CEBS95
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



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Shadowys
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why can't you just plug the whole thing into your calculator?

CEBS95
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because in the exam they already give you the answer, you have to do the procedure

Callisto
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tan60 = \(\sqrt{3}\)
cos45 = \(\frac{1}{\sqrt2}\)
tan30 = \(\frac{1}{\sqrt3}\)
sin30 = 1/2
These are all special angles and you should remember them.

CEBS95
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yes i know them but i just continue to get 6 instead of 0

Shadowys
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I'm getting zero just fine though. note the squares.

CEBS95
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tan^2(60) = 3 ?
4cos^2(45) = 4 ?

Callisto
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4cos^2(45) = 4 (1/2) = ...?

CEBS95
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my math base is not good, that's the reason i wrote step by step solution

Callisto
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\[cos45 =\frac{1}{\sqrt2}\]\[cos^245 =(\frac{1}{\sqrt2})^2 = \frac{1}{2}\]\[4cos45 =4(\frac{1}{2}) = ...?\]

CEBS95
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4 (1/2) = 2

Callisto
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Yes.

winterfez
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\[\frac{ \sin ^{2} (60)}{ \cos ^{2}(60)}+4\cos ^{2}(45)9\frac{ \sin ^{2}(30) }{ \cos ^{2}(30) }8\sin ^{2}(30)\]

CEBS95
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so, tan^2(60) = 3

Callisto
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tan^2(60) = 3 < Yes

winterfez
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dw:1355048071288:dw

CEBS95
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i found in my book that cos 45 = \[\frac{ \sqrt{2} }{ 2}\]

Callisto
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\[\frac{1}{\sqrt2}=\frac{1}{\sqrt2}\times \frac{\sqrt2}{\sqrt2}=\frac{\sqrt2}{\sqrt2^2} = \frac{\sqrt2}{2}\]
Just rationalization.

CEBS95
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9tan^2(30) = \[\tan( 30) = \frac{ \sqrt{3} }{ 3 } ; \tan ^{2}(30) = 3/9 = 1/3 \]

CEBS95
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times 9 = 3 right?

Callisto
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Yes.

CEBS95
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8 sin ^2 (30) = 4 ?

CEBS95
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no, i think it 8

CEBS95
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9tan^2(30)  8sin^2(30) = 3  8 = 5, at the end that will give me 5 (5) that equals 10

Callisto
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8 sin ^2 (30)
= 8 (1/2)^2
= 8 (1/4)
= ...

CEBS95
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2

CEBS95
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still, 32 = 1, 51 = 4

winterfez
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dw:1355049445375:dw
look at this one to see how i get 0

Callisto
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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.

winterfez
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dw:1355049643294:dw

CEBS95
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zero! nice thanks calisto

Callisto
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You're welcome :)

Callisto
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 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  \(\sqrt{3}\)

Callisto
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Memorize the table..

CEBS95
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yes, with practice i will memorise the terms