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anonymous
 5 years ago
First I proved that
cos 2Q=1tan^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.
anonymous
 5 years ago
First I proved that cos 2Q=1tan^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.

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay...bear with me...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It 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)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0clearly 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)=1tan^2Q/1+tan^2Q

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You solve the equation for tan^2Q as\[\tan^2Q=\frac{1\cos 2Q}{1+ \cos 2Q}=\frac{1(1/\sqrt{2})}{1+(1/\sqrt{2})}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and then take the square root of tan^2Q

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Take the square root:\[\tan Q= \sqrt {\frac{1+\sqrt{2}}{1\sqrt{2}}}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The radicand needs to be rearranged as

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{1+\frac{1}{\sqrt{2}}}{1\frac{1}{\sqrt{2}}}=\frac{\sqrt{2}+1}{\sqrt{2}1}=\frac{\sqrt{2}+1}{\sqrt{2}1} \times \frac{\sqrt{2}1}{\sqrt{2}1}=\frac{21}{(\sqrt{2}1)^2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0=frac{1}{(sqrt{2}1)^2}

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Therefore, taking the square root of that result gives,\[\tan Q = \sqrt{\frac{1}{(\sqrt{2}1)^2}}=\frac{1}{\sqrt{2}1}=\frac{1}{\sqrt{2}1} \times \frac{\sqrt{2}+1}{\sqrt{2}+1}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[=\frac{1+\sqrt{2}}{21}=\frac{1+\sqrt{2}}{1}=1+\sqrt{2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Note 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).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thanks. 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 2/2sq 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!!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No worries. Stop cheating ;)
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