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.
Stacey Warren - Expert brainly.com
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Okay...bear with me...
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)
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:
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).
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
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!!