Trigonometric Formulas, Identities and Equation

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To find sin 2t when cos t = 4/5, first determine sin t using the Pythagorean theorem, resulting in sin t = 3/5. The double-angle formula sin 2t = 2 sin t cos t can then be applied. Substituting the known values gives sin 2t = 2(3/5)(4/5), simplifying to sin 2t = 24/25. The discussion highlights the importance of understanding trigonometric identities and the need for collaborative problem-solving.
palui123
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Homework Statement


t = tetha

FInd sin 2t when cos t = 4/5

Homework Equations



no idea

The Attempt at a Solution



no. . . . btw what should I do 1st? I don't understand this question
 
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1) You are given that cos t = 4/5. Can you find sin t?
2) Do you know your trig identities? Particularly the double-angle formulas.
 
palui123 said:

Homework Statement


t = tetha

FInd sin 2t when cos t = 4/5

Homework Equations



no idea

The Attempt at a Solution



no. . . . btw what should I do 1st? I don't understand this question
Exactly what part of "Find sin 2t" do you not understand?
 
Formula:
Sin2t = 2 sint costWhat I do is:
Sin2t = 2sint cost

since cost = 4/5

sin2t = 2sint (4/5)

sin2t = 8sint/5

What should I do then..? I know its wrong
 
w8. . . nvm. I know how to this question now.

sin 2t = 2 sint cost

since cost = 4/5 ,

a=4 , h=5 , o=??

P.theorem:
a^2=b^2+c^2
o=3

then sint = 3/5

then subtitude =D . . . solve get the answer. . .I'm talking by myself T-T ask = me , answer = me T-T . AID ME NEXT TIME PLEASE. . . .
 
palui123 said:
I'm talking by myself T-T ask = me , answer = me T-T . AID ME NEXT TIME PLEASE. . . .
If by "aiding you" you mean, show you the work + answer, then NO, WE CAN'T DO THAT. Read the forum rules. Otherwise, I think I gave you sufficient aid in my post (#2).
 
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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