizing that for Q=67.5, 2Q=135 degrees. The cosine of 135 degrees is the same as the negative of the cosine of 45, which is -1/sqrt(2)

3. anonymous

clearly the question wants you to apply the formula you just proved. the angle $\theta=67.5^0=3\pi/8$ $\cos2\theta=\cos3\pi/4=-\cos \pi/4=-\sqrt2/2$ (since it's in the 2nd quarter) now we should solve the following equation for tan^2Q: -1/sqrt(2)=1-tan^2Q/1+tan^2Q

4. anonymous

You solve the equation for tan^2Q as$\tan^2Q=\frac{1-\cos 2Q}{1+ \cos 2Q}=\frac{1-(-1/\sqrt{2})}{1+(-1/\sqrt{2})}$

5. anonymous

and then take the square root of tan^2Q

6. anonymous

Take the square root:$\tan Q= \sqrt {\frac{1+\sqrt{2}}{1-\sqrt{2}}}$

7. anonymous

The radicand needs to be rearranged as

8. anonymous

$\frac{1+\frac{1}{\sqrt{2}}}{1-\frac{1}{\sqrt{2}}}=\frac{\sqrt{2}+1}{\sqrt{2}-1}=\frac{\sqrt{2}+1}{\sqrt{2}-1} \times \frac{\sqrt{2}-1}{\sqrt{2}-1}=\frac{2-1}{(\sqrt{2}-1)^2}$

9. anonymous

=frac{1}{(sqrt{2}-1)^2}

10. anonymous

Therefore, taking the square root of that result gives,$\tan Q = \sqrt{\frac{1}{(\sqrt{2}-1)^2}}=\frac{1}{\sqrt{2}-1}=\frac{1}{\sqrt{2}-1} \times \frac{\sqrt{2}+1}{\sqrt{2}+1}$

11. anonymous

$=\frac{1+\sqrt{2}}{2-1}=\frac{1+\sqrt{2}}{1}=1+\sqrt{2}$

12. anonymous

13. anonymous

Note only the positive root is taken since you're looking at 67.5 degrees, which is in the first quadrant (i.e. where tan is positive).

14. anonymous

thanks. I got my answer, but i did it illegally; I did this 1-(-1/sq root 2)/ 1+(-1/sq root 2) 2+sq root 2/2-sq root 2 rationalize 3+2(2)^1/2 then i cheated (1+sq root 2)^2 1+ sq root 2 + sq root 2 +2 3 +sq root2 then i wrote it back on my paper. but thanks for the correct way!!

15. anonymous

No worries. Stop cheating ;)