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Poweranimals
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Okay, I'm confused as to how you get the cscØ. The only information I have available is the cosØ, which is 1/2 and the tanØ, which is listed as negative.
The square root of 1/2 turns up 0.707106781arildno said:You're right about the identity!
But:
If cosØ=1/2, what is then the square of cosØ?
And, what must therefore the square of sinØ be?
So then the answer would be -1/ [tex]\sqrt{\frac{3}{2}[/tex]?arildno said:BTW, the positive square root of 3/4 may be written as follows:
[tex]\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{\sqrt{4}}=\frac{\sqrt{3}}{2}[/tex]
Certainly, this can be simplified as:Poweranimals said:So then the answer would be -1/ [tex]\sqrt{\frac{3}{2}[/tex]?