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Trigonometry 22 Online
OpenStudy (anonymous):

cos(arctan(-12/5)+arctan(3/4))

OpenStudy (anonymous):

@dan815 @uri

OpenStudy (anonymous):

I'm assuming that you want to solve this in exact form by hand without a calculator.

OpenStudy (anonymous):

In that case, you will need to use the arctan sum formula.

OpenStudy (anonymous):

@math1234 that would be correct

OpenStudy (anonymous):

1/1+x^2 ?

OpenStudy (anonymous):

No, it is \[\tan^{-1} a + \tan^{-1} b = \tan^{-1} \frac{ a+b }{ 1-ab }\]

OpenStudy (anonymous):

Upon combining the inside using the arctan sum formula, you can use your mentioned formula to compute the cos of the arctan.

OpenStudy (anonymous):

so that gves us \[\tan^{-1} \frac{ \frac{ -12 }{ 5 }+\frac{ 3 }{ 4 } }{ 1-\frac{ -12 }{ 4 }*\frac{ 3 }{ 4 } }\]

OpenStudy (anonymous):

Yes, then you plug it into \[\cos (\tan^{-1} x) = \frac{ 1 }{ \sqrt{1+x^2} }\]

OpenStudy (anonymous):

Where x is your fractional expression above.

OpenStudy (anonymous):

\[\cos (\tan^{-1} \frac{ 33 }{ 16 })=\frac{ 1 }{ \sqrt{1+(\frac{ 33 }{ 6 }})^{2} }\]

OpenStudy (anonymous):

idk where to go from here

OpenStudy (anonymous):

That's your answer.

OpenStudy (anonymous):

Just add the denominator.

OpenStudy (anonymous):

56/65 Refer to the attachment below.

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