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tan^2(60) + 4cos^2(45) - 9tan^2(30) - 8sin^2(30) The answer is zero but why!? i need step by step solution please

Trigonometry
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why can't you just plug the whole thing into your calculator?
because in the exam they already give you the answer, you have to do the procedure
tan60 = \(\sqrt{3}\) cos45 = \(\frac{1}{\sqrt2}\) tan30 = \(\frac{1}{\sqrt3}\) sin30 = 1/2 These are all special angles and you should remember them.

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Other answers:

yes i know them but i just continue to get 6 instead of 0
I'm getting zero just fine though. note the squares.
tan^2(60) = 3 ? 4cos^2(45) = 4 ?
4cos^2(45) = 4 (1/2) = ...?
my math base is not good, that's the reason i wrote step by step solution
\[cos45 =\frac{1}{\sqrt2}\]\[cos^245 =(\frac{1}{\sqrt2})^2 = \frac{1}{2}\]\[4cos45 =4(\frac{1}{2}) = ...?\]
4 (1/2) = 2
Yes.
\[\frac{ \sin ^{2} (60)}{ \cos ^{2}(60)}+4\cos ^{2}(45)-9\frac{ \sin ^{2}(30) }{ \cos ^{2}(30) }-8\sin ^{2}(30)\]
so, tan^2(60) = 3
tan^2(60) = 3 <- Yes
|dw:1355048071288:dw|
i found in my book that cos 45 = \[\frac{ \sqrt{2} }{ 2}\]
\[\frac{1}{\sqrt2}=\frac{1}{\sqrt2}\times \frac{\sqrt2}{\sqrt2}=\frac{\sqrt2}{\sqrt2^2} = \frac{\sqrt2}{2}\] Just rationalization.
9tan^2(30) = \[\tan( 30) = \frac{ \sqrt{3} }{ 3 } ; \tan ^{2}(30) = 3/9 = 1/3 \]
times 9 = 3 right?
Yes.
8 sin ^2 (30) = 4 ?
no, i think it 8
9tan^2(30) - 8sin^2(30) = 3 - 8 = -5, at the end that will give me 5- (-5) that equals 10
8 sin ^2 (30) = 8 (1/2)^2 = 8 (1/4) = ...
2
still, 3-2 = 1, 5-1 = 4
|dw:1355049445375:dw| look at this one to see how i get 0
tan^2(60) + 4cos^2(45) - 9tan^2(30) - 8sin^2(30) tan^2(60) = \(\sqrt3^2\) = ... 4cos^2(45) = 4\((\frac{1}{\sqrt2})^2\) = 4(1/2) = ... - 9tan^2(30) = -9 \((\frac{1}{\sqrt3})^2\) = -9 (1/3) = ... - 8sin^2(30) = -8 (1/2)^2 = -8 (1/4) = ... Sum up the four answers.
|dw:1355049643294:dw|
zero! nice thanks calisto
You're welcome :)
| 30 | 45 | 60 ------------------------------------ sin | \(\frac{1}{2}\) | \(\frac{1}{\sqrt2} /\frac{\sqrt2}{2} \) | \(\frac{\sqrt3}{2}\) ------------------------------------ cos | \(\frac{\sqrt3}{2}\) | \(\frac{1}{\sqrt2} /\frac{\sqrt2}{2} \) | \(\frac{1}{2}\) ------------------------------------ tan | \(\frac{1}{\sqrt3} /\frac{\sqrt3}{3} \) | 1 | \(\sqrt{3}\)
Memorize the table..
yes, with practice i will memorise the terms

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