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anonymous
 one year ago
anyone pls help me .....I will become ur fan and give u a medal?????
anonymous
 one year ago
anyone pls help me .....I will become ur fan and give u a medal?????

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0cos (a+45)cos (a45)=1/2 sin^2 a

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Hey Panda :) Is this an identity that we want to prove? Or are we simply trying to solve for a?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ithis is an identity

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\[\large\rm \cos(a+45^o)\cos(a45^o)=\frac{1}{2}\sin^2a\]Hmm there is probably some shortcut for doing this.. I can't seem to remember though. I guess what I would do is apply your Angle Sum Formula to both cosines.\[\large\rm \cos(\alpha+\beta)=\cos \alpha \cos \beta\sin \alpha \sin \beta\]\[\large\rm \cos(\alpha\beta)=\cos \alpha \cos \beta+\sin \alpha \sin \beta\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2So on the left side, applying both identities would give you\[\large\rm (\cos a \cos45\sin a \sin45)(\cos a \cos45+\sin a \sin45)\]Recognize that here we have conjugates being multiplied together. So we'll get the difference of squares,\[\large\rm \cos^2a \cos^245\sin^2a \sin^245=\frac{1}{2}\sin^2a\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Recall that at 45 degrees, you're getting the same value for sine and cosine, sqrt2/2.\[\large\rm \left(\frac{\sqrt2}{2}\right)^2\cos^2a \left(\frac{\sqrt2}{2}\right)^2\sin^2a=\frac{1}{2}\sin^2a\]Which simplifies a bit,\[\large\rm \frac{1}{2}\cos^2a \frac{1}{2}\sin^2a=\frac{1}{2}\sin^2a\]Which is quite a bit closer to where we want to be :)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2What do you think? :) Too confusing doing it this way?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0another q pls...how to solve these q

LynFran
 one year ago
Best ResponseYou've already chosen the best response.2ok we need to express in terms of \[R \cos(x+\alpha)\]to find R\[\sqrt{2+1}\]so R=\[\sqrt{3}\]and to get alpha we find tan of it so \[\tan^{1}\frac{ 1 }{ 2 }=26.565 \]now we can express it in the given form... \[R \cos(x+\alpha)\]\[\sqrt{3}\cos(x+26.565)\]and then to solve the second part we let \[\sqrt{3}\cos(x+26.565)=1\]can u continue from there?

LynFran
 one year ago
Best ResponseYou've already chosen the best response.2i took the amplitude of 2cosx..which is 2...and the amplitude of sinx..which is 1...and then \[\sqrt{2+1}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I CANNOT SOLVE IT........

LynFran
 one year ago
Best ResponseYou've already chosen the best response.2\[\sqrt{3}\cos (x+26.565)=1\]\[\cos(x+26.565)=\frac{ 1 }{ \sqrt{3} }\]we rationalise the denominator \[\cos(x+26.565)=\frac{ \sqrt{3} }{ 3 }\]we take cos inverse..\[(x+26.565)=\cos^{1} \frac{ \sqrt{3} }{ 3 }+2n \pi\]ok since ur using degree and not radians we have\[(x+26.565)=\pm54.736+360\]\[x=26.565\pm54.736+360\]..so can u find the solution from here? now u must look at the given \[0\le x \le 180\]

LynFran
 one year ago
Best ResponseYou've already chosen the best response.2can u solve for x from there?

LynFran
 one year ago
Best ResponseYou've already chosen the best response.2ok so \[x=26.565+54.736\]now the +360 is the period here thats y we continue to add the period for given domain cause ur answer would vary of that domain so.. \[x=28.171\]now if we add the period which is 360 here we get\[x=28.171+360\]\[x=388.171\]now this 388.171 is out of the given domain so thats not a solution so we have 1 solution to this problem for the given domain \[x=28.171\]...now if we check the other \[x=26.56554.736\]\[x=81.301\]...this is out of domain if we add 360 we get \[x=81.301+360\]\[x=278.699\]so it would not be a solution... the only solution is \[x=28.171\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0SORRY FOR THE LATE REPLY....I JUST ATE MY BREAKFAST...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0OOOH NOW I GET IT ..... TQ VERY MUCH....
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