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Trigonometry
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OpenStudy (anonymous):
cos(arctan(-12/5)+arctan(3/4))
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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 ?
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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.
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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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