Question

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.

Answer 1

Okay...bear with me...

Answer 2

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)

Answer 3

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

Answer 4

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

Answer 5

and then take the square root of tan^2Q

Answer 6

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

Answer 7

The radicand needs to be rearranged as

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

\[=\frac{1+\sqrt{2}}{2-1}=\frac{1+\sqrt{2}}{1}=1+\sqrt{2}\]

Answer 12

Your result, Sir.

Answer 13

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).

Answer 14

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!!

Answer 15

No worries. Stop cheating ;)