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UnbelievableDreams
 one year ago
Confusing math question
UnbelievableDreams
 one year ago
Confusing math question

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UnbelievableDreams
 one year ago
Best ResponseYou've already chosen the best response.0Choose the correct expression that completes the identity. sec^2 (x)  tan^2(x)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0im sorry this is above my math intelligence

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.2sec^2 = what identity ?

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.2do you have sin ,cos identities notes ?

UnbelievableDreams
 one year ago
Best ResponseYou've already chosen the best response.0No, I don't have notes. I am so confused.

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.2now u got the answer if you apply identities that would be easy \[\huge\rm tan^2 +1=\sec^2\]

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.2replace sec^2 by tan^2 +1 \[\huge\rm \color{reD}{sec^2(x)}\tan^2(x)\] \[\huge\rm \color{reD}{tan^2+1 }\tan^2\]

dan815
 one year ago
Best ResponseYou've already chosen the best response.0u need to know only the basic ones sin^2(x)+cos^2(x)=1 and sec(x)=1/cos(x) csc(x)=1/sin(x) cot(x)=1/tan(x) there are the basic ones to know

UnbelievableDreams
 one year ago
Best ResponseYou've already chosen the best response.0I am still confused. :(

freckles
 one year ago
Best ResponseYou've already chosen the best response.1do you know the Pythagorean identity(ies)?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[\sin^2(x)+\cos^2(x)=1 \text{ do you know the one mentioned by dan ?}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1Do you know how to graph the equation x^2+y^2=1? This is the circle with center (0,0) and radius 1. Looks like this: dw:1438638836144:dw say we pick a random point (a,b) on the circle: dw:1438638871637:dw now we draw a right triangle dw:1438638917068:dw we know the relationship between the side of those right triangle to be \[a^2+b^2=1^2 \\ \text{ or } \\a^2+b^2=1 \\ \text{ the 1 comes from the radius of the circle being 1 }\] now we could write this same equation in terms of theta by using the drawing

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[\cos(\theta)=\frac{a}{1} \text{ or } \cos(\theta)=a \\ \sin(\theta)=\frac{b}{1} \text{ or } \sin(\theta)=b \\ \text{ squaring both sides gives } \\ (\cos(\theta))^2=a^2 \text{ also written as } \cos^2(\theta)=a^2 \\ (\sin(\theta))^2=b^2 \text{ also written as } \sin^2(\theta)=b^2 \\ \text{ so replacing the items \in } a^2+b^2=1 \\ \text{ we have } \cos^2(\theta)+\sin^2(\theta)=1 \]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1you can get the identity that nnesha mentioned from this if you divide both sides by cos^2(theta)

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[(\cos(\theta))^2+(\sin(\theta))^2=1 \\ \text{ dividing both sides by } (\cos(\theta))^2 \\ \frac{(\cos(\theta))^2+(\sin(\theta))^2}{(\cos(\theta))^2}=\frac{1}{(\cos(\theta))^2} \\ \text{ separating fraction on \left hand side } \\ \frac{(\cos(\theta))^2}{(\cos(\theta))^2}+\frac{(\sin(\theta))^2}{(\cos(\theta))^2}=\frac{1^2}{(\cos(\theta))^2} \text{ note: notice I replace } 1 \text{ with } 1^2 \\ \text{ now using law of exponents } \\ (\frac{\cos(\theta)}{\cos(\theta)})^2+(\frac{\sin(\theta)}{\cos(\theta)})^2=(\frac{1}{\cos(\theta)})^2 \\ \text{ now there were ratios mentioned by dan that were equivalent \to some you see here }\] for example you know \[\frac{\sin(\theta)}{\cos(\theta)}=\tan(\theta) \\ \text{ and } \frac{1}{\cos(\theta)}=\sec(\theta)\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[\text{ also you know } \frac{\cos(\theta)}{\cos(\theta)}=1 \text{ on it's domain } \\ \text{ so we have } 1^2+(\tan(\theta))^2=(\sec(\theta))^2 \\\ \text{ which can be written as } \\ 1+\tan^2(\theta)=\sec^2(\theta)\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1this is the Pythagorean identity you want to apply

UnbelievableDreams
 one year ago
Best ResponseYou've already chosen the best response.0Can you help me another one?

UnbelievableDreams
 one year ago
Best ResponseYou've already chosen the best response.0It is similar to this question. But I like to see step by step. 1tan^2x= ___________

freckles
 one year ago
Best ResponseYou've already chosen the best response.1do you want to write it in terms of sec or use difference of squares

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[a^2b^2=(ab)(a+b) \text{ is difference of squares in factored form }\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1if you want to write it in terms of sec you can use the identity we just mentioned \[1+\tan^2(\theta)=\sec^2(\theta) \\ \text{ subtract } 1 \text{ on both sides } \\ \tan^2(\theta)=\sec^2(\theta)1 \\ \text {replace } \tan^2(\theta) \text{ with } \sec^2(\theta)1 \]

UnbelievableDreams
 one year ago
Best ResponseYou've already chosen the best response.0The answer is 1/ sec^2x

UnbelievableDreams
 one year ago
Best ResponseYou've already chosen the best response.0How do you get sec?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1did you replace tan^2(theta) with sec^2(theta)1 ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[1\tan^2(\theta) \\ \text{ replace } \tan^2(\theta) \text{ with } \sec^2(\theta)1 \\ 1(\sec^2(\theta)1))\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1distribute and combine like terms but this shouldn't give you 1/sec^2(x)

UnbelievableDreams
 one year ago
Best ResponseYou've already chosen the best response.0Sec^2x  1 ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1nope you distribute \[1\tan^2(x) \\ \text{ replace } \tan^2(x) \text{ with } \sec^2(x)1 \\ \text{ which gives } \\ 1(\sec^2(x)1) \\ \text{ now distribute } \\1\sec^2(x)+1 \\ \text{ combine like terms } (1+1)\sec^2(x)\]
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