Homework Help with trig ident

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To find the value of cos(arcsin u + arccos v), the expression can be simplified using the cosine addition formula: cos(A+B) = cosAcosB - sinAsinB. The identities cos(arcsin x) = √(1-x²) and sin(arccos x) = √(1-x²) are also applicable. Clarification on the values of u and v is necessary for a complete solution. The discussion emphasizes the importance of providing full context in homework questions.
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Homework Statement


Find the value of cos(arcsin u + arccos v)?


Homework Equations





The Attempt at a Solution


 
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You're going to have to be a little more specific, what are u and v? Are you looking to simplify this expression? As well, when you preview a post it adds the template to your post again, so remember to delete it.
 
JoshB645 said:

Homework Statement


Find the value of cos(arcsin u + arccos v)?

i suppsoe it should be a full stop instead question mark,
btw Welcome Joshua to PF:smile:

I think you have not written the whole question.
 
Last edited:
Hello! Use cos(A+B)=cosAcosB-sinAsinB

Then, use:
cos[arcsin(x)]=√1-x2
sin[arccos(x)]=√1-x2
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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