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A few more trig problems ( Sin(α + β), etc. )

  • Thread starter rought
  • Start date
  • #1
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Homework Statement



Suppose sinα = 4/5 and tanβ= -3/4 where π/2 < α < β < π (pi)

Find:

Sin(α + β)

Cos(α + β)

Tan(α + β)

Sin2α

Cos2β

Tan2α

The Attempt at a Solution



What i have been doing is figuring out the other trig functions

They gave me sinα=4/5 and tanβ=-3/4

from that i found cosα=3/5 tanα=3/4 and sinβ=-3/5 cosβ=4/5

I tried problem a

sin(α+β)=sinαcosβ+cosαsinβ

sin(α+β)=4/5 * 4/5 + 3/5 * -3/5

which gives me 7/25 but I'm not sure I am doing this right... what is the whole (π/2 < α < β < π) thing about am I supposed to be finding points on the unit circle or something? =/
 

Answers and Replies

  • #2
52
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The whole "π/2 < α < β < π (pi)" thing refers to the quadrant the terminal sides of angles α and β are in. So what quadrant is this? And how does this affect the evaluation of the trig functions? Because none of these:
cosα=3/5 tanα=3/4 and sinβ=-3/5 cosβ=4/5
are correct.


01
 
Last edited:
  • #3
34
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The whole "π/2 < α < β < π (pi)" thing refers to the quadrant the terminal sides of angles α and β are in. So what quadrant is this? And how does this affect the evaluation of the trig functions? Because none of these:

are correct.


01
ahh Ok so it's in the second quadrant which makes the x negative and the y positive

so would that make

cosα=-3/5 tanα=-3/4 and sinβ=-3/5 cosβ=4/5

is this right?
 
  • #4
33,167
4,851
From your given condition, both alpha and beta are in the 2nd quadrant, so how can sin(beta) be negative and how can cos(beta) be positive? If you have these wrong, you are liable to have the others (the tangents) wrong, too.
 
  • #5
34
0
So would it be cosα = -3/5 tanα = -3/4 and sinβ = +3/5 cosβ = -4/5
 
  • #6
52
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Almost. The signs are correct, but double check tan α.


01
 

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